Plus One Malayalam Textbook Answers, Notes, Chapters Summary HSSLive Kerala

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Kerala Plus One Malayalam Textbook Questions and Answers, Notes, Chapters Summary HSSLive

Unit 1 Kinav

Unit 2 Kalca

Unit 3 Ullariv

Unit 4 Uravu

Plus One Malayalam Textbook Answers, Notes, Chapters Summary

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Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

Gas Laws Mole Concept Text Book Questions and Answers

Text Book Page No: 33

→ Complete the table 2.1
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 1
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 2

→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter

→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases

→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases

Mole to mass calculator that was written in Mathematica.

Text Book Page No: 34

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.

→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy

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→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased

→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase

→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.

→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)

→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature

Text Book Page No: 35
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 3

→ Is there any change in the number of molecules?
Answer:
No

→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

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→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.

→ What about decreasing the pressure?
Answer:
Volume increasing

→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.

Text Book Page No: 36
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 4

→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

→ What do you observe?
Answer:
Ink rises through the tube

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→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.

→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

→ Complete the table 2.2
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 5
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 6
Text Book Page No: 37

→ In which unit is the temperature stated?
Answer:
Kelvin (K)

→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.

→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.

→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased

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→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.

→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases

→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.

→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.

Text Book Page No: 38

→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g

→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000

→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.

→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes

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→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12th mass of carbon atom.

Text Book Page No: 39

→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms

→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.

→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 1023 oxygen atoms

→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g

Text Book Page No: 40

→ Complete the table 2.5
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 7
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 8
→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 1023 sodium atoms.

Text Book Page No: 41

→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 1023

2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 1023

→ Complete the table 2.6
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 9
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 10

→ Calculate the molecular mass of glucose (C6 H12 O6) and sulphuric acid (H2 SO4)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g

Text Book Page No: 42

→ Complete the table 2.7
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 11
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 12
→ What is the molecular mass of oxygen?
Answer:
32 g

→ How many GMM is present in 32g oxygen?
Answer:
1 GMM

→ How many molecules are present in it?
Answer:
6.022 × 1023 oxygen molecules

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→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM

→ How many molecules are present in N2?
Answer:
6.022 × 1023 N2, molecules

→ How many GMM is present in 18 gm water?
Answer:
1 GMM

→ How many H2O molecules arf present in it?
Answer:
6.022 × 1023 H2O Molecules

→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM

Text Book Page No: 43

How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?

→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 1023

→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 1023

→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 1023 water molecules

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→ What is its mass?
Answer:
18 g

→How many GMM is present in it?
Answer:
1 GMM

Text Book Page No: 44

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 13

22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole

Text Book Page No: 45

Complete the flow chart given below, related to one mole of substance.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 14
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 15

Gas Laws Mole Concept Let Us Assess

The mass molarity calculator chemistry tool calculates the mass of compound required to achieve a specific molar concentration and volume.

Gas Laws and Mole Concept Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law

Gmm Chemistry Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law

Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 16
a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 17
b. Avogadro’s law

Mole Concept Question 4.
a. Calculate the mass of 112 L CO2 gas kept at STP (molecular mass = 44)
b. How many molecules of CO2 are present in it?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 18

Mole Concept Class 10 Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L

Mole Concept Questions Class 9 Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N2=28g, H2O= 18g)
a. 56g N2
b. 90g H2O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2

b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5

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Mole Concept Questions with Solution Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM

b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10

c. Number Of molecules = Mole × 6.022 × 1023
= 10 × 6.022 × 1023

Mole Concept Notes Class 9 Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O2
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O2 =2 × 16 = 32 g

b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 1023
= 2 × 6.022 × 1023
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O2) =2
∴ total number of atoms = 2 × 6.022 × 1023 × 2
=4 × 6.022 × 1023

Gas Laws Mole Concept Extended activities

Mole Concept Questions Class 11 Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 1023
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 1023
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)

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What is Gmm in Chemistry Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH3 at STP
c. 67.2 L of N2 at STP
d. 1 mol of H2SO4
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 19
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 20

Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 1023 = 5 × 6.022 × 1023 18

b. ∴ Number of atoms = 3 × 5 × 6.022 × 1023 = 15 × 6.022 × 1023

c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H2O molecule? 10
Total electrons in 90 g H2O = 10 × 5 × 6.022 × 1023 = 50 × 6.022 × 1023

Gas Laws Mole Concept Orukkam Questions and Answers

Question 1.
a. Complete the table based on the data given in the box.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 21
b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 22
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 23

b.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 24

Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 25
a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H2SO4 = …………. g ………… Molecules
1/2 Mole of Al2O3= …………. g ………….. Molecules
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 26
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 27

a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 1023 Molecules
5 mole of CaO = 280g, 5 × 6.022 × 1023 Mol-ecules
2 Mole of H2SO4 = 196g, 2 × 6.022 × 1023 Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 1023

Question 3.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 27
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 29
Question 4.
Complete the data.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 30
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 31
Question 5.
Based on the reaction given below, write the answers for the questions.
N2 + 3H2 → 2NH3 ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen

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Nuclear equation calculator How do you write a balanced nuclear equation.

Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH4+O2 → CO2 + H2O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C2H6 + 7O2 → 4CO2 + 6H2O
b. Based on the equation above, How many moles of CO2 is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO2 gas is formed.
When 20 moles methane burn 20
moles of CO2 are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO2 when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g

Question 7.
Based on the given equation write down the answers.
2H2+ O2 → H2O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g

b. 0.5 Mole

Gas Laws Mole Concept Evaluation Questions

Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC1 contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC1 = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol

Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.

Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO3?
b. How many moles of Oxygen present in 1000gms of CaCO3?
Answer:
a. Mass of 1 mole of CaCO3 = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO3 = \(\frac { 1000 }{ 100 }\) = 10 mol

b. No of moles of Ca in 10 moles CaCO3 =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO3 = 10 × 3 = 30mol.

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Combined Gas Law Calculator … This is a combination of three gas laws, which are Boyle’s law , Charles’s law and Gay Lussac’s law.

Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H2O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol

How to Find Molar Concentration of Solute in a Solution.

Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.

b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.

Question 6.
How many moles of Cl2 present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.

Question 7.
Find out the mass of Oxygen atom in 44.8L of CO2 in STP.
Answer:
No of moles in 44.8 litre of CO2 \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO2 contains 1 mole of C and 1 mole of O2
∴ 2 Mole of CO2 contains 2 mole of O2 or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g

Question 8.
Find out the amount of CO2 formed when the burning of one mole of Ethane.
Answer:
2 moles of CO2 is formed when 1 mole of ethane burns.
Mass of 2 moles of CO2 = 2 × 44 = 88g

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Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)

Gas Laws Mole Concept SCERT Questions and Answers

Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 1023
b. i.6.022 × 1023
ii. 2 x 6.022 × 1023
iii \(\frac { 32 }{ 12 }\) × 6.022 × 1023

Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen

b. A, D – 2 × 6.022 × 1023
B,C – 6.022 × 1023

Question 3.
N2 + 3H2 → 2NH3
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 32

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 33
Answer:
a. 1:3
b. a – 2 NH3,
b – l H2,
c – 12H2,
d – 4NH3

Question 4.
2H2 + O2 → 2H2O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O2 molecules are required to react 100 H2 molecules completely?
c. How many water molecules are formed when 1000 H2 molecules are reacted completely
Answer:
a. 2:1
b. 50 O2 molecules
c. 1000 H2O molecules

Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 34
Answer:
a. 6.022 × 1023
b. 6.022 × 1023
c. 71 g
d. 14 g

Question 6.
A sample of substances are given.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 35
Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH3, 128gO3
b. 49 g H2SO4

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Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 1023 chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 1023 H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 1023
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 1023 H2O molecules.

Question 8.
Complete the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 36
Answer:
a. 2 × 6.022 × 1023
b. 1GMM
c. 6.022 × 1023

Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO2 present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO2=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO2 at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 1023

Question 10.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 37
Answer:
a. 2
b. 2 × 6.022 × 1023
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L

Question 11.
CH4 + 2O2 → CO2 + 2H2O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH4?
b. Calculate the amount of CO2 formed when 100g of CH4 is completely burnt?
Answer:
a. 2 mol
b. Amount of CO2 produced by the combustion of 16g CH4 = 44g
Amount of CO2 produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO2 produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g

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Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose

Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 38
a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO2 = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO2 = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 1023

First, the chemical formula calculator makes sure that the equations have the same number of atoms on both sides of the certain equation.

Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H2(g) + O2(g) → 2H2O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g

Question 15.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 39
Hint: (MM – CO2 = 44, CH4 = 16, SO2 = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.

Question 16.
Analyse the following equation
2NO(g) + O2(g) → 2NO2(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO2 formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O2(g) → 2NO2(g) (2 : 1: 2)
No. of moles in 112L O2 = 5 mol
According to equation no. of moles of NO2 obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO2 obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO2 = 10 × 46 = 460g

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Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO3→ CaO + CO2
(HintMM: CaCO3 – 100, CaO – 56, CO2 – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO3 → CaO + CO2
100g 56g 44g
I I : I
Amount of CaCO3 required to get 56g of CaO = 100g
Amount of CaCO3 required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO3 required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 1023

Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g

Gas Laws Mole Concept Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 1023 atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g

Question 2.
Number of molecules of substance is 3.011 × 1024.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 1024 molecules.
Answer:
a. 6.022 ×1023
b. Number of moles of molecules = \(\frac { Number of molecules }{ NA }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)

Short Answer Type Questions (Score 2)

Question 3.
Fill the patterns.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 40
Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass

Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct

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Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, PV = a constant
Therefore, P1 V1 = P2 V2
Here, P1 = 2atm V1 = 20L P2 = 3atm V2=?
∴ 2 × 20 = 3 × V2
Thus, V2 = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L

The Charles’ law calculator is a simple tool which describes the basic parameters of an ideal gas in an isobaric process.

Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 41

Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g

Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L

Question 9.
Look at the balanced equation given.
2NaOH + CO2 → Na2 CO3 + H20
a. Find out the mass of NaOH needed for 264 g CO2 to react completely.
b. Find out the total number of moles of water molecules when CO2 reacts.
Answer:
a. GMM of CO2 = 44 g
∴Number of moles in the molecule of 264 g CO2 = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO2= 2 moles
∴ NaOH needed for the reaction of 6 moles CO2 = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g

b. H2O formed when lmole CO2 reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO2 reacts = 6 moles

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Question 10.
2C4H10 + 13O2 → 8CO2 + 10H2O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO2 in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C4H10) = 14 kg = 1400g
GMM of C4H10 = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO2when 2 moles of C4 H10 ignites = 8 moles of C4H10 ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO2 formed in STP
= 965.52 × 22.4 L = 21627.65 L

Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.

Question 12.
The molarity of250 ml of Na2CO3 solution is 0.5 M. Find the mass of Na2CO3.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 42
Question 13.
63 g HNO3 is in the dilute solution of 200 ml HNO3 (Nitric acid). Find the molarity.
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 43

Short Answer Type Questions (Score 3)

Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. Pv = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct

b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law

Question 15.
In 100 g of CaCO3
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 44

Question 16.
Fill the blanks in the given table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 45
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 46
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 47
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 48

Long Answer Type Questions (Spore 4)

Question 17.
See CO2 gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO2 molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)

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Question 18.
The molecular formula of ammonium sulfate is (NH4)2SO4.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH4)2SO4
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 49
Question 19.
Fill in the blanks of the table given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 50
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 51

Question 20.
See the diagram given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 52
Answer:
a. 196g
b. 2 × 6.022 × 1023
c. 2 GMM
d. 2 × 6.022 × 1023

Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 53
Question 22.
Certain compounds and its masses are given,
i) 68 g NH3
ii) 28 g N2
iii) 9 g H2O
iv) 128 g O2
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 54
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 55

Question 23.
368 g NO 2gas is given. Find the answers of each one given below,
a. GMM of NO2
b. Number of moles of molecules of 368 g NO2
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO2
= 14 + 2 × 16 = 46 g

b. Number of mole = \(\frac { 368 }{ 46 }\) = 8

c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 1023

d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 1023 = 24 × 6.022 × 1023

e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L

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Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H2)
b. 88.75 g Chlorine (Cl2)
c. 4 g Calcium (Ca2) ,
d. 7.75 g phosphorus (p4)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 1023
Total number of atoms = 2 × 10 × 6.022 × 1023
= 20 × 6.022 × 1023

b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 1023
Total number of atoms = 2 × 1.25 × 6.022 × 1023
= 2.5 × 6.022 × 1023

c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 1023
Total number of atoms = 1 × 0.1 × 6.022 × 1023
= 0.1 × 6.022 × 1023

d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 1023

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

You can Download Periodic Table and Electronic Configuration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

Periodic Table and Electronic Configuration Text Book Questions and Answers

Text Book Page No: 7

→ What is the basis of classification of elements in the periodic table?
Answer:
Atomic Number

Text Book Page No: 8

→ Atomic number of sodium is 11 Electronic configuration – 2,8,1
GroupNumber — …………..
Period number — …………
Answer:
Group Number — 1
Period number — 3

→ Is the group 1 element a metal or a nonmetal?
Answer:
Metal

→ Write the electronic configuration of sodium and argon and complete the Table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 24
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 25

Text Book Page No: 9

→ How many electrons are present in the M shell, the outermost shell of argon?
Answer:
8

→ What is the maximum number of electrons that can be accommodated in the M Shell?
Answer:
18

→ The ‘K’ shell, which is the first shell, has 1 subshell. The next ‘L’ shell has 2, and so on. What will be the number of subshells in the ‘M’ shell and ‘N’
M = ……………… , N = ……………….
Answer:
M = 3, N = 4

→ Which subshell is common to all shells?
Answer:
S

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Text Book Page No: 10

→ Complete the Table 1.3
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 26
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 27

→ Complete the Table 1.4
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 28
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 29

→ What is the maximum number of electronics that can be accommodated in the ‘s’?
Answer:
2

→ What may be the maximum number of electrons to be filled in the ‘p’ subshell?
Answer:
6

Text Book Page No: 11.

The atomic number of hydrogen is 1(1H)

→ How many electrons are present?
Answer:
1

→ In which shell is the electron filled?
Answer:
‘K’ shell

→ In which subshell?
Answer:
S

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→ How many electrons are present in helium (2He)?
Answer:
2

Text Book Page No: 12

→ Complete the subshell electronic configuration?
Answer:
1s2

→ Write the electronic configuration of Lithium (3Li)
Answer:
1s2 2s1

→ Complete the electronic configuration of beryllium?
Answer:
Be[Z=4] -1s2 2s2

→ Write the electronic configuration of Boron
Answer:
B[Z=5] -1s1 2s2 2p1

→ Write the electronic configuration of Carbon
Answer:
C[Z=6] – 1s2 2s2 2p2

→ Complete the Table 1.6
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 30
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 31

Text Book Page No: 13

→ How was the shell wise electronic configuration of potassium written?
Answer:
2, 8, 8, 1

→ Compare the energies of Is and 2s subshells. Which one has lower energy?
Answer:
1s < 2s

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Question 14.
Among the 3s & 3p subshells which has higher energy?
Answer:
3s < 3p

→Among the 3d & 4s subshells which has higher energy?
Answer:
4s < 3d

→ Write down the subshells in the increasing order of their energies.
Answer:
1s <2s <2p <3s <3p <4s <3d <4p

→ Write the subshell wise electronic configu-ration of potassium.
Answer:
1s2 2s2 2p6 3s2 3p6 4s2

→The electronic configuration of scandium (2lSc) is
Answer:
s2 2s2 2p6 3s2 3p6 3d1 4s2

Text Book Page No: 14

→ Write the electronic configu ration of 22Ti, 23V, the two elements after Sc.
Answer:
22Ti — 1s2 2s2 2p6 3s2 3p6 3d2 4s2
23V — 1s2 2s2 2p6 3s2 3p6 3d3 4s2

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→Which is the noble gas preceding sodium (11Na)?
Answer:
Neon(Ne)

→ Write its subshell electronic configuration.
Answer:
10Ne – 1s2 2s2 2p6

→ Subshell electronic configuration of sodium?
Answer:
11Na – 1s2 2s2 2p6 3s1

Text Book Page No: 15

→ Using the symbol of neon, write the subshell electronic configuration of sodium?
Answer:
[Ne] 3s1

→ Complete the Table 1.7
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 32
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 33

→ Write the subshell electronic configuration of 24Cr
Answer:
24Cr – 1s2 2s2 2p6 3s2 3p6 3d5 4s1

→ On the basis of this, identify the correct electronic configuration of 29Cu from those given below:
Answer:
1s2 2s2 2p6 3s2 3p6 3d9 4s2 – False
1s2 2s2 2p6 3s2 3p6 3d10 4s1 – True

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Text Book Page No: 16

If the subshell wise electronic configuration of an atom is 1s2 2s2 2p6 3s2, find answers to the following:

→ How many shells are present in this atom?
Answer:
3

→Which are the subshells of each shell?
Answer:
K — Is, L — 2s, 2p, M — 3s

→Which is the subshell to which the last electron was added?
Answer:
3s

→ What is the total number of electrons in the atom?
Answer:
12

→ What is its atomic number?
Answer:
12

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→ How can the subshell electronic configuration be written in a short form?
Answer:
[Ne]3s2

Text Book Page No: 17

→ Complete the Table 1.8
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 34
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 35

→ Which is the subshell of lithium to which the last electron was added?
Answer:
S

→ What about the subshell to which the last electron of nitrogen was added
Answer:
p

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→ What is the relation between the subshell to which the last electron was added and the block to which the element belongs?
Answer:
The subshell in which the last electron enters represent the block in which the element belongs.

→ Write the subshell electronic configuration of the following elements and find the blocks to which they belong.
a. 4Be: ………………..
b. 26Fe……………..
c. 18Ar: ……………
Answer:
a. 4Be : 1s2 2s2 — s block
b. 26Fe : Is2 2s2 2p6 3s2 3p6 3d6 4s2 — d block
c. 18Ar : 1s2 2s2 2p6 3s2 3p6 — p block

Visit us to know more about the lithium oxide formula, its properties and its uses.

Text Book Page No: 18

→ Complete the Table 1.9
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 36
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 37

→ Complete the Table 1.10
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 38
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 39

→ What is the relation between number of electrons present in the last ‘s’ subshell and their group number?
Answer:
The number of electrons in the outermost ‘s’
subshell = The group number

Text Book Page No: 20

→ When the s block elements react, do they donate or accept electrons?
Answer:
They donate electrons.

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→ Which type of chemical bond is usually formed?
Answer:
Ionic bonds

→ How many electrons are donated by the first group elements in chemical reactions ?-
Answer:
One

→ How many electrons are donated by the second group elements in chemical reaction?
Answer:
Two

→ Complete the table 1.11
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 40
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 41

→ ‘s’ block elements are present at the extreme left side of the periodic table. Relating to their position, what other characteristics can be listed out?
Answer:

  • More metallic character s
  • Less ionization energy
  • Less electronegativity
  • Lose of electrons in chemical reaction
  • Compounds are mostly ionic
  • Oxides and hydroxides are basic in nature

Text Book Page No: 21

→ Which are the group included in the p block
Answer:
13, 14, 15, 16, 17, 18

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→ In which subshell did the filling of the last electron take place?
Answer:
p subshell

→ Complete the table 1.13.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 42
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 43

Text Book Page No: 22

The outermost subshell wise electronic configuration of an element Y (Symbol is hot real) is 3s2 3p4.

→ To which period and group does this element belongs to?
Answer:
Period = 3, Group = 16

→ Write down the outermost subshell electronic configuration of the element coming just below it in the same group?
Answer:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

→ Find out examples of elements in such different states with the help of the periodic tables?
Answer:
Solid – Li, Be, B, C, Na, Mg, Al, Si
Liquid – Br
Gas – H, He, N, O, F, Ne

→ Which element has the highest ionization energy in each period?
Answer:
Group 18 elements.

Text Book Page No: 23

→ The elements having the highest electronegativity is in the p block. Find its name and position?
Answer:
Fluorine F, Period – 2, p block, Group 17

→ Analyze the general characteristics of the p block elements and prepare a note on this?
Answer:

  • The outermost p subshell of the p block elements contains 1 to 6 electrons.
  • Elements showing positive oxidation state and negative oxidation state are members of this block.
  • There are metals and nonmetals in these blocks.
  • Elements in the solid, liquid and gaseous states are present in p block.

→ Complete the table 1.14
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 44
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 45

→ Which element has a valency 1?
Answer:
Y

→Which element shows metallic character?
Answer:
X

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→ Which element has the highest ionization energy?
Answer:
Y

→ Write the chemical formula of the compound formed by the combination of X and Y and label the oxidation states?
Answer:
Compound: X Y2
Oxidation state: X2+, Y1-

→ Where is the position of d block elements in the periodic table?
Answer:
3rd Group to 12th Group

→ From which period onwards does the d block begin?
Answer:
4

Text Book Page No: 24

→ Put a tick mark ✓’ against the statements below, which are applicable to d block elements.
Answer:
1. ‘✓’ These are metals.
2. ‘✓’ The last electron is filled in the penultimate shell.
3. ‘✗’ In the case of these elements in the 4th period, the last electron is filled in 4s.
4. ‘✓’ These are found in groups 3 to 12 of the periodic table.

Text Book Page No: 25

→ Complete the table 1.16
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 46
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 47

→ How does Fe change to Fe2+?
Answer:
By losing 2 electrons from 4s valence subshell.

Valence electrons calculator with steps – Easy to use … A valence electron calculator is an online tool.

Text Book Page No: 26

→ Write down the subshell electronic configuration of Fe21.
Answer:
1s2 2s2 2p6 3s2 3p6 3d6

There is only a small difference of energy between the outermost s subshell and the penultimate d subshell of transition elements.

→ If so, which will be the subshell from which iron loses the third electron?
Answer:
From 3d sub-shell

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→ Write the electronic configuration of Fe3+ on the basis of this.
Answer:
1s2 2s2 2p6 3s2 3p6 3d5

→ Write the subshell electronic configuration of Manganese (Mn).
Answer:
1s2 2s2 2p6 3s2 3p6 3d5 4s2

→ Complete the table 1.17
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 48
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 49

Text Book Page No: 27

→ Examine these compounds available. Find more colored compounds and extend the list.
Answer:

  • Copper sulfate CuSO4.5H2O – blue,
  • Copper nitrate Cu(NO3)2.6H2O – pink.
  • Potassium permanganate KMnO4 – violet.
  • Ferrous sulfate FeSO4.7H2O – Green,
  • Ferrous nitrate (Fe(NO3)2.6H2O) – light green

Text Book Page No: 28
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 50

→ List out the dements of the s block?
Answer:
A, B

→ Which elements shows+2 oxidation state?
Answer:
B, C, D

→ Which elements contains 5 electrons in the outermost shell?
Answer:
E

→ Which is the element that has 5 p electrons in the outermost shell?
Answer:
G

→ Which are the elements in which the last electron enters the d subshell?
Answer:
C, D

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→ Which element has the highest ionization energy?
Answer:
H

→ Which is the highly reactive nonmetal?
Answer:
G

→ Which elements show -2 oxidation state?
Answer:
F

Text Book Page No: 29

The outermost electron configuration of an element in this is 2s2 2p6

→ Which is the element?
Answer:
H

→ Write down the complete subshell electronic configuration?
Answer:
Is2 2s2 2p6

→ Write any two characteristics of this element?
Answer:

  • Noble element / gases.
  • The outermost shell is completely filled

→ Write the chemical number of questions, the answer of which is an element in the table
Answer:
A G

Periodic Table and Electronic Configuration Let Us Assess

Question 1.
Based on the hints given, find out the atomic number and write down the subshell electronic configuration of elements (Symbols used are not real).
i. A – period 3 group 17
ii. B – period 4 group 6
Answer:
A17 — 1s2 2s2 2p6 3s2 3p5
B24 — 1S2 2s2 2p6 3S2 3p6 3d5 4S1

Question 2.
When the last electron of an atom was filled in the 3d subshell, the subshell electronic configuration was recorded as 3d8 Answer the questions related to this atom.
1. Complete subshell electronic configuration
2. Atomic number
3. Block
4. Period number
5. Group number
Answer:
1. Complete subshell electronic configuration:
1 s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2
2. Atomic number: 28
3. Block : d
4. Period number: 4
5. Groupnumber : 8 + 2 = 10

Question 3.
Pick out the wrong ones from the subshell electronic configuration given below.
a. 1s2 2s2 2p7
b. 1s2 2s2 2p2
c. 1s2 2s2 2p5 3s2
d. 1s2 2s2 2p6 3s2 3p6 3d2 4s2
e. 1s2 2s2 2p6 3s2 3p6 3d2 4s2
Answer:
Wrong electronic configuration
a. 1s2, 2s2, 2p7
(2p maximum 6 electrons only)

c. 1s2, 2s2, 2p5, 3s1 (electrons are filled in 3s only after filling 6 electrons in 2p)

d. 1s2, 2s2, 2p6, 3s2, 3p6, 3d2, 4s1 (electrons are filled in 3d only after filling 2 electrons in 4s)

Question 4.
The element X in group 17 has 3 shells. If so,
a. Write the subshell electronic configuration of the element.
b. Write the period number,
c. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
Answer:
a. Three shells are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s2, 2s2, 2p6, 3s2, 3p5

b. Period-3

c. Y – Third period
∴ shells – 3
1 electron in p – subshell
Total electrons in valence shell 2+1=3 (2 electrons in s + 1 electron in p)
Valency of x – 1(1 electron is recieved – electro negative atom)
Valency of y – 3 (3 electrons are lost – electro positive atom)
Therefore they combine to form compounds with chemical formula YX3
(Symbol of electropositive element first followed by electro negative element).

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Question 5.
The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2.
a. Write down the subshell electronic configuration of this ion.
b. Can this element show variable valency? Why?
c. Write down the chemical formula of one compound formed when this element reacts with chlorine (17CI).
Answer:
a. 29Cu — 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1
Cu2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d9

b. Yes. One electron can be lost from 4s subshell and can’exist as Cu+ ion, It is a d-block element.

c. Copper react with chlorine to form two compounds Cu+, Cu2+ ions react with chlorine to form CuCl and CuCl2 respectively.

Question 6.
Certain subshells of an atom are given below. 2s, 2d, 3f, 3d, 5s, 3p
a. Which are the subshells that are not possible?
b. Give the reason.
Answer:
a. Not possible sub-shells are 2d, 3f

b. d – subshell is not possible in 2nd shell
f – subshell is not possible in 3rd shell

Periodic Table and Electronic Configuration Extended Activities

Question 1.
Prepare the comprehensive table which indicates the name, symbol, electron configuration, subshell configuration of elements having atomic number 1 to 36?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 51

Question 2.
Some information related to the elements of the p bllock in the 17th group of the periodic table are given in the table below. Complete the table and analyze the following questions?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 52

a. What is the family names of elements belonging to the 17th group?
Answer:
Halogen

b. What is their common valency?
Answer:
1

c. Which element has the highest electro negativity ?
Answer:
F

d. Which element has the highest ionization energy?
Answer:
F

e. List out the name and chemical formula of the compounds formed by these elements with block elements?
Answer:

  • sodium chloride – NaCl
  • potassium chloride – KCl
  • magnesium chloride- MgCl2
  • calcium chloride- CaCl2
  • magnesium fluoride- MgF2
  • calcium fluoride – Ca F2
  • sodium iodide – Nal
  • potassium iodide – KI
  • potassium bromide – KBr
  • potassium fluoride – KF

Periodic Table and Electronic Configuration Orukkam Questions and Answers

Question 1.
Complete the table of details about shells and subshells.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 53
a. No of electrons in KLMN shell.
b. No of electrons in each shell.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 54
c. Which subshell is common to all sub-shells?
d. Write names of subshells in accordance with increasing energy level,
e. Identify the incorrect subshell electronic configuration.
– 1s2
– ls2 2p6
– ls2 2s2 2p6
– 1s2 2s2 2p6 3s2 3p2
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 55

a, K – 2 ; L – 8 ; M – 18 ; N – 32

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 56

c. s- Subshell

d. 1s <2s <2p <3s <3p <4s <3d <4p <5s <4d <5p <6s <4f <5s

e. 1s2 2p6

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Question 2.
Atomic number of iron is 26. It exhibits Fe2+, Fe3+ oxidation state. Write the subshell electronic configuration.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 57
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 58

Question 3.
Manganese, a d-block element exhibits I different oxidation state. Why?
a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 59

b. Write the oxidation number and subshell electronic configuration K, Cl and O.
Answer:
Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions.

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 60

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 61

Question 4.
Find out atomic number, group, block period using subshell electronic configuration and then complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 62
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 63

Question 5.
Write down the characteristics of s,d,p, f block elements
Answer:
s-block elements:
Elements in which last electron enters into s-subshell are called s-block elements. It contains group I elements (Alkali metals) and group II elements (Alkaline earth metals).

1st group elements lose one electron during chemical combination. Therefore its oxidation state is +1.

2nd group elements lose two electrons from valence shell during chemical combination and their oxidation state is +2.

The highest shell number in a sub-shell electronic configuration is the period number of that element.

1. Group number characteristics = no.of electrons in valence sub-shell.
2. s block ionization energy & electro negativity decreases downwards.
3. Metallic character & reactivity increases downwards.
4. Lose electrons during chemical combination j and they form ionic compounds.
Their oxides and hydroxides are basic.
Their atomic radii are high in a period.

p-block elements:

  • Last electron enters into p-subshell.
  • Group 13 -18 elements.
  • Highly reactive elements are non-metals – group 17,
  • These are elements with positive and negative oxidation state.

Group number of p-block elements = electrons in last p-subshell + 12

d-block elements:

  • Last electron enters into penultimate d-subshell
  • Known as transition elements.
  • Metals
  • Shows similarity in group and period.
  • Variable oxidation states.
  • Form coloured compounds.

Group = electrons in ‘d’-subshdl + electrons in s-subshell.

f-block elements:

  • Last electron enters into antepenultimate f sub-shell.
  • Contains Lanthanoids and Actinoids.
  • Variable oxidation state.
  • Most of the Actinoids are radioactive.
  • Most of the elements are artificial.
  • U, Th, Pu are used in nuclear reactors.
  • Some elements are used as catalyst in pet-roleum industry.

Periodic Table and Electronic Configuration Evaluation Questions

Question 1.
Write down subshell electronic configuration of Cu1+ and Cu2+
Answer:
Cu1+ – 1s2 2s2 2p6 3s2 3p6 3d10
Cu2+ – 1s2 2s2 2p6 3s2 3p6 3d9

Question 2.
How many ‘s’ subshell electrons are in 1s2, 2s2, 2p6, 3s2, 3p2
Answer:
6 Electrons

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Question 3.
11,17,10 are the atomic number of elements X, Y, and Z.
a. Write down their subshell electronic configuration, group, block, period,
b. Write the molecular formulae of the compound formed when any two of the above elements are combined.
c. Write down the oxidation numbers of the elements in those compounds. Write the subshell electronic configuration of both ions.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 64

b. X Y

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 65

Question 4.
Element ‘X’ is having atomic number 28, it gives two electrons to element ‘Y’.
a Write down the electronic configuration of ‘X’ and its ion
b. In which block ‘X’ belongs?
c. Write down the characteristics of that block
Answer:
a. X28 – 1s2 2s2 2p6 3s2 3p6 3d8 4s2
X2+– 1s2 2s2 2p6 3s2 3p6 3d8

b. d block Compound

c. 1. It exhibits variable oxidation states
2. Forms colored compounds
3. Last electron enters d subshell

Question 5.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 66
a. Write down the group and period of each element.
b. What are the use of writing electronic configuration this fashion?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 67

b. Group and period of the element can be identified easily. In the Same way long electron configuration can be avoided.

Question 6.
24Cr – [Ar] 3d5 4s1
29Cu – [Ar] 3d10 4s1
Why chromium and copper exhibits such electronic configuration ?
Answer:
Half filled and completely filled subshells are most stable. Change in the electronic configuration of 24Cr &, 29Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability.

Periodic Table and Electronic Configuration SCERT Questions and Answers

Question 1.
The electronic configuration of the elements A, B, C, Dare given below.
A – 1s2 2s2 2p6 3s2 3p4
B – 1s2 2s2 2p6 3s2
C – 1s2 2s2 2p6 3s2 3p5
D – 1s2 2s2 2p6 3s1
a. Which of these elements show +2 oxidation state?
b. Which metal belongs to 17th group?
c. Which is the period number of the element A ? What is the basis of your findings?
d. Which of these elements can form basic Oxides?
Answer:
a. B

b. C

c. Period number: 3, Period number = No.of shells

d. B, D

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Question 2.
Two compounds of iron are jpven below.
FeSO4 Fe2(SO4)3
(The oxidation state of sulfate radical is-2)
a. Which ofthese compounds show +2 oxidation state for Fe?
b. Which compounds has Fe3+ ion?
c. Write the subshell electronic configuration of Fe3+ ion.
d. Why do transition elements show variable oxidation states?
Answer:
a., FeSO4
b. Fe2(SO4)3
c. Fe3+ – 1s2 2s2 2p6 3s2 3p6 3d5

d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction.

Question 3.
Identify the incorrect electronic configurations and correct them.
i) 1s2 2s2 2p3
ii) 1s2 2s2 2p6 3s1
iii) 1s2 2s2 2p6 2d7
iv) 1s2 2s2 2p6 3s2 3p6 3d4
Answer:
iii). 1s2 2s2 2p6 3s2 3p5 .
iv). 1s2 2s2 2p6 3s2 3p6 4s2 3d2

Question 4.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 68
Answer:
a. – 2
b. 1
c. 17,
d. – 1
e. 12
f. +12

Question 5.
a. Two compounds XY2, XZ4 are given. The oxidation state of Z is 1. What will be the oxidation state of Y ?
b. Write the molecular formula of the compound formed by Y when it combines with aluminum (Al) having oxidation state +3.
Answer:
a. Y= – 2 (oxidation state of X is +4)
b. Al2 Y3

Question 6.
Pick out the statements which suit to f-block elements.
a. All of them are naturally occurring elements.
b. Uranium and Thorium are f block elements.
c. Last electrons is filled in the shell pre-ceding the outermost shell.
d. last electrons are filled up in the antepenultimate shell.
e. Includes some radioactive elements.
f. Some of them are used as catalyst in petroleum industry.
Answer:
b, d, e, f

Question 7.
The atomic number of four elements are given below. (The symbols ore not real)
A – 8
B – 10
C – 12
D – 18
a. Write the sub-shell electronic configuration of the elements,
b. Which of them are inert gases?
c. Write the chemical formula of the compound formed by two elements other than inert gases.
Answer:
a. A – 1s2 2s2 2p4
B – 1s2 2s2 2p6
C – 1s2 2s2 2p6 3s2
D – 1s2 2s2 2p6 3s2 3p6

b. B, D

c. CA, (C2 A2 is simplified and written as CA)

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Question 8.
The subshell electronic configuration of two elements ends as follows. (Symbols are not real)
P – 3s2 Q – 3p4
a. Write the complete subshell electronic configuration.
b. Find out the oxidation state of each element.
c. The chemical formula of the compound formed by these elements is PQ. Is this statement correct? Justify your answer.
Answer:
a. P – 1s2 2s2 2p6 3s2
Q – 1s2 2s2 2p6 3s2 3p4

b. P = +2, Q = – 2 :

c. Right, valency of both P and Q is ‘2’

Question 9.
Match the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 69
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 70

Question 10.
The atomic number of two elements are given below.
Si – 14 Ni – 28
a. Write the subshell electronic configu-ration of these elements.
b. Find out the group and period of each element.
Answer:
a. Si – 1s2 2s2 2p6 3s2 3p2
Ni – 1s2 2s2 2p6 3s2 3p6 3d8 4s2
b. Si – Period Number – 3, Group number – 14 Ni – Period Number – 4, Group number – 10

Question 11.
The element ‘X’ has 4 shells and its 3d subshell has 6 electrons. (Symbol is not real)
a. Write the complete electronic configu-ration of the element.
b. What is its group number? Which is the block?
c. Write any two characteristics of the block to which element X belongs to.
d. From which subshell the electrons are lost when the element X shows +2 oxidation state.
Answer:
a. 1s2 2s2 2p6 3s2 3p6 3d6 4s2
b. Group number – 8, Block – d
c. All of them are metals
d – block elements are placed in group 3 to group 12
d. s – Sub shell

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Question 12.
The outermost electronic configuration of the element A is 2s2 2p2. (Symbol is not real)
a. Find out the group number and block of the element.
b. Write the chemical formula of the compound formed by A when it combines with chlorine.
c. Write the complete electronic configuretion of the element just below ‘A’ in the j periodic table.
Answer:
a. Group number – 14, Block – P
b. ACl4
c. 1s2 2s2 2p6 3s2 3p2

Question 13.
The figure of an incomplete periodic table is given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 71
a. Which one of these elements shows -2 oxidation state?
b. Which of these elements have 3 electrons in their outermost p subshell?
c. Which element has the highest atomic radius? Which one has the least?
d. Which of these elements show variable oxidation state?
e. Which of these elements has the highest ionization energy?
Answer:
a. G

b.F

c. The element having highest atomic radius – A
The element having lowest Atomic radius – H

d. D, C

e. H

Question 14.
Examine the given electronic configurations.
A – 1s2 2s2 2p6 3s2 3p6 3d10 4s2
B – 1s2 2s2 2p6 3s1
C – 1s2 2s2 2p1 3s2 3p6
D – 1s2 2s2 2p6 3s2
E – 1s2 2s2 2p6 3s2 3p6 4s2
a. Which of these elements belongs to 4th period?
b. Which elements belongs to the same group ?
c. Which element doesn’t participate in chemical reactions generally ?
d. Which element has highest metallic character ?
Answer:
a. A, E
b. B, E
c. C
d. E

Question 15.
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl2, YCl2, YCl3 are formed.
a. What is the specialty of the oxidation number of Y, compared to that of X?
b. Explain the reason for this, on the basis of the subshell based electronic configuration.
Answer:
a. Element X has constant oxidation state. Y shows variable oxidation states.
b. X20 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2
X26 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6
Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part.

Periodic Table and Electronic Configuration Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Arrange the following sub-shells in the in-creasing order of energy 5p, 2s, 4f, 3s, 4s, 3d, 6s
Answer:
2s < 3s < 4s < 3d < 5p < 6s < 4f

Question 2.
Last electron in f-block elements goes to
a. Which shell? Outer shell/Penultimate shell /Antepenultimate shell
b. Which sub-shell? Outer f-subshell Penultimate f-subshell/Antepenulti mate f-subshell.
Answer:
a. Antepenultimate shell
b. Antepenultimate f-sub-shell

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Question 3.
Sub-shell electronic configuration of X is given below.
1s2, 2s2, 2p5
a. The element Y is coming just below the element in same group. Then write the sub-shell electronic configuration of Y.
b. Write the sub-shell electronic configuration of the element next to X in same period.
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p2
b. Is2, 2s2, 2p6

Question 4.
A compound of vanadium pentoxide (V20;) is used as catalyst.
a. What is the oxidation state of vanadium in this compound?
b. How vanadium ion is represented?
c. Write the sub-shell electronic configuration of this ion (V – 23)
Answer:
a. +5
b. V5+
c. 1s2, 2s2, 2p6, 3s2, 3p6

Short Answer Type Questions (Score 2)

Question 5.
Find the wrong electronic configurations from the following. What is wrong in these?
a. 1s2,2s2,2p6,3s2,3p6,3d9,4s2
b. 1s2
c. 1s2, 2s1, 2p6
d. 1s2, 2s2, 2p6, 3s2, 3p2
Answer:
(a) and (c) are wrong electronic con figurations.

In (a) one electron from 4s is to be trans-ferred to 3d since completely filled configu-rations are more stable. So the correct electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

In (c) electrons are filled in 2p only after filling electrons in 2s.

Question 6.
Group and period number of two elements are given.
P – group 17, period – 3
Q – group 2, period – 3
a. Write the sub-shell electronic configuration of each.
b Write the chemical formula of the compound formed by their combination.
Answer:
Answer:
a. P – 1s2, 2s2, 2p6, 3s2, 3p5
Q – 1s2, 2s2, 2p6, 3s2

b. Q is electropositive. P is electro negative;
∴Chemical formula QP2

Question 7.
Write the reason for the statement given
below.
a. d-block elements in the same period show similarity.
b. Transition elements show variable oxidation state.
Answer:
a. Valence shell electrons of d-block elements in same periods are almost same. Valence shell electrons are entering in chemical reaction. Therefore they shows similarity.

b. Energy of electrons in s-subshell and inner d- subshells are almost same. Therefore s- electrons or s and d electrons take part in chemical reaction and show variable oxidation state.

Short Answer Type Questions (Score 3)

Question 8.
Write the sub-shell electronic configuration of following elements. Predict the block, group and period. (Symbols are not real)
a. M – 27 b. N – 19 c. P – 15
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p6, 3d7, 4s2
block – d; group – 9; period – 4.
b. 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
block – s; group – 1; period – 4
c. 1s2, 2s2, 2p6, 3s2, 3p3
block – p; group – 15; period – 3

Question 9.
Observe the model of periodic table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 72
a. Which element is having S electrons in valence shell?
b. Which elements are having 2 electrons in valence sub-shell?
c. Which element is having last electron in3p?
d. Which element ends with electronic configuration 4d5, 5s1 ?
Answer:
a. B;
b.A, C;
c. C, D;
d. E

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Question 10.
Calculate oxidation state of transition elements in the following compounds.
Answer:
a. KMnO4 – Mn – 7+
b. Cr2 O3 – Cr – 3+
c. K2Cr2O7 – Cr – 6+

Question 11.
Atomic number of some elements are given. A – 15, B – 8, C – 11, D – 18, E – 20, F – 34, G – 10
a. Which are the elements in same period?
b. Which are the elements in same group?
Answer:
A -1s2, 2s2, 2p6, 3s2, 3p3
(group -15 period – 3)
B – 1s2, 2s2, 2p4 (group –16 period – 2)
C – 1s2, 2s2, 2p6, 31 (group – 1 period – 3)
D – 1s2, 2s2, 2p6, 3s2, 3p6 (group – 18 period – 3)
E – 1s2, 2s2, 2p6, 3s2, 3p6, 4s2.(group – 2 period – 4)
F – 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p4 (group – 16 period – 4)
G – 1s2, 2s2, 2p6 (group – 18 period – 2)
a. A, D same period ;
B, G same period 1
b. B, F same group ;
D, G same group

Long Answer Type Questions (Score 4)

Question 12.
Electronic configuration of some elements are given. Write answers to the following questions.
i. [Ne] 3s2
ii. [Ar ] 3d2,4s2
iii. [Xe] 6s2
iv. [Ne]3s2
v [Ne] 3s2,3p5
a. Which metal is having high reactivity?
b. Which is having possibility of formation of colored compounds?
c. Which is the non-metal?
d Which element shows the possibility of +2 oxidation state?
Answer:
a. [Xe] 6s1
b. [Ar ] 3d2, 4s2
c. [Ne] 3s2, 3p5
d. [Ne] 3s2,[Ar] 3d2, 4s2

Question 13.
Pick the wrong statement from the following.
a. Elements with atomic number 5 belong to group 15.
b. Electronic configuration of scandium (Atomic number 21) is 2,8,8,3.
c. d-block elements are known as transition elements.
d. All s-block elements are metals.
Answer:
a. Wrong. It belongs to group 13.
b. Wrong. Electronic configuration
2,8,9,2 (1s2, 2s2, 2p6, 3s2, 3p6,3d1, 4s2)
c. Correct .
d. Correct

Question 14.
Look at the Bohr model of X-atom.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 73
a. Write the sub-shell electronic configuration of this atom.
b. Mention the compounds in which d-subshell electrons are taking part in chemical reaction during their formation.
XCl2, XO2, X2O7
c. Write the electronic configuration of X ions in the above three compounds.
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2

b. XCl2 – ion X2+ (electrons in 4s only)
XO2 – ion X+4 (2 electrons in 4s and 2 electrons in 3d)
X2O7 – ion X7+ (2 electrons in 4s and 5 electrons in 3d)
XO2, X2O7 d – subshell electrons are taking part in chemical reaction during the formation of X207

c. X2+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
X4+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d3
X7+ – 1s2, 2s2, 2p6, 3s2, 3p6

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Question 15.
Select the suitable one from the following columns A, B, C.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 74
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 75

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 6 Nomenclature of Organic Compounds and Isomerism in Malayalam

Students can Download Chemistry Chapter 6 Nomenclature of Organic Compounds and Isomerism Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 6 Nomenclature of Organic Compounds and Isomerism in Malayalam

Nomenclature of Organic Compounds and Isomerism Text Book Questions and Answers

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 6 Nomenclature of Organic Compounds and Isomerism in Malayalam 1

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alkene reactions cheat sheet summary for organic chemistry reactions.

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Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

You can Download Chemical Reactions of Organic Compounds Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

Chemical Reactions of Organic Compounds Text Book Questions and Answers

Chapter 7 Chemistry Question 1.
Complete stages 2, 3 and 4 in the respective order.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 1
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 2

Organic Chemistry Class 10 SCERT Question 2.
What are the compounds formed when CH3-CH3 (ethane) undergoes substitution reaction with chlorine? Write them.
Chemistry Chapter 7 Test Answer:
CH3 – CH2 Cl, CH3 – CHCl2,
CH3 – CCl3, CH2Cl – CCl3,
CHCl3 – CCl3, CCl3 – CCl3

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alkyne reactions cheat sheet summary for organic chemistry reactions.

Chapter 7 Chemical Reactions Question 3.
Write down the structural formulae of ethane and ethene.
Answer:
CH3 – CH3 – Ethane
CH2 = CH2 – Ethene

Methyl Propanoate Question 4.
What is the peculiarity of the Carbon- Carbon bond in ethene?
Answer:
In ethene, There is carbon – carbon double bond

Text Book Page No: 121

Question 5.
What do we get as the product ?
Answer:
Ethane

Question 6.
Which hydrocarbon is the reactant here? ………..
Answer:
Unsaturated propene

Ethene Structural Formula Question 7.
Is the product saturated or unsaturated ?
Answer:
Saturated Compounds

Question 8.
Complete table 7.1
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 3
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 4

Text Book Page No: 123

Substitute of Polyvinyl Alcohol Question 9.
Complete table 7.2 Suitably.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 5
Answer:

Monomer Polymer Uses
Vinyl Chloride PVC Pipe, Helmet
Ethene Polyethane Carry bags
Isoprene Natural rubber (Poly Isoprene) Tire
Tetra Fluro ethene Teflon Nonstick pan

Text Book Page No: 124

Daily Chemicals Question 10.
Can you write the balanced chemical equation for the combustion of the fuel butane (C4H10) ?
Answer:
2 C4H10(g) + 13O2(g) → 8CO2 + 10H2O + heat

Text Book Page No: 125

Question 11.
Complete Table 7.3 and 7.4 contain¬ing chemical reactions of hydrocarbons.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 6
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 7

Question 12.
Match Columns A, B, and C suitably.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 8
Answer:

Reactants (A) Products (B) Name of Reaction (C)
CH3 – CH3 + Cl2 CH3 – CH2 Cl + HCl Substitution Reaction
C2H6+O2 CO2 + H2O Combustion
n CH2 = CH2 [CH2 – CH2]n Polymerisation
CH3– CH2 – CH3 CH2 = CH2 + CH4 Thermal Cracking
CH = CH + H2 CH2 = CH2 Addition Reaction

Question 13.
CH2 – OH, CH3 – CH2 – OH
Can you write the IUPAC names of these two compounds?
Answer:
CH3 – OH – Methanol
CH3 – CH2 – OH – Ethanol

Text Book Page No: 126

Question 14.
Complete the following word web including more uses of ethanol.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 9
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 10

Question 15.
List out the uses of ethanoic acid.
Answer:

  • In the manufacture of rayon
  • In the rubber and silk industry.
  • Vinegar – Impart Sour taste for food item.
  • Used as preservative.

Text Book Page No: 129

Question 16.
Examine the given structural formulae and select the esters. You may also identify the chemicals required for their preparation.
1. CH3 – CH2 – COO – CH3
2. CH3 – CH2 – COOH
3. CH3 – CH2 – CO – CH3
4. CH3 – OH
5. CH3 – CH2 – CH2OH
6. CH3 – COOH
7. CH3 – COO – CH2 – CH2 – CH3
Answer:
1. CH3 – CH2 – COO – CH3
7. CH3 – COO – CH2 – CH2 – CH3 are esters
1. CH3 – CH2 – COO – CH3
CH3 – CH2 – COOH + OH – CH3 \(\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }\) CH3 – CH2 – COO – CH3+ H2O
7. CH3 – COO – CH2 – CH2 – CH3
CH3 – COOH + OH – CH2 – CH2 – CH3 \(\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }\) CH3 – COO – CH2 – CH2 – CH3 + H2O

Text Book Page No: 130

Question 17.
Take 10 mL distilled water in a test tube and take the same volume of hard water in another test tube. Add a few drops of soap solution to both the test tubes and shake well. Do both the test tubes contain the same quantity of foam? Which test tube contains more foam? What do you infer?
Answer:
No. Both the test tube does not contain same quantity of foam. Distilled water taken test tube contains more foam. Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather.

Question 18.
Take 10 mL each of hard water in two test tubes. Add a few drops of soap solution in the first test tube and add the same amount of detergent solution in the second one. Shake both the test tubes well. What do you observe? Which test tube contains more foam?
Answer:
Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather. But detergents do not give insoluble components on reaction with these salts. Hence detergents are more effective than soaps in hard water.

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Question 19.
List out the merits and demerits of detergents, compared to soaps.
Answer:
Merit:
Detergents are more effective than soaps in hard water.
Detergents are effective in acidic solutions.
Demerits:
excessive use of the detergents causes environmental problems. The microorganisms in water cannot decompose the components of detergents. Hence the detergents released into water lead to the destruction of aquatic life. For example, the detergents which contain phosphate increases the growth of algae and limits the quantity of oxygen. Therefore, it dej creases the quantity of oxygen for the breath of the organisms in water and causes their destruction.

Let Us Assess

Question 1.
Given below are two chemical equations.
a. CH2 = CH2 + H2 → A
b. \(\mathrm{A}+\mathrm{Cl}_{2} \quad \stackrel{\text { sunlight }}{\longrightarrow} \mathrm{B}+\mathrm{HCl}\)
Identify the compounds A and B. Name these reactions.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 11

Question 2.
Name the important chemical reactions of hydrocarbons. Give one example for each.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 12
e. Thermal cracking:
CH3 – CH2 – CH2 -CH3 → CH2 = CH2 + CH3 – CH3

Question 3.
Write chemical formula of propane. Write the names and structural formulae of two compounds, that may be formed during its substitution reaction with chlorine.
Answer:
CH3 – CH3 – CH3 Propane
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 13

Question 4.
Complete the equation for the following chemical reaction. Name this reaction.
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}+\ldots \ldots \ldots \ldots \mathrm{O}_{2} \rightarrow–\mathrm{-}+\)
Answer:
CH3 – CH2 – CH2 – CH3 + 13/2 O2
→ 4CO2 + 5 H2O , Combustion.

Question 5.
Which of the given molecules can form po¬lymers ? Butane, Propane, Propane, Methane, Butene.
Answer:
Propene, Butene

Extended Activities

Question 1.
You are familiar with different chemical reactions of hydrocarbons. Identify the situations in daily life in which these are used.
Answer:
a. Substitution Reaction : Chloroform, CCl4 preparation
b. Addition Reaction : Conversion of unsaturated compounds into saturated.
c. Combustion: Preparation of polymers like PVC.
d. Thermal Cracking: Butane (LPG)can be prepared from higher hydrocarbons

Question 2.
List out the different uses of ethanol. Pre-pare an essay on its adverse effects on human body and the related social issues when it is used as a beverage.
Answer:
Uses of ethanol :

  • Fuels
  • Medicines
  • Preservatives
  • Preparation of organic compounds

Health problems :

  • Reason for aneamia .
  • Increases possibility of cancer
  • Problems related with heart

Social problems:

  • Reason for spoiling family relationships
  • May cause financial crisis

Question 3.
You know how to make soap, don’t you? Try to prepare soaps of different colors and fragrance.Prepare a short note on chemistry of soaps.
Answer:
Fats and oils are esters. They react with alkalies such as NaOH, KOH to form Sodium/ Potassium Salts of their carboxylic acids and glycerol.ester formed from fatty acids + NaOH / KOH → Soap + glycerol. Fatty acids such as palmitic acid, stearic acid react with alcohol, glycerol to form esters. Oils and fats are esters formed by the reaction between glycerol with fatty acid and stearic acid. Soaps are the salts formed when these react with alkalies.

Chemical Reactions of Organic Compounds Orukkam Questions and Answers

Question 1.
After completing the chemical reactions write down to which category they belong.
a. CH2Cl + Cl2 → …….. + HCl
b. CH = CH+H2 →…………
c. CH4 + 2O2 → …….. + H2O
d. CH3 – CH2 – CH3 → ……….
Answer:
a. CH2Cl + Cl2 → CH2Cl2 + HCl Substitution reaction
b. CH = CH + H2 → CH2 = CH2 Addition Reaction
c. CH4 + 2O2 → CO2 + H2O Combustion
d. CH3 – CH2 – CH3 → CH2 = CH2 + CH4 Thermal cracking

Question 2.
Rearrange the table suitably.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 14
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 15
Question 3.
Methane is reacting with Cl in presence of sunlight. Complete equation of that reaction.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 16
a. Write down the reaction of C3H8 with chlorine.
b. What type of reaction is this?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 17
C3HCl7+Cl2 → C3Cl8 + HCl
b. Subsititution Reaction

Question 4.
Examples of additon reaction are given below, complete the equation.
a. CH2 = CH2 + H2 → ………
b. CH2 = CH + Cl2 → ……..
c. CH = CH + H2 → ……….
d. CH = CH + Cl2→ ………
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 17a

Question 5.
a. Examples for combination of Hydrocarbon are given below complete the equation and balance it.
CH4 + O2 → ……. + ……….
C2H6 + O2 → ……. + ………
C3H8 + O2 → …….. + ………
b. Products formed on combustion of Hydrocarbon are ………….
Answer:
a. CH4 + O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
b. Carbon dioxide and Water

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Question 6.
a. Name the product and what type of reaction is this?
nCH2 = CHCl → ………..
b. Write down the names of monomer in it.
c. Give examples for natural polymers
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 18
Polyvinyl chloride, polymerization
b. Vinyl chloride
c. Polyisoprene, Protein

Question 7.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 19
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 20

Question 8.
Arrange the points in two separate columns write column heading also.
a. CO and H2 are reacted in presence of a catalyst to form compound.
b. Sugar cane juice is fermented.
c. It is known as wood spirit.
d. It is used to make paint & varnish.
e. Used for drinking.
f. It is used for adding in industrial spirit.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 21
Question 9.
a. Ethanol has very large industrial utility. When it enter into our body it creates large amount of problems in our body as well as in our society. List out the probl-em happening in our body and in the society.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 22
b. In industrial ethanol always Methanol is added to prevent misuse by humans. Name the process and what are the side effects formed after consuming it?
Answer:
a.

In human body In society
Liver problems Economic problems
Cliolestrol Family issues
Kidney problems Loses personality

b. Denatured Spirit:

  • Loses eyesight permanently
  • Can lead to death
  • Vomiting

Chemical Reactions of Organic Compounds SCERT Questions and Answers

Question 10.
Analyze the reactions and answer the following questions.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 23
a. Identify A, B.
b. Write the name of the compound ‘a’.
c. Write the name of the reaction by which ‘b’ is formed.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 24
b. Polythene
c. Addition reaction.

Protein Molecular Weight Calculator. Protein Molecular Weight accepts a protein sequence and calculates the molecular weight.

Question 11.
Some reactions of propane are given.
i. Hydrogen atoms are substituted one by one, in presence of sunlight.
ii. When heated in the absence of air, it decomposes to hydrocarbons with lesser molecular mass.
iii. Combines with oxygen to give C02 and H2O.
a. Identify the type of reaction in each case.
b. Write the chemical equation of the reaction (ii).
Answer:
a. i. Substitutional reaction
ii. Thermal cracking
iii. Combustion
b. CH3 – CH2 – CH3 → CH2 = CH2 + CH4

Enter a valid molecular formula and press the calculate button to determine the correct molar mass calculator with steps.

Question 12.
Analyse the reactions and answer the following questions.
\(\text { i. } \mathrm{CH}_{3}-\mathrm{OH}+\mathrm{CO} \stackrel{\text { Catalyst }}{\longrightarrow} \ldots \mathrm{A}\)
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 25
\(\text { iii. } \mathrm{A}+\mathrm{B} \stackrel{\mathrm{Con} . \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \quad \ldots . \mathrm{C} \ldots .+\mathrm{H}_{2} \mathrm{O}\)
a. Identify A, B, C.
b. What is the general name/ class to which product ‘C’ belongs? Write the IUPAC name.
Answer:
a. A — CH3 – COOH
B – Methanol/CH3 – OH
C – CH3 – COO – CH3
b. Esters, Methyl ethanoate

Question 13.
Analyze the given reactions and answer the following questions.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 26
a. Identify A and B
b. What is the name of reaction by which ‘B’ is formed?
Answer:
a. A – CH2 = CH2, B – CH3 Cl
b. Substitution reaction

Question 14.
Some reactions regarding the production of ethanol are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 27
a. Identify A and B.
b. Write the name of the ester formed when the product B reacts with propanoic acid,
c. Write the chemical equation for the formation of the ester.
Answer:
a. A- C6 H12 O6 B – C2H5 – OH
b. Ethyl Propanoate
c. CH3 – CH2 – COOH + HO – CH2 – CH3
CH3 – CH2 – COO – CH2 – CH3 + H2O

First, the chemical formula calculator makes sure that the equations have the same number of atoms on both sides of the certain equation.

Question 15.
Acetylene (ethyne) is prepared in the laboratory when calcium carbide reacts with water. Write the chemical equations of the reactions for converting it to ethane.
Answer:
CH = CH + H2 \(\stackrel{N i}{\longrightarrow}\) CH2 = CH2
CH2 = CH2 + H2 \(\stackrel{N i}{\longrightarrow}\) CH3 – CH3

Question 16.
Complete the table
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 28
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 29

Question 17.
a Write the structure of the organic comp¬ound with molecular formula QH.
b. What is the name of the compound formed when one hydrogen atom of benzene is replaced with methyl radical?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 30
b. Methyl benzene (Toluene)

Question 18.
Two equations are given below.
i. CH = CH + HCT → ……… A ………
ii. nA → B
a. Identify A and B.
b. Identify the type of reaction (i)?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 31
b. Addition reaction

Question 19.
Three equations are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 32
a. Identify P, Q, R
b. Identify the name of the chemical reaction (ii) and (iii).
c. Write the IUPAC name of R.
Answer:
a. P – CH2 = CH2
Q – CH3 – CH3
R – CH3 – CH2Cl
b. ii. Addition reaction
iii. Substitution Reaction
c. Chloroethane

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Question 20.
Ethanol is an industrially important compound.
a. What is the name of 8-10% solution of ethanol?
b. How is it converted into rectified spirit?
c. What is denatured spirit?
Answer:
a. Wash
b. Fractional distillation of wash
c. Product obtained by adding poisonous (methanol, pyridine) substances.

Question 21.
Uses of some important organic compounds are given. Pick out the suitable compounds from the box.
Power alcohol, Teflon, Polythene,
Ethanoic acid, Ethanol
a. For the preparation of rayon,
b. For making the coating of inner surface of non-stick cookware,
c. Solvent in paint industry,
d. As fuel in motor vehicles
Answer:
a. Ethanoic acid
b. Teflon
c. Methanol
d. Power alcohol

Question 22.
Some reactants, products, and names of reactions are given in the table. Complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 33
Answer:
a. Substitution reaction
b. CH = CH2
c. HBr
d. Addition reaction
e. H2O
f. Combustion
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 34
h. Polymerization

Question 23.
Pick out the suitable compounds from the box for the following reactions.
CH4, C2H4, C3H8, CH3Cl
a Thermal cracking
b. Addition Reaction
Answer:
a. C3H8
b. C2H4

Chemical Reactions of Organic Compounds Exam Oriented Questions & Answers

Very Short Answer Type Questions (Score 1)

Question 24.
Equation of some chemical Reactions are given below. Complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 35
Answer:
a. A → CH3 – CH3
B → CH3 – CH2 – Cl
C → HCl
b. A → CH2 = CH2
B → CH3 – CH2 – Cl

Question 25.
Teflon used as nonstick polymer.
a. Write the structure of the monomer of this. Write the IUPAC name.
b. Write the reaction equation for the preparation of the monomer.
Answer:
a. CF2 = CF2 Tetra fluroethene
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 36

Question 26.
Recognize and write A, B, C from following equation.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 37
Answer:
A. CH3 – CH2 – OH
B. CH3 – COOH – OH
C. H2O

Short Answer Type Questions (Score 2)

Question 27.
Some chemical equations are given below. Write each type of chemical reaction.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 38
CH3 – CH = CH2 + CH3 – CH3
Answer:
a. Polymerization
b. Substitution reaction
c. Addition reaction
d. Combustion
e. Thermal cracking

Question 28.
Look at the following reactions on heating pentane.
i. In the absence of air
ii. In the presence of air
a. Write the name of the reaction (i), (ii).
b. Write the chemical equation for the reaction.
Answer:
a. Reaction (i) – Thermal Cracking
Reaction (ii) – Combustion
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 39
ii. for writing the combustion equation for hydrocarbons we can use the equation.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 40

Question 29.
Write each of the following.
a. Molasses
b.Wood spirit
c. Vinegar
d. Esters
Answer:
a. Molasses is mother liquor left after the crystallization of sugar from sugar cane juice.
b. Poisonous chemical methanol (CH3 – OH) is known as wood spirit.
c. 5-8% ethanoic acid (CH3 – COOH) is known as vinegar.
d. esters are salts formed by the reaction ale ‘ whole and organic acids. They have the smell of fruits and flowers.

Short Answer Type Questions (Score 3)

Question 30.
Chloroform can be prepared from Methane.
a. What is the chemical formula of chlor of or m?
b. What is the name of reaction when chloroform is prepared from methane?
c. Write the chemical equation for the reaction.
Answer:
a. CHCl3
b. Substitution Reaction
\(\mathrm{c.} \mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text {sunlight}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\)
CH2 Cl + Cl2 → CH2Cl2 + HCl
CH2 Cl2 + Cl2 → CHCl3 + HCI

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Question 31.
Methyl Ethanoate is an ester,
a. Write the structural formula.
b. Write the structural formula of alcohol and carboxylic acid needed for the preparation.
c. Write the reaction equation for the preparation of this ester.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 41

Question 32.
Fill in the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 42
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 43

Question 33.
Write one use of each of the following,
a. Methanol
b. Power alcohol
c. Butane
Answer:
a. Used as solvent in the preparation of paint
b. Fuel in motor vehicle.
c. Cooking gas (LPG)

The enthalpy calculator lets you find the enthalpy of any endothermic or exothermic reaction.

Question 34.
Write the reason for the following statements.
a. Hydrocarbons like butane are used as fuel.
b. Drinking denatured spirit is harmful
c. Esters are used in perfumes and fruit juice.
Answer:
a. Combustion of hydrocarbon produces plenty of heat.
b. Ethanol on addition with poisonous material is called denatured spirit.
c. Esters have the pleasant smell of flowers and fruits.

Question 35.
Write structural formula of following com-pounds!
a. Benzene
b. Phenol
c. Toluene
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 44

Long Answer Type Questions (Score 4)

Question 6.
Ethyne is a compound that belongs to the class of alkynes.
a. Write chemical formula of ethyne.
b. Write the chemical equation for the preparation of following compounds from ethyne.
i. PVC
ii. 1, 2 – dichloroethane
iii. Chloro ethane
Answer:
a. CH = CH
b. i. CH = CH + HCl →
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 45
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 46

Question 37.
Molecular formula of some compounds are given in the box.
C2H4 C6H14 CH3 – CH2 – CI
CH3 – COOH C6H6
a. Which is aromatic compound?
b. Which can be prepared by substitution reaction?
c. Which monomer is used in preparation of polythene?
d. Which compound can be used in food?
Answer:
a. C6H6
b. CH3 – CH2 – Cl
c. C2H4
d. CH3COOH

Question 38.
Answer the following:
a. Name the reaction for the conversion of sugar solution into ethanol.
b. Which enzymes are used in this reaction
c. Chemical involved in grape spirit.
d. Name the monomer of P VC.
e. Products formed during cracking of propane.
Answer:
a. Fermentation
b. Invertase, Zymase
c. Ethanol (CH3 – CH2 OH)
d. Polyvinyl chloride (CH2 = CH – Cl)
e. Ethene(CH2 = CH2), Methane (CH4)

In this post, you’ll learn how to calculate bond order using molecular orbital theory, Lewis structure, and a variety of formulas and examples.

Plus Two Physics Notes Chapter 12 Atoms

Students can Download Chapter 12 Atoms Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 12 Atoms

Introduction
What is the arrangement of +ve charge and the electrons inside the atom? In other words, what is the structure of an atom?

Alpha-particle Scattering And Rutherford’s Nuclear Model Of Atom
Rutherford’s scattering experiment:
Plus Two Physics Notes Chapter 12 Atoms - 1

Experimental arrangement:
α particles are incident on a gold foil (very small thickness) through a lead collimator. They are scattered at different angles. The scattered particles are counted by a particle detector.

Observations:
Most of the alpha particles are scattered by small angles. A few alpha particles are scattered at an angle greater than 90°.

Plus Two Physics Notes Chapter 12 Atoms

Conclusions

  1. Major portion of the atom is empty space.
  2. All the positive charges of the atom are concentrated in a small portion of the atom.
  3. The whole mass of the atom is concentrated in a small portion of the atom.

Rutherford’s model of atom

  1. The massive part of the atom (nucleus) is concentrated at the centre of the atom.
  2. The nucleus contains all the positive charges of the atom.
  3. The size of the nucleus is the order of 10-15m.
  4. Electrons move around the nucleus in circular orbits.
  5. The electrostatic force of attraction (between proton and electron) provides centripetal force.

1. Alpha-particle trajectory and Impact parameter:
The impact parameter is the perpendicular distance of the initial velocity vector of the a particle from the centre of the nucleus.
Plus Two Physics Notes Chapter 12 Atoms - 2
It is seen that an α particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the α particle rebounds back. For a large impact parameter, the α particle goes nearly undeviated and has a small deflection.

2. Electron orbits (Rutherford model of atom):
In Rutherford atom model, electrons are revolving around the positively charged nucleus. The electro-static force of attraction between the positive charge and negative charge provide centripetal force required for rotation.
For a dynamically stable orbit,
Centripetal force = Electrostatic force of attraction
Fc = Fe
Plus Two Physics Notes Chapter 12 Atoms - 3
Thus the relation between the orbit radius and the electrons velocity,
Plus Two Physics Notes Chapter 12 Atoms - 4
Total energy of electron of Hydrogen atom (Rutherford model atom):
From eq. (1), we get
Plus Two Physics Notes Chapter 12 Atoms - 5

Plus Two Physics Notes Chapter 12 Atoms
∴ Kinetic energy of electron
KE = \(\frac{1}{2}\)mv2 ……….(3)
Substituting eq.(2) in eq. (3) we get
KE = \(\frac{e^{2}}{8 \pi \varepsilon_{0} r}\) …………(4)
The electrostatic potential energy of hydrogen atom
\(\frac{e^{2}}{8 \pi \varepsilon_{0} r}\)
u = \(\frac{-e^{2}}{4 \pi \varepsilon_{0} r}\) ………..(5)
∴ The total energy E of the electron in a hydrogen atom
E = K.E + Potential energy (U)
Plus Two Physics Notes Chapter 12 Atoms - 6
The total energy of the electron is negative. This implies that the electron is bound to the nucleus.
If E is positive, the electron will escape from the nucleus.

Atomic Spectra
There are two types of spectra

  1. Emission spectra
  2. Absorption spectra

1. Emission spectra:
When an atomic gas or vapor is excited, the emitted radiation has a spectrum which contains certain wavelength only. A spectrum of this kind is termed as emission line spectrum. It consists of bright lines on a dark background.

Absorption spectra:
When white light passed through a gas, the transmitted light has spectrum contain certain wavelength only. A spectrum of this kind is termed as absorption line spectrum. It consists of dark lines on a bright background.

1. Spectral series:
Plus Two Physics Notes Chapter 12 Atoms - 7
The frequencies of the light emitted by a particular element exhibit some regular pattern. Hydrogen is the simplest atom and therefore, has the simplest spectrum, the spacing between lines of the hydrogen spectrum decreases in a regular way. Each of these sets is called a spectral series.

The first such series was observed by a Johann Jakob Balmer in the visible region of the hydrogen spectrum. This series is called Balmer series. Balmer found a simple empirical formula for the observed wavelengths.
Plus Two Physics Notes Chapter 12 Atoms - 8
where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107m-1. This equation is also called Balmer formula.

Other series of spectra for hydrogen were discovered. These are known, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:
Lyman series:
Plus Two Physics Notes Chapter 12 Atoms - 9

Plus Two Physics Notes Chapter 12 Atoms
Balmer series:
Plus Two Physics Notes Chapter 12 Atoms - 10
Paschen series:
This series is in the infrared region. For this series the electron must jump from higher orbit to the third orbit.
Plus Two Physics Notes Chapter 12 Atoms - 11
Bracket series:
This series is the infrared region, for this the electron must jump from higher energy level to fourth orbit.
Plus Two Physics Notes Chapter 12 Atoms - 12
P-fund series:
This series is in the infrared region.
Plus Two Physics Notes Chapter 12 Atoms - 13

Bohr Model Of Hydrogen Atom
Limitations of Rutherford model:
1. Circular motion is an accelerated motion, an accelerated charge emit radiations. So that electron should emit radiation. Due to this emission of radiation, the energy of the electron decreases. Thus the atom becomes unstable.

2. There is no restriction for the radius of the orbit. So that electron can emit radiations of any frequency.

Bohr postulates:
Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

  • Electrons revolve round the positively charged nucleus in circular orbits.
  • The electron which remains in a privileged path cannot radiate its energy.
  • The orbital angular momentum of the electron is an integral multiple of h/π.
  • Emission or Absorption of energy takes place when an electron jumps from one orbit to another.

Radius of the hydrogen atom:
Consider an electron of charge ‘e’ and mass m revolving round the positively charged nucleus in circular orbit of radius ‘r’. The force of attraction between the nucleus and the electron is
Plus Two Physics Notes Chapter 12 Atoms - 14
This force provides the centripetal force for the orbiting electron
Plus Two Physics Notes Chapter 12 Atoms - 15

Plus Two Physics Notes Chapter 12 Atoms
According to Bohr’s second postulate, we can write
Angular momentum, mvr \(=\frac{n h}{2 \pi}\).
ie. v = \(\frac{n h}{2 \pi m r}\) _____(4)
Substituting this value of ‘v’ in equation (2), we get
Plus Two Physics Notes Chapter 12 Atoms - 16
Energy of the hydrogen atom:
The K.E. of revolving electron is
K.E\(=\frac{1}{2}\) mv2 ______(6)
Substituting the value of equation (3) in eq.(6), we get
K.E = \(\frac{1}{2} \frac{e^{2}}{4 \pi \varepsilon_{0} r}\) ______(7)
The potential energy of the electron,
P.E = \(\frac{-e^{2}}{4 \pi \varepsilon_{0} r}\) _______(8)
ie. The Total energy of the hydrogen atom is,
T.E = Ke + PE
Plus Two Physics Notes Chapter 12 Atoms - 17
Substituting the value of equation (5) in equation (9) we get
Plus Two Physics Notes Chapter 12 Atoms - 18

Plus Two Physics Notes Chapter 12 Atoms

1. Energy levels
Ground state (E1):
Ground state is the lowest energy state, in which the electron revolving in the orbit of smallest radius.
For ground state n = 1
∴ Energy of hydrogen atom E1 = \(\frac{-13.6}{n^{2}}\) = -13.6 ev.

Excited State (E2):
When hydrogen atom receives energy, the electrons may raise to higher energy levels. Then atom is said to be in excited state.

First Excited state:
For first excited state n = 2
∴ Energy of first excited state E2 = \(\frac{-13.6}{2^{2}}\) = -3.04ev
Similarly energy of second excited state
E3 = \(\frac{-13.6}{3^{2}}\) = -1.51ev

Energy difference between E1 and E2 of H atom:
The energy required to exist an electron in hydrogen atom to its first existed state.
∆E = E2 – E1 = 3.4 – 13.6 = 10.2eV.

Ionization energy:
Ionization energy is the minimum energy required to free the electron from the ground state of atom. (ie. n = 1 to n = ∞)
The ionization of energy of hydrogen atom = 13.6 ev

2. Energy level diagram of hydrogen atom:
Plus Two Physics Notes Chapter 12 Atoms - 19
Note:
An electron can have any total energy above E = 0ev. In such situations electron is free. Thus there is a continuum of energy states above E = 0ev.

The Line Spectra Of The Hydrogen Atom
According to the third postulate of Bohr’s model, when an atom makes a transition from higher energy state (ni) to lower energy state (nf), photon of energy hvif is emitted.
ie. hνif = Eni – Enf

Get and Sign Counting Atoms Calculator Form.

Plus Two Physics Notes Chapter 12 Atoms

De Broglie’s Explanation Of Bohr’s Second Postulate Of Quantization
Louis de Broglie argued that the electron in its circular orbit, behalf as a particle wave. Particle waves can produce standing waves under resonant conditions.
The condition to get standing wave,
2πrn = nλ
n = 1, 2, 3……..
The quantized electron orbits and energy states are due to the wave nature of the electron.

DeBroglie’s Proof for Bohr’s second postulate:
According to De Broglie, the electron in a circuit orbit is a particle wave. The particle wave can produce standing waves under resonant conditions. The condition for resonance for an electron moving in nth circular orbit of radius rn,
2πrn = nλ______(1)
n = 1, 2, 3………
If the speed of electron is much less than the speed of light, wave length
Plus Two Physics Notes Chapter 12 Atoms - 20

Plus Two Physics Notes Chapter 12 Atoms
Note:
The quantized electron orbits and energy states are due to the wave nature of the electron.

Limitations of Bohr atom model:

  1. The Bohr model is applicable to hydrogenic atoms. It cannot be extended to many electron atoms such as helium
  2. The model is unable to explain the relative intensities of the frequencies in the spectrum.
  3. Bohr model could not explain fine structure of spectral lines.
  4. Bohr theory could not give a satisfactory explanation for circular orbit.

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals

You can Download Compounds of Non-Metals Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals

Compounds of Non-Metals Text Book Questions and Answers

Text Book Page No: 79

SSLC Chemistry Chapter 5 Question 1.
Take a little ammonium chloride (NH4Cl) in a watch glass and add a little calcium hydroxide (Ca(OH)2) to it. Stir well. Can you sense any smell?
Answer:
Irritating smell is experienced.

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compounds of nonmetals Question 2.
Show wet blue and red litmus papers over the watch glass one by one. Which litmus paper shows a color change ?
Answer:
Red litmus turns to blue

5 nonmetals Question 3.
Is the gas acidic or basic?
Answer:
It is basic

Text Book Page No: 80

Question 4.
Why ammonia gas is passed over quick lime (CaO)?
Answer:
To remove the water content in the ammonia gas.

Question 5.
What may be the reason for collecting ammonia in this manner ?
Answer:
Ammonia is lighter than air.

Question 6.
What is your inference about the density of ammonia from this ?
Answer:
The density of ammonia is lighter than air

Question 7.
Arrange the apparatus as shown in figure (Figure 5.2).
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 1
What do you observe?
Answer:
Water in the beaker lifts up in the flask which is filled with ammonia.

Non-metals in chemistry Question 8.
What inference can be made about the solubility of ammonia in water? Why does water rush into the flask?
Answer:
Ammonia dissolves abundantly in water. The pressure in the flask decreases with more ammonia dissolves in a water. So more quantity of water lifts up in the flask.

Barium sulfate Question 9.
Why does water entering the flask change its color?
Answer:
The ammonium hydroxide formed by dissolving ammonia in water is an alkali.

The color of water chapter 5 Question 10.
Which property of ammonia is responsible for this change in color?
Answer:
The property of base/alkali

Question 11.
Complete the chemical equation given below and find the product obtained when ammonia is dissolved in water.
Answer:
NH3 + H2O → NH2OH

Question 12.
Tick ✓ which is applicable to ammonia
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 2
Answer:

Colour Yes/No colour
Smell Pungent smell / No smell
Property Basic / Acidic
Solubility in water More soluble /Less soluble
Density Less than air / More than air

Question 13.
When an Ammonia tanker leaks, water is sprayed to reduce its intensity. What is the reason for this?
Answer:
Ammonia dissolves abundantly in water. Ammonia gas is released when the tanker leaks, it dissolves in water and can minimize the spreading of gas into atmosphere.

Ques. 14.
Some uses of ammonia are given below.
Answer:

  • For the manufacture of chemical fertilizers like ammonium sulfate, ammonium phosphate, urea, etc.
  • As a refrigerant in ice plants.
  • To clean tiles and window panes.
  • Ammonia is also used as a refrigerant gas, for purification of water sup¬plies, and in the manufacture of plastics, explosives, textiles, pesticides, dyes and other chemicals.
  • It is found in many household and industrial-strength cleaning solutions.

Text Book Page No: 82

Question 15.
Take some ammonium chloride (NH4Cl) in a boiling tube and heat it. Don’t you sense a peculiar smell?
Answer:
Yes. Pungent smell

Question 16.
Which is the gas evolved here?
Answer:
Ammonia.

Question 17.
Show a wet red litmus paper on the mouth of the boiling tube. What change can you observe?
Answer:
Turns blue

Question 18.
Keep the litmus paper for some more time at the mouth of the boiling tube and then observe its color change. What is the change occurred?
Answer:
The wet litmus paper has changed again to red color due to the presence of the hydrogen chloride (HCl) gas. When ammonium chloride (NH4Cl) is heated. Lighter NH3 comes out first then the denser HCl comes out.

Question 19.
Write the chemical equation of this reaction.
Answer:
NH4Cl(S) → NH3(g) + HCl (g)

Text Book Page No: 83

Question 20.
A glass rod dipped in cone. HCl is, shown into the jar which is filled with ammonia. What do you observe?
Answer:
White smoke is formed.

Question 21.
Complete the equation and find out the product?
Answer:
NH3 + HCl → NH4C1
NH4Cl is the product. The formation of NH4Cl is the reason for the white smoke.

Question 22.
What happens to the white powder on heating?
Answer:
It disappears

Text Book Page No: 84

Question 23.
Examine the chemical equations given below and write the forward and backward reactions in each.
1. N2 (g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
2. 2SO2 (g)+ O2 (g) \(\rightleftharpoons\) 2SO3(g)
3. H2(g) + I2 (g) \(\rightleftharpoons\) 2HI (g)
Answer:
1. N2 (g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Forward reaction:
N2 (g) + 3H2(g) → 2NH3(g) 1
Backward reaction:
2NH3(g) → N2 (g) + 3H2(g)

2. 2SO2 (g)+ O2 (g) \(\rightleftharpoons\) 2SO3(g)
Forward reaction:
2SO2 (g)+ O2 (g) → 2SO3(g)
Backward reaction:
2SO3(g) → 2SO2 (g)+ O2 (g)

3. H2(g) + I2 (g) \(\rightleftharpoons\) 2HI (g)
Forward reaction:
H2(g) + I2 (g) → 2HI (g)
Backward reaction:
2HI (g) → H2(g) + I2(g)

Text Book Page No: 85

Question 24.
What happens to the rates of forward and backward reactions as time progresses?
Answer:
As the time passes, rate of forward reaction decreases and rate of backward reaction increases.

Question 25.
Identify the point at which the rates of both forward and backward reactions become equal?
Answer:
A

Text Book Page No: 86

Percent concentration by mass is defined as the mass of solute divided by the total mass of the solution and multiplied by 100%.

Question 26.
The rate of which reaction increases when the concentration of nitrogen is increased? Forward reaction/ Backward reaction.
Answer:
Forward reaction

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Question 27.
What happens if the concentration of ammonia is increased?
Answer:
Rate of backward reaction increases

Question 28.
What will be the effect of removing ammonia continuously from the system?
Answer:
Rate of forward reaction increases.

Question 29.
Complete the table writing the effect of change in concentration in the system at equilibrium.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 3
Answer:

Action Change of concentration Change in rate
More hydrogen is added Increase the concentration of reactant Rate of forward reaction increases
More ammonia is added Increases the concentration of the product Rate of backward reaction increases
Ammonia is removed Decreases the concentration of the product Rate of forward reaction increases
More nitrogen is added Increases the concentration of the reactant Rate of forward reaction increases

Question 30.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
In this equation what is the total number of moles of the reactant molecules?
Answer:
Total number of moles of the reactant = 4

Text Book Page No: 87

Question 31.
What about the products?
Answer:
2

Question 32.
Forward reaction : 4 mole reactant molecule → 2 mole product molecules ( volume decreases)
Backward reaction : …. (a) …..mole product molecules → ….(b) ….. mole reactant molecules (volume ……(c) …….)
Answer:
a. 2
b. 4
c. increases

Question 33.
In the manufacture of ammonia, the reaction in which direction results in the decrease in the number of molecules?
Answer:
Forward reaction.

Question 34.
What happens when the pressure of system is decreased?
Answer:
Rate of forward reaction increases.

Question 35.
What if the pressure of the system is decreased?
Answer:
Rate of backward reaction increases.

Question 36.
In the manufacture of ammonia, why is a pressure of 150-300 atm used?
Answer:
When pressure is increased, rate of forward reaction increases. As a result, there will be an increase in the yield of product.

Question 37.
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
What is the total number of moles of reactants?
Answer:
2

Question 38.
What about the products?
Answer:
2

Text Book Page No: 88

Question 39.
N2 (g) + 3H2 (g) – 2NH3(g) + Heat
Which is the endothermic reaction in this ?
Forward reaction / Backward reaction
Answer:
Backward reaction.

Text Book Page No: 89

Question 40.
Sulphuric acid is formed also by the direct dissolution of sulfur trioxide in water. Still, sulfur trioxide is not directly dissolved in water. Why?
Answer:
The dissolution of sulfur trioxide in water is an exothermic process. It may turn sulphuric acid initially formed into fine fog-like particles (smog) which will hinder further dissolution.

Text Book Page No: 90

Question 41.
Complete the flow chart.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 4
Answer:
a. SO3
b. H2O

Question 42.
Take 5mL water in a test tube and slowly add concentrated sulphuric acid to it. Touch the bottom of the test tube. What do you feel? Is the reaction exothermic or endothermic?
Answer:
Feels hot. It is a exothermic reaction.

Text Book Page No: 91

Question 43.
What are the constituent elements of sugar?
Answer:
Carbon, Hydrogen, and Oxygen

Question 44.
Which is the black substance in the product formed?
Answer:
Carbon

Question 45.
What is the ratio of hydrogen and oxygen in sugar?
Answer:
2: 1

Question 46.
Which is the substance that absorbed hydrogen and oxygen from sugar in the ratio as in water?
Answer:
Conc. H2SO4

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Question 47.
Complete the table by involving the activities given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 5
Answer:

No. Activity Observation
1 Dropping Con. H2SO4 on a cotton cloth. Clothes will get burnt. It will extract water out of your clothes and all will be left will be carbon alongside some other things.
2 Adding Con. H2SO4 to glucose taken in a small beaker. The sulfuric acid removes water from the sugar in a highly exother­mic reaction, releasing heat, steam, and sulfur oxide fumes. The white sugar turns into a black carbonized tube that pushes itself out of the beaker
3 Adding Con. H2SO4 to a watch glass in which CuSO4 crystals are taken. Blue copper sulfate crystals it was decolorised. Sulphuric acid is an strong dehydrating acid therefore, it removes Oxygen.

Question 48.
Why is concentrated sulphuric acid not used as a drying agent in the preparation of ammonia?
Answer:
Ammonia is a base. Ammonia reacts with sulphuric acid and the salt ammonium sulfate is produced.
2NH3 + H2SO4 → (NH2)2SO4

Text Book Page No: 92

Question 49.
Add concentrated sulphuric acid to a test tube containing a small quantity of carbon. Heat it. What do you observe?
Answer:
Gases are formed. CO2 and SO2 are the gases.

Question 50.
Analyze the chemical equation and find the reason for your observation.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 6
a. What is the oxidation state of elemental carbon?
Answer:
Zero

b. What about the carbon in carbon dioxide?
Answer:
The oxidation state of carbon in carbon dioxide is +4.

c. Was carbon oxidized or reduced in this reaction?
Answer:
Carbon is oxidized.

d. Which is the oxidizing agent?
Answer:
Sulphuric acid is the oxidizing agent.

Question 51.
See the reaction between concentrated sulphuric acid and copper.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 7

a. Is copper oxidized or reduced in this case?
Answer:
Copper is oxidized.

b. Which is the oxidizing agent in this reaction? Which is the reducing agent?
Answer:
Copper is the reducing agent.
Sulphuric acid is the oxidizing agent.

Text Book Page No: 93

Question 52.
Analyze the given chemical equation.
Na2SO4 + BaCl2 → BaSO4 + 2NaCl

a. Which substance is soluble in water among the products?
Answer:
NaCl

b. Which substance is the white precipitate?
Answer:
Barium sulphate (BaSO4).

c. Does the white precipitate dis-solve when dilute hydrochloric acid is added to it?
Answer:
No. It is not soluble in dilute HCl

Question 53.
Write down the observation in the table given below, when lmL Barium chloride solution is added to the solutions given in the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 8
Answer:

No. Solution By adding BaCh solution When dilute HCl is added to this
1. MgSO4 Barium sulfate (BaSO4) formed. A white precipitate. Insoluble in dil HCl. Precipitate not disappear.
2. ZnSO4 Barium sulfate (BaSO4) formed. A white precipitate. Insoluble in dil HCl. Precipitate not disappear.

Let Us Assess

Question 1.
In which of the following reversible reactions does change in pressure not influence equilibrium? What is the reason?
i. H2(g) + I2(g) \(\rightleftharpoons\) 2HI
ii. N2 + 3H2(g) \(\rightleftharpoons\) 2NH3
Answer:
i. H2(g) + I2(g) \(\rightleftharpoons\) 2 Hg(g)
In this reaction, the number of moles of the reactants and the number of moles of the products are same.

Question 2.
What is the use of applying high pressure during the formation of ammonia from nitrogen and hydrogen?
Answer:
In this reaction, the number of moles of the products is less than the number of moles of the reactants. So if the pressure is increased, rate of forward reaction is increased. Thereby the yield of the product will be more.

Question 3.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 9
a. Identify the reactants and products.
b. Products are frequently removed from the system? Explain the reason.
Answer:
a. Reactants : C(s), H2O(g)
Products: CO(g), H2(g)
b. When the products are frequently re moved from the system, rate of forward reaction increases. Hence more products will be formed.

Question 4.
2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g) + heat
In this reaction how do the following changes influence the amount of the product?
a. Decrease in temperature
b. Increase in pressure
c. Increase in concentration of oxygen
Answer:
a. When the temperature is decreased, rate of forward reaction increases. Hence more products is formed.
b. When pressure is increased, rate of forward reaction increases. Hence more product is formed.
c. When the concentration of oxygen is increased the rate of forward reaction increases. Hence more product is formed.

Question 5.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) + heat
a. What change in pressure is required for the maximum yield of the product?
b. What is the change in concentration required for increasing the rate of the forward reaction?
Answer:
a. On increasing the pressure, yield of the product will be maximum.
b. Increase the concentration of reactants nitrogen or hydrogen. Removal of product ammonia from the system.

Question 6.
The chemical equation of one of the different stages of manufacturing sulphuric acid by contact process is given below. Find out the influence of the following factors in the reaction given below.
2SO2 (g) + O2 (g) \(\rightleftharpoons\) 2SO3(g) + heat
1. Increase the amount of oxygen
2. Pressure is increased
3. Catalyst vanadium pentoxide (V2O5)is added
4. SO3 is removed
Answer:
1. Rate of forward reaction increases.
2. Rate of forward reaction increases
3. Attains the equilibrium quickly.
4. Rate of forward reaction increases

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Question 7.
Calcium oxide (CaO) is used as drying agent in the preparation of Ammonia in laboratory. Can concentrated H2SO4 be used as drying agent instead of CaO? Justify your answer.
Answer:
Sulphuric acid can not be used instead of CaO. it is because sulphuric acid reacts with ammonia and the salt ammonium sulfate is formed.

Question 8.
Which property of sulphuric acid is shown in the following situations.
a. During the preparation of chlorine, the gas is passed through concentrated H2SO4
b. Wooden cupboards appeared to be burnt when concentrated sulphuric acid happened to fall on it.
Answer:
a. It’s character as a drying agent
b. It’s character as a dehydrating agent.

Extended Activities

Question 1.
The graph for the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)+ heat is given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 10
a. Identify and write the reactions C and D
b. What happens to the position of point A in the graph when a catalyst is used? Redraw the graph.
Answer:
a Reaction C – Forward reaction,
Reaction D – Backward reaction,
b. Equilibrium is attained quickly. Point A shift to the left side.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 11

Question 2.
It is often said that the production of sulphuric acid is a bench mark of the industrial development of a country. Prepare a note based on the various uses of sulphuric acid.
Answer:
Sulphuric acid is an important industrial chemical which is used in the manufacturing processes of many goods over a wide range of applications. Sulfuric acid used in pulp and paper industry for chlorine dioxide generation, tall oil splitting and pH-adjustments.Sulfuric acid is a strongly acidic, oily liquid that may be clear to cloudy in appearance. Concentrated sulfuric acid acts as both an oxidizing and dehydrating agent.

Sulfuric acid was once known as oil of vitriol. Here are some of the growing number of end-users and applications using sulfuric acid is Agricultural chemicals Aluminum Sulfate, Batteries, Cellophane, Detergents, Explosives, Fertilizers, Gasoline, Herbicides, Iron and steel pickling, Jet Fuel, Kerosene, Leather, Lubricating Oils, Medicinal processes, Oil additives, Paper, Rayon and rubber, Sugar, Synthetic fibers, Veterinary drugs, Water softener regeneration, Water treatment, Yellow pigments.

Question 3.
Fill half of a beaker of capacity 50 mL with sugar. Add concentrated sulphuric acid so that the sugar is immersed in it. Observe the changes. What are the products formed? Which property of sulphuric acid is revealed here?
Answer:
The sulphuric acid removes water from the sugar in a highly exothermic reaction, releasing heat, steam, and sulfur oxide fumes. Aside from the sulfurous odor, the reaction smells a lot like caramel. The white sugar turns into a black carbonized tube that pushes itself out of the beaker. Dehydrating properties of H2SO4 are shown in the experiment. Concentrated sulphuric acid has the ability to absorb chemically combined water, or hydrogen and oxygen from substances in the ratio corresponding to that of water. This process is known as dehydration. Concentrated sulphuric acid is a strong dehydrating agent.

Compounds of Non-Metals Orukkam Questions and Answers

Question 1.
a. Find out the relation between Temperature and rate of a reaction?
Select the materials from the list above
Sodium the Sulphate, Test tube, dil HCl, Cu, Mg ribbon, Beaker, Water, Spirit lamp, boding Tube
b. Prepare a write up for finding the relation between temperature and rate of a reaction?
c. Why rate of a reaction increase when temperature increases?
Answer:
a. Essential Materials:- Sodium Thiosu- phate, Hydrochloric Acid, Water, Boiling Tube, Spirit Lamp. Prepare a solution of diluted sodium Thiosulphate in a beaker. Take equal amount of this solution in two Boiling Tubes. Heat one of the boiling tubes for some time. Add equal amount of diluted HCl to both the boiling tubes.
b. The Essential material/Materials Required Mg Ribbon, Cone HCl, dil HCl
Write up
Take Magnesium Ribbons of equal mass in two test tubes. Add dil HC1 to one test tube and dilute HC1 to the other in equal volumes.
c. Energy as well as the speed of molecules increases when reactants are heated. That is as temperature increases the number of molecules with three hold energy increases. As a result, the number of effective collisions increases and these rate of reaction also increases.

Question 2.
Some chemical reactions are given below.
1. Zn + 2HCI → ZnCl2 + H2
2. 2Mg + O2 → 2MgO
3. NH4Cl \(\rightleftharpoons\) NH3 +HCl
a. What are the peculiarities of first two reactions
b. Conduct an experiment for viewing the dissociation and association taking place in the third equation.
c. In the three reactions, reactants turned into product and products are converted into reactants, is it true? d What type of reactions are they all represent?
e. Write down the characteristics of the re-action.
Answer:
a. In these reactions, the reactants change to products but the products cannot be changed back to reactants.
b. Heat some Ammonium chloride in a boiling tube.
c. No
d. First two Reactions are irreversible reactions. Third Reaction is a reversible Reaction.
e. Characteristics of Irreversible Reaction:- Reactants become products but products cannot be changed back to Reactants.
Characteristics of Reversible Reaction:- Reaction takes place in both directions.

Question 3.
Fe(NO3)3 + 3KCNS → Fe(CNS) + 3KNO3 This balanced chemical equation is wrote on the black board when the teacher is going to conduct an experiment on chemical equilibrium.
a. In the above reaction, which chemical has red color.
b. Fe(NO3)3, KCNS combined together, put it on the test tube stand, is there any col our change? Is the color is diminishing while it kept in test tube stand?
c. Convert the solution reared into four be a kers dilute each with equal amount of water.
In the first beaker add Fe (NO3)3, in the second one KCNS, in the third KNO4 like wise. Compare the color change with the fourth beaker. Find out the reason.
d. Point out the characteristics of equilibrium based on the experiment done.
e. In minute level chemical equilibrium is Kinetic energy why?
f. How and when a reversible reaction attains chemical equilibrium.
g. In the graph given below, when the reactant and product attain the level A? What are the characteristics of the point A?
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 12
Answer:
a. Fe (CNS), (Ferric Thiocyanate)
b. Yes, Red color increases
c.

Activity Performed Observations                                  Reason
Added Fe (NO3)3 Red color of solution inten­sifies Fe (NO3)3 reacted with the remaining KCNS to form the product
Added KCNS The red color of solution intensifies KCNS reacted with remain­ing in the solution to form more product
Added KNO3 Intensity of the red color of the solution is reduced significantly The product Fe (CNS), is reacted with KNO3 to form reactants.

d. At equilibrium both the reactants and products coexist.

  • The rates of forward and backward reactions become equal at equilibrium.
  • Chemical equilibrium is dynamic at the molecular level.
  • The system can attain equilibrium even when the concentration of the reactants and products are not equal. Once equilibrium has been attained, there will be no change in the concentration of reactants as well as that of products.
  • Chemical equilibrium is attained in closed system.

e. Even in equilibrium state reactants, molecules form products and products form reactants. Hence in minute level chemical equilibrium is kinetic energy.
f. The concentration of reactants and products do not change a reversible reaction attain chemical equilibrium,
g. A is the point at which the rates of both forward and backward reactions become equal.
At equilibrium both the reactants and products coexist.

Question 4.
N2 + 3H2 \(\rightleftharpoons\) 2NH3
H2+I2 \(\rightleftharpoons\) 2Hl
N2O4(g) \(\rightleftharpoons\) 2NO3
2SO2 + O2 \(\rightleftharpoons\) 2SO3
Write down detail how amount of the prod¬ucts increases in the above reactions (based on Le chandelier principle)
Hints: reference must be given on each of the following, concentration, pressure, temperature, catalyst
Answer:

  • Increase the concentration of reactants or anyone reactants.
  • Increase the temperature – Increases the rate of endothermic reactions.
  • Decrease the temperature- Increase the rate of exothermic reactions. Increase the pressure- Reaction takes place faster in the direction where the molecules are less.
  • Decrease the pressure- Reaction takes place faster in the direction where molecules are more.
  • Catalysts are added at the beginning of a chemical reaction.

Compounds of Non-Metals SCERT Questions and Answers

Question 5.
The graph showing the progress of the reaction N2 + 3H2 \(\rightleftharpoons\) 2NH3 is given
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 13
a. Identify the reactions represented by B and C?
b. What is the significance of the state A?
c. Is there any change in the concentration, as time passes after attaining the stage A? Explain.
Answer:
a. B – Forward reaction, C – Backward reaction
b. Equilibrium
c. No At equilibrium rate of both forward and backward reactions are equal.

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Question 6.
Coldwater is taken in one test tube an hot water in another one. Mg ribbon with same size is dropped in each of the test tubes,
a. In which test tube, hydrogen is formed with greater speed?
b. Which factor influences the rate of reaction? Explain the reason.
Answer:
a. Test Tube which contains hot water.
b. Temperature, When temperature increases the number of molecules with thresh-old energy increases. As a result number of effective collisions increases.

Question 7.
The chemical equation of the industrial preparation of ammonia is given below.
N2 + 3H2 \(\rightleftharpoons\) 2NH3 + Heat
Suggest the methods to get more NH3.
Answer:
Increase the Pressure and Temperature.

Question 8.
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) + Heat
The following circumstances influence the reaction
a. Increase the concentration of H2.
b. Increase the pressure
c. Increase the temperature.
Answer:
a. Increases the rate of forward reaction
b. No effect
c. Increases the rate of backward reaction

Question 9.
The formation of SO3 in the industrial preparation of sulphuric acid is given below.
2SO2 + O2 \(\rightleftharpoons\) 2SO3+ Heat
a. Explain the effect of concentration of 02 to get maximum yield of SO3? State rea son.
b. Identify the law related to it. State it.
Answer:
a. Increase the concentration of oxygen, which increases the rate of forward reaction,
b. When the concentration, pressure or tern perature of a system at equilibrium is changed, the system will readjust itself so as to nullify the effect of that change and attain a new state of equilibrium. This is Le Chatelier’s principle.

Question 10.
The chemical equation of a stage in the industrial preparation of sulphuric acid is given below.
2SO2 + O2 \(\rightleftharpoons\) 2SO3+ Heat
a. Which is the catalyst used in this reaction?
b. What is the influence of the catalyst in equilibrium?
Answer:
a. Vanadium pentoxide / V2O5
b. Catalyst increases the rate of both the forward and backward reactions to the same extent.

Question 11.
The chemical equation of the industrial preparation of ammonia is given below.
N2 + 3H2 \(\rightleftharpoons\) 2NH3+ Heat
a. Temperature is to be decreased to get maximum yield of ammonia, according to the Le Chatelier principle. Why?
b. What is the reason for taking an optimum temperature in this reaction?
Answer:
a. When the temperature decreases exothermic reaction increases ie. forward reaction increases.
b. At low-temperature rate of forward and backward reaction is slow.

Question 12.
Analyse the following equations and answer the questions
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 14
a. Which of these reactions are affected by change in pressure? What are the reasons?
b. How the increase in pressure influence the reaction which you have already identified ?
Answer:
a. iii. N2 + 3H2 \(\rightleftharpoons\) 2NH3
In first reaction NH4Cl is solid compound In second reaction number of reactants and product molecules are equal.
b. When pressure increases, rate of forward reaction increases.

Question 13.
Catalysts are substances which influence the rate of chemical reactions.
Explain how the catalysts influence the rate of reversible reaction?
Answer:
The catalyst increases the rate of both the forward and the backward reactions to the same extent. As a result, the system reaches equilibrium at a faster rate.

Question 14.
Some features of a reversible reaction are given below.
1. Product formation increases when the temperature is increased.
2. There is no effect when the pressure is increased.
Explain the reason for above inferences.
Answer:
1. Forward reaction is an exothermic.
2. Number of reactant and product molecules

Question 15.
A + B + Heat \(\rightleftharpoons\) 2C + D
This reversible reaction is in equilibrium. What happens to the amount of products under the following conditions?
a. C is removed from the system
b. B is added in excess
c. Temperature is increased
d. A suitable catalyst is added.
Answer:
a. Increase the amount of product
b. Increase the amount of product
c. Decrease the amount of product,
d. Increase the amount of product

Compounds of Non-Metals Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 16.
Select the correct statements which are related to the influence of catalyst in a reversible reaction.
a. Forward reaction takes place when a cata lyst is used in a reversible reaction, h, Attains equilibrium faster,
c. Catalyst does not help to form more pro-duct
d The catalyst increases the rates of both the forward and the backward reactions to the same extent.
e. Increases the speed of backward reaction.
f. Does not helps to produce more product
Answer:
Statements b, c, d are correct.

Question 17.
2X(g) + heat \(\rightleftharpoons\) Y(g) + Z(g) Explain the influence of pressure and temperature in this reaction.
Answer:
Pressure — Pressure has no effect here. Because the number of molecules in the reactants and products side are equal.
Temperature — Here the forward reaction is endothermic. So increasing temperature increases the rate of forward reaction. Decreasing temperature decreases the rate of forward reaction.

Short Answer Type Questions (Score 2)

Question 18.
Observe the equation of the chemical reactions given below.
i. NaOH (aq) + HCl (aq) → 2NaCl (aq) + H2O(l)
ii. N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
iii. Zn(s) + ZHCl → ZnCl2(ql) + H2(g)
iv. 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
a In which chemical reaction pressure can influence its speed ?
b. In this reaction what changes will be made in pressure for increasing forward reaction ? Why ?
Answer:
Reaction (iv)
2SO3(g) + O3(g) \(\rightleftharpoons\) 2SO3(g)
b. In this reaction, the number of moles of the products is less than the number of moles of the reactants. So if the pressure is increased, rate of forward reaction is increased. Thereby the yield of the product will be more.

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Question 19.
A graph given below deals with the reversible reaction.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 15
a What does A, B, C indicate?
b. What inference can be drawn about the concentration of the reactants and products at the point D and E?
Answer:
A – Forward reaction
B – Backward reaction
C – Equilibrium
b. There is no change in the concentration, of reactants and products at the points D and E. Because at the point ‘C’ the process attains equilibrium.

Question 20.
Some chemicals are given below. Sodium chloride, Ammonium hydroxide, Nitric acid, cone. Sulphuric, Sodium hydroxide.
a. Which are the substances needed to produce hydrogen chloride?
b. Suggest a method to identify hydrogen chloride gas.
Answer:
a. Sodium chloride
b. Show a glass rod dipped in ammonia, just above the jar in which hydrogen chloride gas is taken. A white smoke of NH34Cl is produced.

Question 21.
When ammonia was leaked two solutions were arised.
i) Spray water
ii) Spray HCl
Which method you adopt? Justify your answer.
Answer:
Spray water. Ammonia dissolves abundantly in water. So we can prevent the easy spreading of ammonia. When HCl is used the white smoke of NH4Cl spread in the atmosphere.

Short Answer Type Questions (Score 3)

Question 22.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3 (g) + heat
How do the following factors influence the forward reaction?
a. One of the products is removed
b. Increase in pressure
c. More N2 is added.
Answer:
a. Speed of forward reaction increases
b. Speed of forward reaction increases
c. Speed of forward reaction increases

Question 23.
A graph given below deals with the reversible reaction.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 16
a. What happened to the forward and backward reactions as time passes?
b. In which minute does the system attain equilibrium?
c. What change occurs to the equilibrium, when a catalyst is used?
Answer:
a. As time passes the speed of forward reaction decreases and backward reaction increases.
b. 25th-minute
c. Attain equilibrium fast ie. before 25th minute.

Question 24.
Ammonia is an industrially useful compound of nitrogen.
a. Name the industrial production of ammonia
b. Write the equation of the reaction,
c. Write any two uses of ammonia
Answer:
a. Haber process
b. N2 + 3H2 → 2NH3
c. i. For the manufacture of the fertilizers
ii. As a coolant in ice plants.

Question 25.
Ammonia is a pungent-smelling gas.
a. How does ammonia convert into
i. liquor ammonia
ii. liquid ammonia
b. Write the color, smell, solubility in water and density of ammonia.
Answer:
a. i. Prepare a concentrated solution of ammonia by dissolving it in water. It is liquor ammonia.
ii. The gaseous ammonia can be converted into liquid by applying pressure. It is called liquid ammonia.
b. Ammonia has no color and irritating smell. It dissolves abundantly in water. It has density less than air.

Question 26.
Sulphuric acid is industrially manufactured by using contact process. Write the equation of this process.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 17

Question 27.
Write equations for the following chemicals to prepare from sulphuric acid.
a. Hydrogen chloride
b. Oleum
c. Sodium sulfate
Answer:
a. NaCl + H2SO4 → NaHSO4 + HCl
b. H2SO4 + SO3 → H2S2O7
c. 2NaOH + H2SO4 → Na2SO4 + 2H2O

Question 28.
Observe the following chemical equation
Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
a Which one has oxidation state as 0?
b. What is its oxidation state after the chemical reaction?
c. Is the change oxidation or reduction?
d. Which chemical nature of sulphuric acid is found here?
Answer:
a. Cu
b. 2+
c. Oxidation
d. Nature of oxidation

Question 29.
Litmus has colored in the solution.
A. When BaCl2 solution was added to this.
B. Solution, a white precipitate insoluble in HCl is produced.
1. Write the chemical formula of A.
2. Write the chemical name and chemical formula of the white precipitate produced when it was added with BaCl2.
c. Complete the equation
C + 2(A) →…….. + …… +……..
Answer:
a. H2SO4
b. Barium sulphate (BaSO4)
c. C + 2H2SO4 → CO2 + 2H2O + SO2

Question 30.
Analyze the table given below and complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 5 Compounds of Non-Metals 18
Answer:
a. White precipitate is formed
b. Ba2CO3
c. Clear solution
d. White precipitate is formed
e. BaSO4
f. No change
g. No substance is formed

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Question 31.
Sulphuric acid is a drying agent as well as a dehydrating agent.
a. Find out the difference between these two activities.
b. Write one example each for these nature of sulphuric acid.
Answer:
a. Dehydration is the process of absorbing hydrogen and oxygen from substances in the ratio of water.
Drying activity is the process of absorbing water content present in a substance.
b. Sugar bums to black when concentrated sulphuric acid is added in it. It is an example for dehydration.
Concentrated H2SO4 is used to remove water particles from HCl gas during its manufacture – drying agent.

Question 32.
Some incomplete equations are given below:
i. MgSO4 + BaCl2 → …….. + MgCl2
ii. K2CO3 + BaCl2 → ……… + 2KC1
iii. KCl + AgNO3 → …….. + KNO3
a. Complete the equations.
b. Find out the white precipitate in each of the reactions.
Answer:
a. i. MgSO4 + BaCl2 → BaSO4 + MgCl2
ii. K2CO3 + BaCl2 → BaCO3 + 2KCl
iii. KCl + AgNO3 → AgCl + KNO3
b. i. BaSO4
ii. BaCO3
iii. AgCl

Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light

You can Download Reflection of Light Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light

Reflection of Light Text Book Questions and Answers

Text Book Page No. 80

Light Chapter Class 10 Question 1.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 1

→ Which is the incident ray?
Answer:
AO

→ Which is the reflected ray?
Answers:
QB

→ Is there any relation between the angle of incidence and the angle of reflection?
Answer:
The angle of incidence and angle of reflection are equal.

→ Are the incident ray, reflected ray and normal to the mirror at the point of incidence in the different planes?
Answer:
In the same plane.

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Reflection of Light Class 10 Question 2.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 2

→ What difference is seen between the surfaces of the two objects?
Answer:
In the first figure the surface is smooth, in the second figure surface is rough.

→ Fig. 4.2(b) are the rays of light travelling parallel after reflection?
Answer:
No. When light falls on a rough surface, it undergoes an irregular reflection. This is scattered reflection.

Reflection of Light Class 10 Question 3.
In Fig. 4.2(a) regular reflection is depicted. Can you give a definition for such reflections by observing the figure?
Answer:
When light falls on a smooth surface, it undergoes an regular reflection, the rays of light travelling parallel after reflection This is regular reflection.

Text Book Page No. 81

Light Reflection and Refraction Question 4.
Record in your science diary the following features about the images formed here.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 3

→ The distance from the mirror to the object and the image from the mirror.
Answer: Equal

→ Is the image real or virtual?
Answer: Virtual

→ The size of the image.
Answer:
Same size as object.

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Light Reflection and Refraction Class 10 Question 5.
How many images can we see at a time busying two mirrors?
Answer:
The number of images changes accordance with the relation between the angle between the mirrors.

Text Book Page No. 82

Class 10 Physics Light Reflection and Refraction Question 6.
Table 4.1.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 4

Answer:

Angle
(0)

Number of images
(n)

45 7
60 5
90 3
120 2
180 1

→ How many images can be seen when viewed from A and B?
Answer:
3

→ What if viewed from other positions in between the mirrors?
Answer:
3

→ How much is the angle between the mirrors?
Answer:
90°

→ What is the relation between the angle between the mirrors and the number of images?
Answer:
Number of images = \(n=\frac{360}{\theta}-1\)

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Reflection of Light Class 10 Question 7.
Table 4.2.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 5
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 6

Text Book Page No. 83

Light Chapter of Class 10 Question 8.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 7
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 8

Text Book Page No. 84

Reflection and Refraction of Light Class 10 Question 9.
Draw a straight line on a table as shown in the figure. At one end of the line, place a concave mirror of focal length 20 cm. Mark principal focus (F) and center of curvature (C) on the line.

Fix a burning candle on the principal focus in such a way that it is at a slight distance from the center of curvature. Arrange a screen in such a way that a clear image is obtained on the screen.

→ What is the position and features of the image?
Answer:
Between F and C, small, real, inverted.

→ Observe the change in position of image and features on changing the position of the candle
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 9

Light and Reflection Class 10 Question 10.
Table 4.4.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 10
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 11

Class 10 Physics Light Reflection and Refraction Question 11.
Aren’t the focal length of \(\frac{u v}{u+v}\) the mirror and the average value of obtained from the table the same?
Answer:
Yes

Text Book Page No. 85

Light Lesson for Class 10 Question 12.
Record the measurements shown in the figure (4.6) using the New Cartesian Sign Convention.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 12

→ Distance to the object from the mirror (u) =
Answer:
Negative.

→ Distance to the image from the mirror (v) =
Answer:
Negative.

→ Height of object (OB) =
Answer:
Positive.

→ Height of image (IM) =
Answer:
Negative.

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Text Book Page No. 86

Reflection and Refraction Class 10 Question 13.
The given figure shows the image formation by a concave mirror. Analyse the figure and write down different measures using New Cartesian Sign Convention.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 13
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 14

Light Class Question 14. An object is placed in front of a concave mirror 20 cm away from it. If its focal length is 40 cm, locate the position of image and its nature.
Answer:
f = – 40 cm, u = – 20 cm
f = \(\frac { uv }{ u+v }\)
– 40 = \(\frac { – 20 }{ – 20 + v }\)
– 40 ( – 20 + v ) = – 20 v
– 20 + v = \(\frac { – 20 }{ – 40 }\) = \(\frac { 2 }{ v }\)
– 20 = [/latex] = \(\frac { v }{ 2 }\) – v = – \(\frac { v }{ 2 }\)
∴ – \(\frac { v }{ 2 }\) = – 20
v = 40 cm
Features of image:
At behind the mirror, very large, virtual, direct.

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Class 10 Physics Chapter 4 Question 15. Find out the height of object (ho), height of image (hi), position of object(u) and position of image (v) using New Cartesian Sign Convention and tabulate them. (The height of image hi can be directly measured by fixing a graph paper on the screen).
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 15

Text Book Page No. 87

Question 16.
\(\frac{h_{i}}{h_{o}}\) is magnification. Does it have any relational with the value of \(\frac{v}{u}\) ?
Answer:
\(m=\frac{-h_{i}}{h_{o}}=\frac{-v}{u}\)

Text Book Page No. 88

Question 17.
An object is placed 8 cm away in front of a concave mirror of focal length 5 cm. Find out the position of image and magnification. Find out whether the image is inverted or erect by drawing the ray diagram on a graph paper.
Answer:
f = – 5 cm, u = – 8 cm
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 16
Features of images : Beyond C, big, real, inverted
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 17
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Question 18.
What are the features of an image that is obtained from magnification?
Answer:

  • When magnification is 1, the size of the image and the size of the object are equal.
  • When magnification is more than 1, the size of the image is greater than the size of the object.
  • When magnification is less than 1, the size of the image is smaller than the size of the object.
  • When the magnification is positive, image is real and inverted.
  • When the magnification is negative, image is virtual and erect.

Question 19.
Observe the given figures and complete the table using New Cartesian Sign Convention.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 18
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 19

Text Book Rage No. 89

Question 20.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 20
From the above table, find out which mirror always gives an erect and diminished image and write it down.
Answer:
The image formed by a convex mirror is always erect and diminished.

Question 21.
Why it is written on rearview mirrors that “Objects in the mirror are closer than they appear”.
Answer:
The image formed by a convex mirror is always erect and diminished. Hence the driver who sees the image of vehicles on the mirror develops a feeling that the vehicles coming from behind are at a greater distance. This may turn out to be dangerous.

Reflection of Light Let Us Assess

Question 1.
A dental doctor uses a mirror of focal length 8 cm. To see the teeth clearly what should be the maximum distance between the teeth and the mirror? Justify your answer. Which type of mirror has been used by the doctor?
Answer:
The dental doctor uses a concave mirror. Effect and enlarged image can be obtained using the mirror. Such images are formed when the object is kept in between the main focus and pole of a concave mirror. So the minimum distance between the mirror and the teeth must be in 8 cm to view the teeth clearly.

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Question 2.
Imagine that a spherical mirror gives an image magnified 5 times at a distance 5 m. If so determine whether the mirror is concave or convex. How much will be the focal length of the mirror?
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 21
Image formed is larger than the object, so the used mirror is concave.

This actual yield calculator will answer your questions about how to calculate the actual yield and help you find the true yield.

Question 3.
A motorcyclist observes a car coining from behind with a magnification 1/6. If the actual distance between the car and the bike is 30 m calculate the radius of curvature of the mirror.
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 22

Question 4.
A shaving mirror of focal length 72 cm is kept in a beauty clinic. A man uses it standing 18 cm away from the mirror. At what distance will the image be formed? Is the image real or virtual? What is the magnification of the image?
Answer:
f =-72 cm, u = -18cm
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 23
Image formed on the 24 cm away from the mirror. Virtual image.
Magnification.
m = \(\frac{-v}{u}=-\frac{24}{18}\) = – 1.33
This is a concave mirror.

Question 5.
Wrap a rubber ball of diameter 12 cm completely with an aluminum foil and make the surfaces smooth. Where will be the image of an object kept 12 cm away from the center of the ball? Is the image real or virtual?
Answer:
Surface of a rubber ball is similar to a convex mirror.
Diameter = 12 cm
⇒ Radius = 6 cm
2f = R
f = \(\frac { R }{ 2 }\) = \(\frac { 6 }{ 2 }\) = 3 cm
u = \(\frac { 12 }{ 2 }\) = 6 cm
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 24
Mirror is a convex, so image is virtual.

Question 6.
We are able to read a book since light falling on a surface gets reflected from the book and reaches the eye. But on such occasions, we cannot see our images like that from a mirror. Explain why?
Answer:
Irregular reflection takes place in the book. So images not formed.

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Question 7.
Is the image formed by a plane mirror real or virtual? Write an instance when such a mirror gives an inverted image.
Answer:
Image formed by a plane mirror is always direct and virtual. The image will be at the same distance at the object.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 25
If the objects placed erect as shown in the figure then the image will be inverted.

Reflection of Light Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
1. Classify the following statements as to those related to concave mirrors and convex mirrors and tabulate them accordingly.
a. to view the face.
b. as makeup mirror.
c. as reviewer mirrors in vehicles.
d. in solar concentration.
e. in periscopes.
f. as shaving mirror.
Answer:
Concave mirror:

  • In solar concentration.
  • Makeup mirror.
  • Shaving mirror.

Convex mirror:

  • In rearview mirrors of vehicles.
  • In Searchlights.

Plane mirror:

  • In periscopes.
  • To see face.

Question 2.
Find out the relation and All in the blanks.
Small, virtual and direct images:
Convex mirror.
Size same as object, virtual and direct image :………………
Answer:
Plane mirror.

Question 3.
Which mirror gives rectifiable whatever the distance between the object and mirror is
i. Plane mirror.
ii. Convex mirror.
iii. Concave mirror.
iv. Plane or convex mirror.
Answer:
Plane or convex mirror.

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Question 4.
Relation between angle of incidence (i) and angle of reflection (r) of light is………..
i. i > r
ii. i < r
iii. i = r
iv. None of these
Answer:
iii. i = r

Question 5.
Find the odd one in the group.
More scattered reflection, virtual image, direct image, large image
Answer:
Large image.

Very Short Answer Type Questions (Score 2)

Question 6.
Radius of curvature of a concave mirror is 24 cm. Then find the focal length of the mirror?
Answer:
R = 24 cm
f = \(\frac{R}{2}=\frac{24}{2}\) = 12 cm

Question 7.
Find the radius of curvature of a convex mirror whose focal length is 0.6 m.
Answer:
f = \(\frac{\mathrm{R}}{2}\)
∴ R=2f
0.6 = \(\frac{R}{2}\)
R = 2 × 0.6 = 1.2 m

Question 8.
Write the characteristics of the image formed by an object placed at the center of curvature of a concave mirror?
Answer:
Position: At the center of curvature at the same side
Size: Same as the size of the object
Nature: Real, Inverted

Question 9.
i. Which are the different types of mirror?
ii. What are the peculiarities are images obtained in the face mirror?
Answer:
i. Convex, concave and plane mirror.
ii. Image will be formed behind the mirror and will be equal distance from the mirror to the object. It will be direct and virtual.

Question 10.
State whether the below-given statement are true or false. If false correct them.
i. Real images are always produced by concave mirror.
ii. The image formed from a convex mirror is direct and enlarged.
Answer:
i. False. If the position of object is between focus and pole, virtual image is produced by concave mirror.
ii.False. The image formed by convex mirror will be smaller than the object and erect.

Very Short Answer Type Questions (Score 3)

Question 11.
Write three differences of real image and virtual image which is made by spherical mirrors.
Answer:
Real images:

  • Inverted.
  • Can be shown on a screen.
  • Can measure the length of the image and distance to the image.

Virtual images:

  • Cannot be taken on a screen.
  • Not able to be measured direct.

Question 12.
Write the uses of concave mirror?
Answer:

  • Used as a shaving mirror.
  • Used as a makeup mirror.
  • Doctors used as head mirrors.
  • On film projectors.

Question 13.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 26
a. Examine the figure and find the magnification.
b. What is the height of the object if height of the image is 4 cm when the object is placed on the same position in front of the mirror.
Answer:
a. Magnification = \(=\frac{\mathrm{hi}}{\mathrm{ho}}=\frac{– 10}{5}=– 2\)

Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 27

Question 14.
Write the uses of convex mirror?
Answer:

  • As reflectors in street lights.
  • As rear view mirrors in vehicles.
  • In searchlights.

Question 15.
Examine the position of the object given in the figure and table the following peculiarities.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 28
a. Position of image.
b. Size of image.
c. Features of image.
Answer:
a.Behind the mirror.
b.Larger than obj etc.
c. Direct, virtual.

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Question 16.
When an object of height 4 cm is placed in front of a concave mirror an image of height 8 cm is formed. Find the magnification.
Answer:
Magnification
hi = 4 cm, ho =  – 8 cm
Magnification m = \(\frac { hi }{ h0 }\) = \(\frac { –8 }{ 4 }\) = – 2

Very Short Answer Type Questions (Score 4)

Question 17.
Complete the table.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 29
Answer:
a. Passes through principal focus.
b.Seem to come from the principal focus.
c, d. Returns parallel to the principal axis.
e, f. Returns through the same path.
g, h. Reflects in the same angle of incident ray.

Question 18.
Complete the ray diagram.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 30
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 31
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 32
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 33

Question 19.
Complete the table.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 34
Answer:
a. 30°
b. 40°
c. 50°
d. 60°

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Question 20.
An object of 5 cm is kept in front of a convex mirror having radius of curvature 30 cm at a distance of 10 cm. Calculate the position, size, and feature of object.
Answer:
u = – 10 cm
v = ?
Focal length = \(\frac{r}{2}=\frac{30}{2}\) = + 15 cm
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 35
The image is formed 60cm behind the concave mirror. The’ image will be virtual and direct.
Kerala Syllabus 10th Standard Physics Solutions Chapter 4 Reflection of Light 36

Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management

You can Download Energy Management Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management

Energy Management Textual Questions And Answers

Textbook Page No. 147

Energy Management Textbook Question 1. It is said that energy can neither be created nor destroyed; then how does energy crisis occur?
Answer:
When energy is transformed from one form to another, some part of it gets lost in other forms. Such a loss is the main cause for energy crisis.

Textbook Page No. 148

Energy Management Syllabus Question 2. List down the different forms of energy you use for various purposes from the time you wake up till you reach your school.
Answer:

  • Muscular energy – for different physical activities
  • Chemical energy – for cooking
  • Mechanical energy – for moving
  • Sound energy – to call friends.

Physics Chapter 7 Question 3. From which sources are you getting these forms of energy?
Answer:
From different sources like sun, fuels and power stations.

Textbook Page No. 149

Question 4.
Complete the table 7.1
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 1
Answer:

Solid Liquid Gas
Firewood Kerosene Biogas
Coke Petrol Methane
Wood charcoal Diesel LPG
Coal Fuel oil Cool gas
Power kerosene C.N.G
Ethanol Hydrogen

Question 5.
Take three papers of the same size. Keep one stretched. Crumble the next. Make the third paper wet using water. Burn each of them over a candle flame using pincers. Compare the burning of each.
Complete the table 7.2
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 2.
Physics Chapter 7 Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 3

Textbook Page No. 150

Energy Management Products Question 6. Write down the situations/specialties for partial combustion.
Answer:

  • Partial dryness
  • Insufficient availability of O2
  • Lack of facilities for the removal of oxygen.

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Question 7.
What are the drawbacks of partial combustion?
Answer:

  • Fuel loss
  • Wastage of time
  • Economic loss
  • Atmospheric pollution
  • More smoke is produced
  • CO2, CO (carbon monoxide) are produced as byproducts.

Tidal Energy Animation Question 8. What are the advantages of using smokeless choolahs at home? Note them down in the science diary.
Answer:
Makes home neat, lung disease can be reduced, does not affect the oxygen-carrying capacity of blood. Reduces wastage of time and fuel loss.

Question 9.
Visit a nearby pollution testing center, interact with the staff there and prepare a note on the permissible pollution rate.
Answer:
The carbon monoxide produced as a result of combustion of fuels causes environmental pollution. More carbon monoxide may be produced if the vehicles are not working properly. Smoke testing is conducted to know what quantity of carbon monoxide is present in the smoke coming out of vehicle.

Question 10.
Which are the fuels that are used in vehicles and industries?
Answer:
Petrol, Diesel, LPG, CNG etc

Question 11.
Tabulate the category to which these fuels belongs to.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 4
Answer:

Coal Petroleum Natural Gas
Coke Petrol LNG
Coal gas Kerosene Methane
Coal tar Diesel Propane
Ammonia Gasoline Ethane

Textbook Page No. 151

Question 12.
Which are the products obtained from fractional distillation of petroleum?
Answer:
Petrol (or gasoline), naptha, kerosene, diesel oil, lubricating oil, fuel oil, grease wax, and some residue.

Question 13.
Which is the cooking gas that we get in cylinders for domestic use?
Answer:
LPG

Question 14.
How will you know if there is leakage in a LPG cylinder?
Answer:
The smell of LPG is felt

Question 15.
Never switch on or switch off electricity when there is a leakage of LPG Why?
Answer:
It is usually advised not to switch on or off any of the electric switches if you detect a gas leakage. It is because the fumes of gas are highly flammable and even smallest of sparks can ignite a huge fire.

Question 16.
If there is a leakage of LPG does it rise up or come down in the atmosphere? Why?
Answer:
Another reason why care should be taken during storage is that LPG vapor is heavier than air, so any leakage will sink to the ground and accumulate in low lying areas and may be difficult to disperse. LPG expands rapidly when its temperature rises

Question 17.
If there is leakage of LPG it is mandatory to open the doors and windows. Why?
Answer:
Immediately open all the doors and windows to your house so that the gas can escape. Never open electrical fans or even an exhaust fan. Let the gas escape naturally. Once you do that, go outside the house and isolate the main electric supply. Be sure to evacuate yourself and others from the area.

Question 18.
What precautions are to be taken to avoid accidents due to LPG leakage? Discuss and record in the science diary.
Answer:

  • Examine the rubber tube at regular intervals and ensure that it does not have a leakage.
  • Turn on the knob of stove only after the regulator is turned on.
  • Always buy LPG cylinders from authorized franchisees only
  • Check that the cylinder has been delivered with the company seal and safety cap intact, do not accept the cylinder if the seal is broken.
  • Please look for the due date of test, which is marked on the inner side of the cylinder stay plate and if this date is over, do not accept the cylinder
  • Disconnect LPG regulator and affix safety cap on the cylinder when your gas stove is not in use for prolonged period.
  • Always store the LPG cylinder in an upright position and away from other combustible and flammable materials.
  • Check for gas leaks regularly by applying soap solution on cylinder joints and Surakshapipes.

Textbook Page No. 153

Question 19.
If a gas leak is suspected or if the fire spreads on a cylinder, what else could be done? Think it over.
Answer:
If you are convinced that there is a gas leak, disconnect electricity from outside the home (switch off the main switches). Switch off the regulator and shift the cylinder to an empty space. Keep the windows and doors open. Request help from the Fire Force by calling in the toll-free number 108.

Well trained rescue operators can put out the fire by covering the top end of the cylinder with wet sack to prevent the contact with oxygen. If the fire is in flat or the top story, then one should not try to escape using lifts. Only staircase should be used. Cover the nose and the mouth with soft cloth to avoid the intake of smoke or gases.

Question 20.
Haven’t you seen bio-wastes dumped in public places? Don’t you experience a putrid smell, when you pass them by? Which are the gases responsible for this smell?
Answer:
Methane gas is responsible for the putrid smell.

Question 21.
Besides air pollution, what are the problems that may arise when garbage is heaped? Discuss and record.
Answer:

  • Soil contamination
  • Air contamination
  • Water contamination
  • Bad impact on human health
  • Impact on animals and marine life.
  • Disease-carrying pests
  • Adversely affect the local economy
  • Missed recycling opportunities.

Textbook Page No. 154

Question 22.
Some of you may be using LPG in your houses. What is the weight of the LPG filled in the cylinders supplied to your homes?
Answer:
14.2 kg

Question 23.
Using this quantity of LPG for how many days can you cook?
Answer:
This must be sufficient for a month.

Question 24.
How many days can you cook using firewood of the same weight?
Answer:
14.2 kg firewood would be sufficient for a few a days only.

Question 25.
What difference do you feel in the efficiency of these two fuels?
Answer:
The LPG with more calorific value has more fuel efficiency.

Textbook Page No. 155

Question 26.
Based on its calorific value, which fuel can be considered as the most efficient?
Answer:
Hydrogen

Question 27.
Which are the instances when hydrogen is used as fuel?
Answer:
Hydrogen used as fuel in rockets.

Question 28.
Why is hydrogen not used as a domestic fuel?
Answer:
The combustion rate is very high for hydrogen, the possibility of explosion is also more. And it is also difficult to store hydrogen.

Textbook Page No. 156

Question 29.
What is the energy conversion in a generator?
Answer:
Mechanical energy → Electrical energy

Question 30.
From where do we get energy required for the working of a generator?
Answer:
The mechanical energy required can be provided by engines operating on fuels such as diesel, petrol, natural gas, etc. or via renewable energy sources such as a wind turbine, water turbine, solar-powered turbine, etc

Question 31.
Power stations can be classified based on the nature of the source providing the energy required to operate the generator.
Answer:
Flowing water -Hydroelectric power station.
Nuclear energy – Nuclear power station
Coal – Thermal power station

Textbook Page No. 157

Question 32.
On the basis of notes and discussions, complete the table :
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 5
Answer:

Power stations Energy conversions
Hydroelectric power station Moolamattom Kuttiadi Pallivasal Sabarigiri Potential energy → kinetic enegy → mechanical energy → electri­cal energy
Thermal power station Neyveli Kayamkulam Ramagundam Ennora Chemical energy → heat energy mechanical energy → electrical energy

Textbook Page No. 158

Question 33.
We get different forms of energy from the Sun. Attempts are underway now to utilize solar energy of its maximum. What are the devices used for this?
Answer:

  • Solar panel
  • Solar water heater
  • Solar cooker
  • Solar water heater
  • Solar thermal power plant

Textbook Page No. 159

Question 34.
What is the energy transformation that takes place in a solar cell?
Answer:
In a solar panel light energy is converted into electrical energy.

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Question 35.
There are certain situations in which a solar panel cannot be put to use. Which are they?
Answer:
When the sky is cloudy, during the rain and night time we can not use solar panels.

Question 36.
What are the situations where solar panel alone is depended on?
Answer:
In space stations, satellite, and in remote islands where there is no electricity, etc solar panels are used.

Question 37.
Take two conical flasks. Paint the outer surface of one with black paint and the other, with white paint. Fill both of them with water and expose them to direct sunlight for the same period of time. Which one gets heated first? What may be the reason?
Answer:
The conical flask which is painted black seems to be more hot when kept under the sun for an hour because the black body absorbs more heat and do not let it out from it. It is the property of black body to absorb heat and there is no chance of heat to escape.

Textbook Page .159

Question 38.
List down the specialties of a solar cooker by examing Fig. 7.7.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 6
1. A box with blackened interior
2. A glass cover for the box
3. A mirror outside the each? What is the function of each?
Answer:
1. A box with blackened interior:
The black color of the vessel lets it to absorb more heat as dark colors absorb more heat.
2. A glass cover for the box:
The glass lid will allow the light and heat energy of the sun to come inside but the light energy will go out but the heat will be trapped inside the cooker to cook food.
3. A mirror outside the each?:
The plane mirror will reflect the maximum light of the Sun inside the cooker.

Question 39.
Find out the working of solar cookers other than the box type and record them in the science diary.
Answer:
Panel solar cookers are inexpensive solar cookers that use reflective panels to direct sunlight to a cooking pot that is enclosed in a clear plastic bag.

Textbook Page .160

Question 40.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 7
Observe the figure. Discuss and record how hot water is formed in the tank of solar heater.
Answer:
Solar water heating (SWH) is the conversion of sunlight into heat for water heating using a solar thermal collector. A sun-facing collector heats a working fluid that passes into a storage system for later use. SWH are active (pumped) and passive (convection-driven). They use water only, or both water and a working fluid. They are heated directly or via light-concentrating mirrors. They operate independently or as hybrids with electric or gas heaters. In large- scale installations, mirrors may concentrate sunlight into a smaller collector.

Textbook Page .162

Question 41.
‘Ocean as a source of energy: its possibilities and limitations’ Prepare a seminar paper on this.
Answer:
Hints: Oceans cover 70 percent of the earth’s surface and represent an enormous amount of energy. Although currently under-utilized, Ocean energy is mostly exploited by just a few technologies: Wave, Tidal, Current Energy and Ocean Thermal Energy.
Limitations:

  • Few limited sites/places where this wave, tidal or ocean thermal energy can be obtained.
  • The cost of construction of plants is very; high
  • The efficiency of producing energy is also low.

Question 42.
Why is it said that geothermal power plants are not possible in Kerala? Discuss and record.
Answer:
There are some minor environmental is-sues associated with geothermal power. Geothermal power plants can in extreme cases cause earthquakes. There are heavy upfront costs associated with both geothermal power plants and geothermal heating/cooling systems. Very location-specific (most resources are simply not cost-competitive). Some countries have been blessed with great resources – Iceland and Philippines meet nearly one-third of their electricity demand with geothermal energy. If geothermal energy is transported long distances by the means of hot water (not electricity), significant energy losses has to be taken into account. Geothermal power is only sustainable (renewable) if the reservoirs are properly managed.

Textbook Page .164

Question 43.
What are the different methods by which energy is produced from the nucleus?
Answer:
Nuclear fusion, nuclear fission.

Question 44.
Even if the matter converted is very small, the energy produced is very large. What is the reason?
Answer:
According to Einstein’s equation E = mc2. The amount of energy produced is high, even though the transformed mass is very less. Here’ represents the converted mass, c is the speed of light (3 x 108 m/s) and E is the amount of energy obtained.

Question 45.
What is the reason for an uncontrolled fission reation ending in an explosion?
Answer:
Nuclear fission is the process in which the nuclei of greater atomic mass are split into lighter nuclei using neutrons. The mass of nucleus produced in such a process is less than of its parent nuclei. Thus there will be loss of matter in fission process. The mass lost during fission converts into energy. The two or three neutrons produced during the process, bombards with other nucleus and fission process continue rapidly and ends in big explosion.

Textbook Page . 165

Question 46.
Complete the Table 7.5
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 8
Answer:

Natural Man-made
1. Cosmic rays from outer space 1. isotopes in the medical field.
2. Radiations from radioactive materials on the Earth 2. Wastes from nuclear reactors.
3. Radiations from radioactive materials on the soil, water, and vegetation. 3. Televisions. Medical X-rays, Smoke detectors, Lantern mantles Nuclear medicine, Building materials
4. Internal radiations
Potassium – 40, Carbon -14, Lead – 210
4. During nuclear experiments

Textbook Page .166

Question 47.
Classify the energy from the follow-ing sources as green energy and bown energy. Solar cells, atomic reactors, tidal energy, hydroelectric power, diesel engines, windmills, thermal power stations.
Answer:

Green Energy Brown Energy
Solar cells Atomic reactors
Windmills Diesel engines
Tidal energy Thermal power stations
Hydroelectric power

Question 48.
What must be done to ensure maximum utilization of green energy while constructing a house?
Answer:

  • Sufficient sunlight should be available in the rooms during day time.
  • Comfortable warmth, coolness and air circulation must be available without the help of electricity.
  • Using the sun for heating through south facing windows during the winter lowers heating costs. Shading those same windows in summer lowers cooling costs.
  • Grid-tied solar photovoltaic (PV) panels currently provide the most cost- effective form of renewable energy for a zero energy home.
  • Fresh filtered air and moisture control are critical to its success. This need for ventilation has a silver lining:

Let Us Assess

Question 1.
In a way most of the important energy sources of today can be said to be solar. Which of the following does not belong to the solar energy?
a. fossil fuel
b. energy from the mind
c. nuclear energy
d. biomass
Answer:
Nuclear energy

Question 2.
Which of the following is a green energy?
a. coal
b. naptha
c. biogas
d.petroleum gas
Answer:
Biogas

Question 3.
Write down the advantages and limitations of solar cooker.
Answer:
Advantages:

  • Eco-friendliness
  • Maintain better air quality indoors, reduce carbon monoxide emissions
  • Less expensive
  • Don’t cause any environmental pollution

Limitation:

  • Cooking with solar cookers obviously requires sunlight, which makes it difficult to use during winter months and on rainy days.
  • Cooking also takes a significantly longer time compared to conventional methods.
  • Solar cookers are not as efficient at retaining heat as conventional cooking devices. Factors such as wind,  rain, and snow can seriously hinder operation, and in such weather conditions, even after the food is cooked, it  will lose its warmth very quickly.

HSSLive.Guru

Question 4.
Kerala has a long coastal land. Still, ocean is not considered as a major source of energy. Why?
Answer:
In Kerala usually, waves are of low. So they have less energy. Also during tides Kerala sea coasts, the sea attitude does not increasingly rise.

Question 5.
The graph regarding the calorific value of certain fuels is given below. Analyze the graph and answer the following Questions.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 9
a. Which fuel has the highest calorific value? Which has the lowest?
b. How many kilogram of dried cow- dung is to be burnt to obtain the same amount of heat produced when 1 kg of LPG burns?
c. From the graph find out the most suitable fuel for household pur-pose. Justify your answer.
Answer:
a. Highest – Hydrogen
Lowest – Cowdung (Dried)
b. Energy released when 1kg LPG bums = 54000 kJ.
Dried cow-dung is to be burnt to obtain the same amount of heat \(\frac { 54000 }{ 6000 }\) = 9 kg
c. Biogas, because it is a renewable energy. Comparatively high calorific value.

Extended Activities

Question 1.
Find out the scope of hydrogen as a fuel with a high calorific value and prepare an essay.
Answer:
The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value. Hydrogen has high calorific value and also possess high combustion rate too. Hydrogen does not create much atmospheric pollution like other gases. Therefore seek method to combust it in moderate rates and use fuel cells.

Question 2.
Visit a hydroelectric power station and
try to understand different stages of the production of electricity. Make use of this principle and find out the scope of mini hydroelectric power project.
Answer:
There are only some primary expenses to build a hydrogen-electric power station. Hydroelectric power stations must be constructed only in areas of heavy rainfall and good streamflow. If these required conditions are satisfied, the production of electricity from such power station is very profitable one.

Question 3.
Visit a biogas plant and explore the possibility of establishing a community biogas plant in your region.
Answer:
Generally, the biomass required to construct a biogas plant may not be available from a single house and it is not profitable to construct separate biogas for each house. We know that the social cleanliness is equally important as personal cleanliness, due to several reasons, like.economic advantages, it is necessary to run a social biogas plant. And it will be useful to more people.

Question 4.
Write a short play to make the public aware of the need for making use of solar energy.
Answer:
Sun is the major source of energy available in earth, during photosynthesis the light energy is converted into chemical energy and is stored in plant cells. Since this plant reaches in animals as food, the stored energy reaches in animals too. The remainings of the animals hurried in earth transforms under high pressure and temperature and becomes fuels. Write drama using these hints.

Question 5.
Solar energy has an incredible future in the field of transportation. We are in its infant stage. Write an essay on the topic “Prospects of solar energy”.
Answer:
Solar energy is the ultimate energy source. All energy sources are related and connected with solar energy in one or the other way. Expand these points and complete the essay.

Question 6.
Find out the advantages and disadvantages of main energy sources and tabulate them.
Answer:

Energy source Advantages Disadvantages
Solar energy Fuel profit High temperature
Solar power plant low pollution rate It is difficult con­struct a plant ca­nnot be used all the time
petrol It provides more energy in the useful manner More expense creates more po­llution. It gets ov­er very soon.
coal It provides more energy in the useful form More expense creates more po­llution. It gets ov­er very soon
Biogas It can be renewed less environmental pollution More expense space limitation

Question 7.
A nuclear reactor is about to be established in Kerala. What is your reaction to this proposal? Justify.
Answer:
Kerala is a highly populated state in India. Therefore there are not much vacant places available for the construction a nuclear reactors here. Since the radiations given out from such power plant are very harmful to human life, it may cause unpredictable effects. I strongly disagree with this opinion.

Question 8.
A man pointing at a car running on petrol says. “This car is running on solar energy”. Write down your responses about this matter.
Answer:
We use fossil fuels like petrol or diesel in cars. The plants which lived millions of year ago have also used solar energy for photosynthesis to make food. Humans and animals grew up by taking these plants and fruits. Then this humans and animals are burned in earth after their death and became fossil fuels. Fuels are formed out of them. This fuel is being used in cars so the ultimate source of energy here is the solar energy itself. The vehicles connected with solar panels uses solar energy in the direct form.

Energy Management Orukkam Questions and Answers

Question 1.
Observe the fast and complete combustion of a flat paper. Understand that combustion of a crumbled paper make more smoke.
a. List the conditions for the complete combustion.
b. Name the gases produced during complete combustion.
c. Name the gases produced during partial combustion.
d. How does the partial combustion causes environmental pollution?
e. How are the fossil fuels formed?
f. Why are they considered as non-renewable?
g. Write the full form of LNG and CNG
h. Which is the main constituent of LNG? What are the uses of CNG?
i. Name the main constituent of CNG.
Answer:
a. Conditions for complete combustion. Dryness, Fast evaporation, Materials should reach at a certain temperature that is required for combustion.
b. Carbon dioxide, Steam, Heat, and light.
c. Carbon monoxide, soot and a little of car-bondioxide.
d. Besides the fuel loss and time loss, the carbon monoxide produced during partial combustion
e. Fossil fuels are formed by the transformation of plants and animals that went under the earths crust millions of years ago. Eg: Coal, petroleum and natural gases.
f. They are not replenished or renewed in proportion to their consumption.
g. LNG – Liquefied Natural Gas CNG – Compressed Natural Gas
h. Methane. CNG is used as fuel in vehicles, thermal power stations.
i. Methane

Question 2.
1. Write the full form of LPG, what is its main constituent?
2. Why is ethyl mercaptan added to LPG?
3. Which element is the main component of coal?
4. Name the different types of coal?
5. Name the process by which the components of coal are separated?
6. Write the products obtained by the distillation of coal.
Answer:
1. Liquefied Petroleum Gas, Butane
2. To identify the leak of LPG.
3. Carbon
4. Peat, Lignite, Anthracite and Bituminous coal.
5. Distillation
6. Ammonia, Coal gas, Coal tar, and Coke.

Work Sheet:

Question 1.
Identify the relation and fill in the blanks
a. LPG: Butane
LNG ………….
b. LNG: Fuel in Vehicles
LPG …………..
Answer:
a. Methane
b. For domestic use.

Question 2.
Name the chemical used to identify the leakage of cooking gas.
Answer:
Ethyl mercaptan

Question 3.
Find the odd one out and explain the reason
(Peat, Anthracite, Bauxite, Lignite)
Answer:
Bauxite, Others are types of coal.

Question 3.
1. Is the heat produced by the combustion of different fuels the same? Discuss
2. What do you mean by fuel efficiency?
3. What is its unit?
4. List the qualities of a good fuel.
5. Compare biomass and biogas
6. Name the device that converts solar energy into electrical energy.
7. What is the reason for electric current in this device?
8. In a Solar Panel Solar energy is converted into electrical energy. What do you mean by a Solar panel?
9. Write down the situation where solar panels are used.
10. What is the energy change in a solar heater?
11. What is the difference between a solar voltaic power plant and solar thermal power plant?
Answer:
1. No, the heat produced by combustion of different fuels is not the same because their calorific value is not the same.
2. We use firewoods, kerosene, LPG etc in our homes as fuels. Each of them liberates different amounts of heat. This is indicated as fuel efficiency.
3. Kilo Joule/ Kilogram
4. Should be easily available Should be of low cost
Should cause minimum atmospheric pollution on combustion.
5. Biogas: Biogas refers to a mixture of different gases produced by the breakdown of organic mater in the absence of oxygen. Biomass: The body parts of plants and animals are known as Biomass.
6. Solar panel
7. The flow of electrons from p – region to n – region.
8. A large number of solar cells are suitably assembled to form a solar panel.
9. In lighting street lamps, In artificial satellites
10. Solar energy to heat energy
11. Both photovoltaic and solar thermal are the two established solar power technologies. Photovoltaics use semi conductor technology to directly convert sunlight into electricity. Solar thermal works by using mirrors to concentrate sunlight.

Question 4.
Watch the animation video of nuclear fission and fusion
a. Which nuclear reaction happens in an atom bomb?
b. E=mc2
E = energy, then m = ………., c = ………….
c. What will you call the nuclear reaction in which small atoms combine?
d. Which nuclear reaction is carried out in stars?
e. What is energy change in a nuclear reactor?
f. Classify the given energy sources into conventional and non-conventional energy.
(Fossil fuels, Solar energy, Nuclear energy, Biomass, Hydro-Electric Power)
g. What will call the energy sources that lead to global warming?
h. What do you mean by green energy?
i. Why is it instructed to control the use of brown energy sources?
j. List the names of renewable energy sources.
k. Write down the causes an remedies of energy crisis.
Answer:
a. Nuclear Fission
b. m = mass, c = velocity of light
c. Nuclear fusion
d. Nuclear fusion
e. Nuclear energy to electical energy
f. Conventional energy sources – Fossil fuels, Biomass, Hydroelectric power Nonconventional energy sources – Solar energy, Nuclear energy
g. Brown energy
h. Green energy is the energy produced from natural sources which does not cause environmental pollution.
i. Brown energy are the sources which cause environmental problems including global warming, so it is essential to control the use of brown energy sources.
j. Solar energy, Wind energy, energy from biomass.
k. Judicious utilization of energy, Maximum utilization of energy Timely repairing of machines

Work Sheet

Question 1.
Classify the given energy sources into renewable and nonrenewable.
(Solar energy, Petroleum, Nuclear energy, Coal, Geothermal energy)
Answer:
Renewable energy resources – Solar energy, Geothermal energy, Coal

HSSLive.Guru

Question 2.
Identify the relation and fill in the blanks.
a. Hydrogen bomb: Nuclear fusion
Atom bomb: …………
b. Solar energy: Green energy
Nuclear energy: ……….
Answer:
a. Nuclear fission
b. Brown energy

Question 3.
Why green energy is called clean energy?
Answer:
Green energy is the energy produced from natural sources which does not cause environmental pollution. Hence it is known as clean energy.

Question 4.
What is the fuel used in a nuclear reactor?.
Answer:
Enriched Uranium

Energy Management SCERT Question Pool Questions and Answers

Question 5.
Though energy is available in many forms we depend mostly on electrical energy.
a. Write down two forms of energy, other than electrical energy, used in daily life.
b. Electrical energy is used much in daily life. Why?
c. Does increase in population and mechanization lead to energy crisis? How?
Answer:
a. Heat energy, Light energy.
b. Electrical energy can be easily converted into many other forms.
c. Small increase in population cause large increase in the use of energy consumption, mechanization will lead to excessive use of sources of energy and leads to energy crisis

Question 6.
The heat energy obtained on burning 2 kg of hydrogen, coal and petrol are given below:
Petrol – 9 × 107 J
Hydrogen- 3 × 108 J
Coal – 6 × 107 J
a. Which of these is the most efficient fuel?
b. How much is the calorific value of hydrogen?
c. Arrange these fuels in the ascending order of their calorific values.
d. Which of the above will you select as a good fuel? What is the basis of your selection?
Answer:
a. Hydrogen
b. 150000 kj/kg
c. Coal, petrol, hydrogen.
d. Petrol.
Easy to handle: less atmospheric pollution, Calorific value is high.

Question 7.
It is advisable to keep stirring heap of wastes while burning them.
a. Write down two essential situations for the complete burning of fuels.
b. How does the stirring help the combustion? Explain.
c. Write down two disadvantages, of partial combustion.
Answer:
a. To increase the availability of oxygen.
i. Increase the surface area exposed to air
ii. Make available the temperature needed for burning.
b. Sufficient oxygen is made available which increases the rate of combustion.
c. Partial combustion pollutes the air by giving excess of smoke, soot, carbon monoxide an carbon dioxide.

Question 8.
a. What is meant by nonrenewable sources of energy?
b. How did such fuels originate in nature?
c. Write down any two examples for it.
Answer:
a. sources that cannot be renewed.
b. The biowastes (bio residue) that went into the earth change into petroleum by undergoing chemical changes due to high temperature, in the absence of air.
c. Petrol, coal.

Question 9.
a. What are fuels? Write down the names of two fuels each from the various category of fuels.
b. How does the excessive use of fuels in-fluence global warming? Explain.
c. English the need for prohibiting diesel vehicles in our country in the context of environmental pollution.
Answer:
a. Fuels are substances that release large quantities of heat on burning; solid: logs: liquid: kerosene.
b. Excessive of fuels produce greenhouse gases which increase the temperature of atmosphere and cause global warming.
c. The amount of CO2 and SO2 released by diesel vehicles is very large. This accelerates global warming.

Question 10.
a. How does the biomass change into biogas in a biogas plant?
b. Of these which is the fuel that is advantageous?
c. What are the advantages of having community biogas plants?
Answer:
a. Biogas is formed in the biogas plants by the action of bacteria on biomass in the absence of oxygen.
b. Biogas has higher calorific value; environmental pollution is less.
c. Community biogas plants help in con-trolling environment pollution, centralised energy source minimizes expense.

This theoretical yield calculator will answer all the burning questions.

Question 11.
a. On what basis are the sources of energy classified as green energy and brown energy?
b. Classify the following into green energy and brown energy Solar cell, nuclear reactor, tidal energy, hydroelectric, diesel engine, windmill, thermal power station
c. Give any two suggestions to use the green energy to the maximum level.
Answer:
a. Green energy – green energy is the energy produced from natural resources that do not cause environmental pollution Brown energy – Energy from petroleum, coal, etc., and nuclear energy.
b. Green energy: solar cell, tidal energy, electricity from water, windmill.
Brown energy: nuclear reactor, diesel engine, thermal power station.
c. i. Install biogas plants
ii. Use solar panels

Question 12.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 10
The figure indicates the fission reaction of a heavy nucleus of Uranium.
a. Write down an activity to make energy available by using lighter nucleii.
b. Calculate the energy obtained if Ig matter changes into energy during nuclear fission reaction (c = 3 x 108 m/s).
c. What will be the result if the reaction shown in the above figure continues?
d. Why are we establishing nuclear reactors despite of the realization that nuclear reactors are controlled nuclear bombs?
Answer:
a. Nuclear fusion
b. E = mc2 m = lg
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 11
c. Continuous fission reaction ends in an explosion (atom bomb).
d. Energy crisis and industrial growth promote production of energy.

Question 13.
Coal is the most abundant fossil fuel on the earth.
a. How did the fossil fuels originate?
b. Which is the main constituent of coal?
c. What are the substances obtained when coal is allowed to undergo distillation in the absence of air?
d. How does the excessive use of fossil fuels cause global warming?
Answer:
a. Plants and animals that went under the earth many years ago changed into fossil fuels by undergoing changes under high pressure, high temperature, and absence of oxygen.
b. Carbon
c. Ammonia, coal gas, coaltar, coke
d. The greenhouse gases given out by the burning of fossil fuels increase the temperature of atmosphere and causes global warming

Question 14.
LNG and CNG are made from natural gases.
a. What is the advantage of LNG over CNG?
b. Compare LNG and CNG from the point of view of fuel efficiency.
Answer:
a. LNG – liquefied natural gases can be conveniently transported by liquefaction. Can be changed into gaseous state at the same temperature and can be distributed through pipes.

Question 15.
Hydrogen is a fuel with a high calorific value.
a. What is the limitation of hydrogen as a fuel?
b. Which is the substance added to hydro-gen to make a hydrogen fuel cell? ‘
Answer:
a. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.
b. Oxygen.

Energy Management Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 16.
Which of the following is a renewable energy source?
(Petroleum, Solar energy, Coal)
Answer:
Solar energy

Question 17.
What is the fuel used in a nuclear reactor?
Answer:
Enriched Uranium

Question 18.
Using the relation from the first pair, complete the other.
LNG – Methane
LPG –
Answer:
Butane

Question 19.
Write the full expansion of LPG.
Answer:
Liquified Petroleum Gas

Question 20.
Which nuclear reaction is carried out in stars?
Answer:
Nuclear Fusion

Question 21.
Which material is used along with hydro-gen to make Hydrogen fuel cells?
Answer:
Oxygen

Question 22.
Using the relation from the first pair, complete the other. Solar cell – Green energy Atomic reactor –
Answer:
Brown energy

Question 23.
Find the odd one in the group and write the reason.
[Coal gas, Coaltar, Butane, Coke]
Answer:
Butane. Others are those materials got from distillation of coal.

Question 24.
Which nuclear reaction takes place in an atom bomb?
a.Nuclear fission
b. Nuclear fusion
c. Radioactivity
d. Combustion
Answer:
a.Nuclear fission

Question 25.
Using the relation from the first pair, complete the other.
Atomic bomb – Nuclear fission
Hydrogen bomb – …………….
Answer:
Nuclear fusion

Short Answer Type Questions (Score 2)

Question 26.
What is meant by Biomass? Give examples for biomass.
Answer:
The body parts of plants and animals are known as biomass. Examples for biomass are Wood, Excreta, etc.

HSSLive.Guru

Question 27.
L.P.G is one of the common fuel for domestic use.
a. Which is the main constituent of LPG?
b. Write full form of LPG.
Answer:
a. Butane b. Liquified Petroleum Gas

Question 28.
Calorific value of Hydrogen is 1,50,000 kj / kg. What does it mean?
Answer:
The amount of heat liberated by the complete combustion of 1 kg of hydrogen is 1,50,000 kj/kg.

Question 29.
a. Write any two qualities of hydrogen as a fuel.
b. What is the limitation of hydrogen as a domestic fuel?
Answer:
a. High calorific value, High availability b. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.

Question 30.
Complete the flowchart.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 12
Answer:
a. Methane
b. LNG

Short Answer Type Questions (Score 3)

Question 31.
Write down the remedies of energy crisis.
Answer:

  • Judicious utilization of energy.
  • Maximum utilization of solar energy
  • Minimizing the wastage of water
  • Making use of public transportation as far as possible
  • Construction and beautifying of houses and roads in a scientific manner.
  • Controlling of the street lamps with LDR.

Question 32.
Classify the following suitably.
Solar cell, Atomic reactor, Tidal energy, Hydroelectric power, Diesel engine, Thermal power station, Windmill, Biogas

Green energy Brown Energy
solar cell Atomic reactor
tidal energy Diesel engine
Hydroelectric power Thermal power station
Windmill LPG
Biogas

Question 33.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 13
Various situations which uses solar energy are given in the diagram. What are the advantages & disadvantages of solar energy list them
Answer:

Advantages Disadvantages
1. Fuel efficient It is difficult to create high temperature
2. less atmospheric ‘ pollution Solar panel can not be made pollution use all the time

Question 34.
a. What are fossil fuels?
b. Why it is called as fossil fuels?
Answer:
a. Coal, Petroleum, Natural gas
b. Fossil fuels are formed by the transformation of animals and plants that went under the earth millions of years ago.

Question 35.
High calorific value is one among the properties that a substance must possess to be considered as a good fuel.
a. What do you mean by calorific value?
b. What are the other properties a substance must have in order to be considered as a good fuel?
Answer:
a. The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value.
b. Low price, more availability, less atmospheric pollution, portability, and easiness to store, etc. are the properties of a good fuel.

Long Answer Type Questions (Score 4)

Question 36.
Firewood, Cowdung cake, Kerosene, Petrol, LPG, etc are fuels.
a. Which among these has high calorific value?
b. In which state it exists?
c. What are the qualities of a good fuel?
d. Complete the table.

Fuel Main constituent
Biogas Methane
LPG ….. 1 ….
LNG ….. 2 …..
CNG …… 3….

Answer:
a. LPG
b. Gaseous state
c. Fuel with high calorific value is considered as good fuel
d. 1. Butane
2. Methane
3. Methane

Question 37.
A few news related to energy crisis are given below. State two reasons of energy crisis and write how we can overcome them.
Kerala Syllabus 10th Standard Physics Solutions Chapter 7 Energy Management 14
Answer:
Reasons: Increase in population, Overexploitation of nonrenewable energy resources, Industrialisation, Urbanisation. Remedial measures:- Population control, Use of renewable energy sources, Industrialisation after finding required energy sources.

Question 38.
Choose the correct answer from the bracket
Fossil fuel, Hydrogen, electrical energy
methane, Butane, Petroleum, Solar panal
a. The most abundant element in sun is ………
b. A solar cell converts solar energy into ………..
c. ……….. is the main constituent of biogas.
d. ……….. provides the power required for artificial satellite.
Answer:
a. Hydrogen
b. Electrical energy
c. Methane
d. Solar panel

HSSLive.Guru

Question 39.
a. Among the energy sources identify which will give green energy.
Energy from fossil fuel. Nuclear energy, Solar energy, Energy from wind
b. What are the precautions to be taken for green energy in household buildings.
c. When fuels v are partially burned it can cause certain hazards. List out some.
Answer:
a. Solar energy, Energy from wind
b. Use solar panels, Use biogas plants.
c. Energy loss, Pollution

Question 40.
a. What are the properties that a good fuel must-have?
b. What is the full form of LNG? List out its characteristics.
Answer:
a. 1. Should be of low cost
2. Should be easily available
3. Should cause minimum atmospheric pollution on combustion.
b. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous format atmospheric temperature and distributed through pipelines.

Question 41.
a. What is the full form of LNG? List out its characteristics.
b. Compare Nuclear fission and Nuclear Fusion.
Answer:
a. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous form atmospheric temperature and distributed through pipelines.
b. Nuclear fusion: The process in which lighter nuclei are combined to form heavier ones. Nuclear Fission: The process by which the nuclei of greater mass are split into lighter nuclei, using neutrons. The mass of small nuclei is less than that of parent nucleus.

 

Kerala Syllabus 8th Standard Maths Solutions Chapter 1 Equal Triangles

You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 1 Equal Triangles

Equal Triangles Text Book Questions and Answers

Textbook Page No. 11

Question 1.
In each pair of triangles below, find all pair of matching angles and write them down.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 1
Solution:
(i) ∠A = ∠R (The angles opposite to 5cm sides)
∠B = ∠P (The angles opposite to the sides of length 4 cm)
∠C = ∠Q (The angles opposite to the sides of length 6 cm)

(ii) ∠L = ∠Y (The angles opposite to the sides of length 10cm)
∠M = ∠Z (The angles opposite to the side of length 4 cm)
∠N = ∠X (The angles opposite to the side of length 8cm)

Question 2.
In the triangles below AB = QR, BC = RP, CA = PQ
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 2
Compute ∠C of ∆ABC and all angle of ∆PQR.
Solution: C = 80° (Use the property that the sum of three angles of a triangle is 180°)
AB = QR
∴ ∠C = ∠P
∠C = 80°
∴ ∠P = 80°

BC = RP
∴ ∠A = ∠Q
∠A = 40°
∴ ∠Q = 40°

CA = PQ
∴ ∠B = ∠R
∠B = 60°
∴ ∠R = 60°
(The angle opposite to equal sides are equal)

Question 3.
In the triangle below.
AB = QR BC = PQ CA = RP
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 3
Compute the remaining angles of both the triangles.
Solution:
AB = QR ∴ ∠C = ∠P
BC = PQ ∴ ∠A = ∠R
CA = RP ∴ ∠B= ∠Q
∠A = 60° ∴ ∠R = 6o°
∠Q = 70° ∴ ∠B = 70°
∠A = 60, ∠B = 70° then ∠C = 180 – (60° + 70°)
∴ ∠C = 50° ∴ ∠P = 50°

Question 4.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 4
Are the angles of ∆ABC and ∆ABDequal in the figure above? Why?
Solution:
The side AB is common to both the triangles in the figure.
The side of ∆ABC are equal to the sides of ∆ABD. So the angles of ∆ABC are equal to the sides of ∆ABC.

HSSLive.Guru

you can go online and type the temperature you want to convert into Google using this method: “350 degrees F to C” or vice versa.

Question 5.
In the quadrilateral ABCD shown below, AB = AD, BC = CD
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 5
Compute all the angles of the quadrilateral?
Solution:
AB = AD, BC = CD
AC is the common side
The sides of the triangles ABC and ADC are equal. So their angles are also equal. AB = DC
∴ ∠ACD = ∠ACB = 50° (Angles opposite to equal sides of a triangle are equal)
BC = CD
∴ ∠BAC = ∠DAC = 30° (Angles opposite to equal sides are equal in a triangle )
∴ ∠D = ∠B = 100°

Textbook Page No. 15

Question 1.
In each pair of triangles below find the pairs of matching angles and write them down.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 6
Solution:
(i) BC = PR (If two sides of a triangle and the angle made by them are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal.)
∴ ∠B = ∠R
∴ ∠C = ∠P
∴ ∠A = ∠Q (The opposite angles of equal sides of two triangles are also equal)

(ii) MN = XY (If two sides of triangle and the angle made by then are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal)
∴ ∠L = ∠Z
∠M = ∠Y
∠N = ∠X (If two sides of a triangle are equal, the angles opposite to these sides are also equal)

Question 2.
In the figure below, AC and BE are parallel lines.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 7
(i) Are the lengths of BC and DE equal. Why?
(ii) Are BC and DE parallel? Why?
Solution:
(i) Given AC and BE are parallel lines.
∴ ∠CAB = ∠EBD
When we consider the triangles ∆CAB, ∆EBD (Corresponding angles)
BC = DE (The two sides of ∆ CAB and the angle made by them are equal to the two sides of ∆ EBD and the angle made by them. So the thirif side of triangle are also equal.)

(ii) Yes, they are parallel.
∠ABC = ∠BDE (The angles opposite to the equal sides of equal triangles are equal) But they are corresponding angles. BC and DE are parallel.

Question 3.
Is ABCD in the figure, a parallelo¬gram? Why?
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 8
Solution:
AC = BD
AB is the common side.
The angles between the sides AC, AB and BD, AB are equal.
∴ BC = AD
The opposite sides of quadrilateral ∆CBD are equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. So the opposite angles in quadrilateral ACBD are also equal.
∴ ACBD is a parallelogram.

Question 4.
In the figure below, M is the midpoint of the line AB. Compute the other two angles of ∆ABC
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 9
Solution:
AM = BM (Given M is the mid point of AB)
CM = CM (common)
∠AMC = 90° = ∠BMC
∴ The two sides in ∆AMC and ∆BMC and the angle made by them are equal.
So the third side and other angles are equal.
∠A = 50° ∴ ∠B = 50°
∠ACM = 40° ∴ ∠BCM = 40°
∴ ∠C = 80°

Question 5.
In the figure below, the lines AB and CD are parallel and M is the mid point of AB.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 10
(i) Compute the angle of ∆AMD, ∆MBC and ∆DCM?
(ii) What is special about the quadrilateral AMCD and MBCD?
Solution:
Given AB = 12 cm and M is the mid-point of AB.
∴ AM = MB = 6 cm
In quadrilateral AMCD,
AM = CD
AB||CD ∴ AM||CD
∴ AMCD is a parallelogram.
∴ ∠AMD = ∠CDM (Alternate interior angles)
∠ADM = ∠CMD (Alternate interior angles)
∠A = ∠DCM = 40° = ∠CMB
∴ ∠MCB = 80° [180 – (60 + 40)]
(i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively.
(ii) Both of them are parallelograms.

Textbook Page No. 21

Question 1.
In each pair of triangles below, find matching pairs of sides and write their names.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 11
Solution:
(i) BC = PQ
AC = PR (The two angles and the side in between them in ∆ ABC are equal to the two angles and the side in between them in ∆PQR. So the third angles of the triangles ∠C and ∠R are also equal. Also BC and PQ, opposite to the 50° angle are also equal. The sides AC and PR opposite to the 70° angle are also equal.

(ii) ∠N = 70°
∠Z= 80°
MN = XZ
∠M = ∠Z(Sides opposite to equal angles are also equal)

Question 2.
In the figure, AP and BQ equal and parallel are lines drawn at the ends of the line AB. The point of inter section of PQ and AB is marked as M.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 12
(i) Are the sides of ∆AMP equal to the sides of ∆BMQ? Why?
(ii) What is special about the position of M on AB.
(iii) Draw a line 5.5 cm long. Using a set square, locate the midpoint of this line.
Solution:
(i) Yes, they are equal
∠P = ∠Q
∠A = ∠B (alternate angles formed by cutting the parallel lines AP and QB by PQ and AB.)
AP = QB
∴ The third angle of ∆APM and ∆BMQ and opposite sides of equal angles are equal.
(ii) AM = BM. So M is the midpoint of AB.
(iii) Draw a line segment of length 5.5 cm. Draw perpendiculars of equal lengths upward at one end of the line and downwards at the other end. Join the ends. This line divides the first one equally.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 13

Question 3.
In the figure, ABCDE is a pentagon with all sides of the same length and all angles of the same size. The sides AB and AE extended, meet the side CD extended at Px and Q.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 14
(i) Are the sides of ∆BPC equal to the sides of ∆EQD? Why?
(ii) Are the sides of AP and AQ of ∆ APQ equal? Why?
Solution:
(i) Yes, this are equal the sides and angles of a pentagon are equal.
∴ BC = DE
∠PBC = ∠PCB (Exterior angles of a regular pentagon)
∠QDE = ∠QED (Exterior angles of a regular pentagon)
∆QDE = ∆QED (If two angles and side of one triangle are equal are equal to two angles and corresponding side of the other triangle then their sides are equal.
BP = EQ and PC = DQ

(ii) AB = AE sides of regular pentagon.
BP = EQ
∴ AP = AQ [AB + BP = AC + EQ]

Question 4.
In ∆ABC and ∆PQR shown below.
AB = QR BC = RP CA = PQ
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 15
(i) Are CD and PS equal? Why?
(ii) What is the relation between the areas of ∆ABC and ∆PQR?
Solution:
(i) AB = QR
BC = RP
∠A = ∠Q
∴ ∆ABC and ∆QRP are equal triangles. Given all sides of ∆ABC are equal to the sides of ∆QRP
∴ CD and PS are equal. (Opposite sides of equal angles

(ii) AB = QR and CD = PS
⇒ 1/2 AB × CD = 1/2QR × PS
∴ The areas of ∆ABC and ∆PQR are equal.

HSSLive.Guru

Question 5.
In the quadrilateral ABCD shown below the sides AB and CD are parallel. M is midpoint of the side BC. The lines DM and AB extended meet at N.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 16
(i) Are the areas of ∆DCM and ∆BMN equal? Why?
(ii) What is the relation between the areas of the quadrilateral ABCD and the triangle ADN.
Solution:
(i) M is the midpoint of the line BC.
∴ CM = MB; BN || DC
∴ ∠DCM = ∠NBM (Alternate angles)
∠DMC = ∠NMB (Vertically opposite angles)
∴ ∆DCM and ∆BMN are equal triangles. So their areas are equal.

(ii) The areas of ∆DCM and ∆BMN are equal and quadrilateral AB, MD common
∴ The area of the quadrilateral
ABCD and the area of ∆ADN are equal.

Question 6.
Are the two diagonals of a rectangle equal. Why?
Solution:
ABCD is a rectangle.
Consider the ∆ABD and ∆ABC
AB = AB, common AD = BC (opposite sides of rectangle); ∠A = ∠B =90°
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 17
Both sides of ∆ABD and ∆ABC and the angle formed by them are equal. So the third sides BD and AC are equal. So the diagonals of the rectangle are equal.

Textbook Page No. 26, 27

Question 1.
Some are equal isosceles triangles are drawn below, In each, one angle is given. Find the other angles.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 18
Solution:
(i) 30°, 75°, 75°
(ii) 40°, 70°, 70°
(iii) 20°, 8o°, 8o°
(iv) 100°, 40°, 40°

Question 2.
One angle of an isosceles triangle is 90°. What are the other two angles?
Solution:
The other two angles are equal. So they are 45°, 45°

Question 3.
One angle of an isosceles triangle is 6o°. What are the other two angles.
Solution:
The other two angles are equal. So they are 60°, 60°

Question 4.
In the figure below, O is the centre of the circle and A, B are points on the circle.
Compute ∠A and ∠B?
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 19
Solution:
OA = OB (radius of circle)
∆AOB is a isosceles triangle; ∴ ∠A = ∠B
∠O = 60°
∴ ∠A = ∠B = 60°

Question 5.
In the figure below, O is the centre of the circle and A, B, C are points on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 20
What are the angles of ∆ABC?
Solution:
∆AOB, ∆AOC, ∆BOC are isosceles triangles. Each triangles are with angles 120°, 30° and 30°.
∴ ∠A = ∠B = ∠C = 30° + 30° = 60°

Text Book Page No. 29

Question 1.
Draw a line of 6.5 centimetres long and draw its perpendicular bisector.
Solution:
Draw a line segment AB of length 6.5 c.m with A and B as centres draw arcs on both sides of AB with equal radii. The radius of each of these arcs must be more the half the length of AB. Let these arcs cut each other at points C and D. Join CD which cuts AB at M
Then AM = BM. Also ∠AMC = 90°
Hence, the line segment CD is the perpendicular bisector of AB as it bisects AB at M and is also perpendicular to AB.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 21

Question 2.
Draw a square, each side 3.75 centimetres long?
Solution:
Draw AB = 3. 75 cm at. A Construct ∠PAB = 90° from AP, cut AD = 3.75 cm
Taking D as centre, draw an arc of radius 3.75 cm and taking B as centre, draw one more are of radius 3.75 cm.
Let the two arcs intersect at point C. Join BC and DC.
Then ABCD is the required square.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 22

Question 3.
Draw an angle of 750 and draw its bisector?
Solution:
Draw a line segment AB of any suitable length with A as centre. Draw an arc of any size to cut AB at D. With D as centre. Draw another arcs of some size to cut the previous arc at C.
Now ∠CAD = 60°. Draw ∠EAB = 90° and bisect ∠EAC.
∴∠PAC = 150 ∠DAC + ∠CAP = 60 + 15 = 75°
∴ ∠BAP = 75°
Then bisect
∠BAP AQ to the bisectors of ∠PAB
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 23

Question 4.
Draw a circle of radius 2.25 centimetres.
Solution:
Draw a line of length 4.5 cm. Draw its perpendicular bisector it meet at point ‘O’.
‘O’ is the centre of the circle and radius = 2.25 cm. Then complete the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 24

Question 5.
Draw ∆ ABC, with AB = 6 cm,
∠A = 22\(\frac{1}{2}\)°, ∠B = 67\(\frac{1}{2}\)°
Solution:
Draw the line AB in 6 cm length. Draw angle A at 45° and draw its bisector. Draw angle 135° at B and draw its bisector. Mark the point as C where bi-sectors meet.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 25

Question 6.
Draw a triangle and perpendicular bisectors of all three sides. Do all these three bisectors intersect at the same point?
Solution:
Draw a triangle ABC. By using compass mark the arcs on both sides from each ends.
Draw the same for all sides. They intersect at same point
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 26

Question 7.
Draw a triangle and the bisectors of the three angles. Do all three bisectors intersect at the same point.
Solution:
Draw a triangle PQR and by using com-pass draw the bisectors of angles. All three bisectors meet at the same point.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 27

Question 8.
Prove that if both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram.
Solution:
When the diagonal of a quadrilateral with equal opposite sides is drawn, we get two equal triangles. The angels opposite to the diagonal in the triangles are equal. That is the opposite sides and angles in the quadrilateral are equal. So the quadrilateral is a parallelogram.

HSSLive.Guru

Question 9.
In the figure, ABCD is parallelogram and AP = CQ
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 28
Prove that PBQD is a parallelogram
Solution:
DC = AB ………. (1)
AD = CB
QC = AP ……. (2) (as ABCD is a parallelogram)
(1) – (2) ⇒ DC – QC = AB – AP; ∴ DQ = PB
When, ∆ APD, ∆ CQB are considered.
AD = CB
AP = QC
∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.
∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.

Question 10.
Prove that if all sides of a parallelogram are equal, them each diagonal is the perpendicular bisector of the other.
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 29
The diagonal BD divides the parallelogram into two isosceles triangles. [The angles opposite to the equal sides in an isosceles triangles are equal.] So the diagonal DB bisect ∠D and ∠B.
Similarly the diagonal AC bisect A and ∠C. 4x +4y = 360° ⇒ x + y = 90°
The four triangles formed by intersec¬ting the diagonals are equal triangles. Each one 90 angle. So each diagonal is the perpendicular bisector of the other. In ∆AMD ∠AMD = 180 – (x – y) = 180 – 90 = 90° ⇒ BD ⊥ AC

Question 11.
In the figure below O is the centre of the circle and AB is the diameter. C is the point on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 30
(i) Compute ∠CAB
(ii) Draw another figure like this with a different number for the size of ∠COB. Calculate ∠CAB
Solution:
(i) ∠BOC = 50°
∴ ∠COA = 180° – 50°= 130° (straight angle)
OA = OC (radii)
∴ AOC is an isosceles triangle.
∴ ∠A = ∠C = \(\frac{180-130}{2}\) = 25°

(ii) ∠O = 70°
∴ ∠COA = 180 – 30 = 150°
OA = OC
∴ ∠CAB = ∠ACO =\(\frac{180-150}{2}=\frac{30}{2}\) = 15°
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 49

Question 12.
In the figure below, O is the centre of the circle and AB is a diameter. C is a point on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 31
(i) Compute ∠ACB
(ii) Draw another figure like this, changing the size of ∠COB and calculate ∠ACB.
Solution:
(i) ∠BOC = 50°
∠AOC = 130°
∆ AOC and ∆ BOC are isosceles triangles.
∠OAC = ∠OCA = \(\frac{180-130}{2}\) = 25°
∠OBC = ∠OCR = \(\frac{180-50}{2}\) = 65°
∠ACB = ∠OCA+ ∠OCB
25° + 65° = 90°

(ii) ∠AOC = 180 – 80 = 100°
∠OAC + ∠OCA = 180 – 100 = 8o°
∴ ∠OAC = ∠OCA = 80 ÷ 2 = 40°
∆ OBC
∠OBC + ∠OCB = 180 – 80 = 100°
∠OBC =∠OCB = 100 ÷ 2 = 50°
∠ACB = ∠OCA + ∠OCB
40° + 50° = 90°

Question 13.
How many different isosceles triangles be drawn with one angle 50° and any one side 7 centimetres.
Solution:
An isosceles triangle can be drawn with one angle 50° as angles either 50°, 50°, 8o° or 50°, 65°, 65°. In both the cases, 7 cm can be taken as equal sides or can be without 7 cm one as side. So there can be 4 ways of drawing diagram.

Question 14.
Draw ∆ ABC with AB = 7 cm, ∠A = 67\(\frac{1}{2}\), ∠B = 15° without using protector.
Solution:
Draw AB with length 7 cm. Extend both the sides. Draw the perpendicular from A. Draw the bisector through the left 90° angle among the 90° angles obtained. Draw an angle as 90° + 45° = 135°, Draw its bisector.
Now ∠A = 67\(\frac{1}{2}\). Draw an angle 60° in B to construct an equilateral triangle. Draw the bisector of its bisector. Then ∠B = 15°. We get ∆ ABC.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 48

Equal Triangles Additional Questions & Answers

Question 1.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 32
O is the centre of the circle in the diagram. If AB = BC,
(a) Then prove that ∠AOB = ∠BOC
(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?
(c) Find out how many equilate¬ral triangles can be drawn in a circle with length of its side is radius.
Solution:
(a) OA = OB = OC, AB = BC
∆ OAB and ∆ OBC are equal triangles.
∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.
(b) If OA = AB then ∆ OAB is an equilateral triangle.
If OB = BC, ∆ OBC is equilateral triangle.
∴ ∠AOB = ∠BOC = 60°
(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.

Question 2.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 33
If AB = AD, BC = CD in the diagram, then prove that ∠ABC = ∠ADC
Solution:
The three sides of triangles ∆ ABC are equal. The angles opposite to the sides are also equal.
AB = AD, BC = DC, AC = AC
AC is the common side. So the angles opposite to this side ∠ADC and ∠ABC are also equal,
i e ∠ABC = ∠ADC

Question 3.
Draw a rhombus with sides and a diagonal as 5 cm.
Solution:
Draw a line of length 5 cm. Draw equilateral triangles on both ends of the line with length 5 cm and line as one side.

Question 4.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 34
In the figure, AB = DE, BC = EF and AC = DF. Can ∠BPD = ∠C? Prove it?
Solution:
The sides of ∆ DEF and ∆ ABC are equal. The angles opposite to equal sides are equal.
∠E = ∠B, ∠F = ∠C, ∠D = ∠A
But ∠C = 180 – (∠B + ∠A)
∠P = 180 – (∠B + ∠D)
∴ ∠D = ∠A
∴ ∠C = ∠P = 180 – (∠B + ∠A)
∴ ∠C = ∠P

Question 5.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 35
In the figure, AB = PQ, AC = PR, BC = QR. PQ is parallel to AB.
(a) Then show that BC is parallel to QR.
(b) Also show that PR is parallel to AC
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 36
AB = PQ, AC = PR, BC = QR
∴ The angles of ∆ ABC and ∆ PQR are equal.
∠A = ∠P, ∠B =∠Q, ∠C = ∠R
(a) AB||PQ, ∴ ∠B = ∠PMN (corresponding angle);
∴ ∠PMN = ∠Q
∴ MN||QR
∴ BC||QR

(b) BC||QR
∠R = ∠MNP = ∠C
∴ NP||AC, PR||AC

Question 6.
Diagonals of three parallelograms with equal areas are given. Draw the parallelograms.
(i) length of diagonal 7 cm
(ii) length of diagonal 6 cm
(iii) length of diagonal 5 cm.
Solution:
(i) Draw a line of length 7 cm. Draw triangles of sides 7 cm, 6 cm, 5 cm at both the ends of the line to get a parallelogram by joining both the triangles.
(ii) Draw a line of length 6 cm and follow the above method.
(iii) Draw a line of length 5 cm and follow the above method.

HSSLive.Guru

Question 7.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 37
O is the centre of circle and AC, BD are diameters in the figure. Prove that AB = CD
Solution:
Consider ∆ODC and ∆OAB.
OD = OC = OA = OB (radi ∠AOB = ∠DOC; Two triangles are equal So the third sides of the triangles AB and CD are equal.)

Question 8.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 38
ABCD in the figure is a parallelogram. P, Q, R and S are the mid points of the sides of the parallelogram. The prove that PQ = RS, and QR = PS.
Solution:
Consider the triangles ∆APS and ∆CRQ
AP = CR, (half of the equal lines AB and CD)
AS = CQ (half of the lines with equal lengths AD and BC.)
∠A = ∠C (opposite angles of the parallelogram are equal)
When two sides and the angle made by them, in a triangle are equal then the third sides are also equal.
∴ QR = PS
Similarly if ∆ DSR and ∆ BQP are considered, PQ = RS is obtained.

Question 9.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 39
P is the midpoint of the sides AB and DF in the figure.
(a) Prove that BD = AF
(b) Is EF parallel to BC? Why?
(c) If D is the midpoint of BC, A is the midpoint of EF and Q is the mid point of DE then can Q be the midpoint of AC? Why?
Solution:
(a) Consider ∆APF and ∆DBP
FP = DP and AP = PB
∠APF = ∠DPB. The sides and the angle made by them in the triangles are equal. So the third sides BD and AF are equal.

(b) FP = PD. So the angles opposite to them are also equal.
∠FAP and ∠DBP are equal.
∴ FA||BD andBC|| EF.

(c) Consider BD = AF, in Question (a)
∴ BC = EF, Consider ∆ AEQ and ∆ DCQ
AE = DC, QE = DQ
∠AEQ = ∠CDQ
Two sides of a triangle and the angle made by them are equal. So the third sides are also equal, ie AQ = QC.
∴ So Q is the midpoint of AC

Question 10.
Draw a parallelogram if one of its diagonal is 8 cm length and one side is 6 cm. and the angle formed by the side and the diagonal is 40.
Solution:
Draw a diagonal of length 8 cm. Draw a line of 6 cm with 40 angle at its one end. Draw the same in its opposite direction.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 40

Question 11.
One angle of an isosceles triangle is 80. Find the other possible angles of the triangle.
Solution:
8o°, 8o°, 20°
8o°, 50°, 50°

Question 12.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 41
O is the centre of the circle in the diagram. Radiaus is 3 cm and ∠AOB = 60°. Find the perimeter of ∆ AOB?
Solution:
∆ OAB is an isosceles triangle.
∴ ∠A =∠B
∠O = 60
∴ ∠OAB is an equilateral triangle so perimeter = 3 + 3 + 3 = 9 cm.

Question 13.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 42
OM is perpendicular to AB in the diagram. Prove that M is the mid point of AB.
Solution:
OA = OB
∴ ∆ OAB is an isosceles triangle.
∴ ∠A = ∠B
When ∆ OMA and ∆ OMB are considered, OM is the common side
∠AMO = ∠BMO = 90 ∠AOM = ∠BOM
One side of the triangle and angles at the ends of sides are equal. So the other two sides are also equal.
∴ AM = MB
∴ M is the midpoint of AB.

Question 14.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 43
In the figure ∠ABC = ∠ADC and AB = AD. Prove that A BCD is an isosceles triangle?
Solution:
AB = AD
∴ ∆ ABD is an isosceles triangle.
∴ ∠ABD = ∠ADB
It is given that ∠ABC = ∠ADC
∴ ∠CBD = ∠CDB
∴ CD = CB
∴ ∆ BCD is an isosceles triangle.

Question 15.
In ∆ ABC, AB = AC = 10 cm. M is the midpoint of BC. If BC = 12 cm, Find AM? Also find the area of ∆ ABC?
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 44

Question 16.
Show that, the triangle obtained by joining the mid points of the sides of an isosceles triangle is also an isosceles triangle.
Can we get an equilateral triangle by joining the mid points of the sides of an equilateral triangle?
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 45
∆ ABC is an isosceles triangle. P, Q, R are the mid points of the sides of the triangle. Consider ∆ PBR and ∆ QRC.
PB = QC
BR = CR
∠B = ∠C
Two sides and the angle made by them are equal. The third sides PR and QR also equal.
∴ ∆ PQR is an isosceles triangle. Similarly the triangle obtained by joining the mid points of the sides of an equilateral triangle is an equilateral triangle.

Question 17.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 46
In the figure ∠B = ∠C = 90. If AB = CE and BE = CD, then find angles of ∆AED?
Solution:
If ∆ ABE and ∆ ECD are considered, AB = EC, BF = DC and ∠B = ∠C. Two sides of the triangle and the angle made by them are equal. The third sides AE and DE are also equal. ∆ADE is an isosceles triangle.
∆BAE = ∆DEC
∠BEA = ∠EDC (Angles opposite to the equal sides are also equal)
∠BAE + ∠BEA = 90
∴ ∠BEA + ∠+ BEA = 90
∴ ∠AED = 90°
∴ ∆AED is an isosceles triangle.
∴ ∠EAD = ∠EDA = 45°

Question 18.
Construct the following triangles by using only scale and compass.
(a) In ∆ ABC, AB = 6 cm, ∠A = 45°, ∠B = 75°
(b) In ∆ PQR, PQ = 7 cm,
∠P = 52\( \frac{1}{2}\)° , ∠ Q = 82\( \frac{1}{2}\)° 2
Solution:
(a) Draw AB = 6 cm
Draw AP making angle 45° with AB.
Draw BQ making angle 75° with AB.
Let AP and BQ intersect at C.
∴ ABC is the required triangle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 47