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Kerala Syllabus 8th Standard Textbooks Solutions Guide

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Kerala State Syllabus 8th Standard Textbooks Solutions

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Plus One Chemistry Notes Chapter 7 Chemical Equilibrium

November 25, 2023November 25, 2023 by HSS Live

Students can Download Chapter 7 Chemical Equilibrium Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 7 Chemical Equilibrium

Introduction
Chemical equilibria are very important in numerous biological and environmental processes. At equilibrium state, the rate of product formed is equal to the rate of reactants formed. The mixture of reactants and products at equilibrium state is called an equilibrium mixture. A equilibrium mixture involving ions in aqueous solutions which is called as ionic equilibrium

Equilibrium In Physical Processes
Phase transformation processes are the familiar example for equilibrium in Physical process.
They are,
Solid \(\rightleftharpoons \) liquid
Liquid \(\rightleftharpoons \) gas
Solid \(\rightleftharpoons \) gas

Solid Liquid Equilibrium
Consider a perfectly insulated thermos flask containing some ice and water at 273 K and normal atmospheric pressure. Since the flask is insulated, there will be no exchange of heat between its contents and the surroundings. It is seen that as long as the temperature remains constant, there is no change in the mass of ice and water. This represents an equilibrium state between ice and water and maybe represented as

We observe there is no change in mass of both ice and water. Since the rate of both reactions are equal.
rate of melting = rate of freezing For any pure substance at 1 atmospheric pressure the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance.

The equilibrium concentration calculator enables us to find how much the reactants and products.

Liquid – Vapour Equilibrium
In order to understand the liquid-vapour equilibrium, let us consider evaporation of water in a closed vessel. Consider a closed vessel connected to a manometer. The water vapour present in the vessel is first removed by placing some drying agent such as anhydrous calcium chloride in it for some time. The drying agent is then removed. Now the level of mercury in both the limbs of the manometer will be same. Introduce some water into the vessel and allow to stay at room temperature. Now water starts evaporating. A Pressure will gradually develop within the vessel due to the formation of water vapours. The change of pressure can be easily measured from the manometer. As evaporation continues, the pressure goes on increasing and the level of mercury in the right limb of the manometer starts rising. After some time it is observed that pressure becomes constant. This shows that the quantity of water vapour is not increasing any more, although liquid water is still present in the vessel. This indicates that a state of dynamic equilibrium has been attained between liquid water and water vapours.
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 2

At equilibrium, both reaction take place at the same rate. Thus at equilibrium,
rate of evaporation = rate of condensation

The pressure exerted by the vapours in equilibrium with the liquid at a particular temperature is called
vapour pressure of the liquid.

It may be noted that the equilibrium between the vapours and the liquid is attained only in a closed vessel. If the vessel is open, the vapours leave the vessel and get dispersed. Hence the rate of conden-sation will never become equal to the rate of evapo-ration.

Solid – Vapour Equilibrium
Consider systems where solids sublime to vapour phase, For example,

Equilibrium involving Dissolution of Solid or Gas in Liquids
Solids in liquids: In a saturated solution, a dynamic equilibrium exits between the solute molecules in the solid state and in the solution: the rate of dissolution of sugar = rate of crystallisation of sugar. Gases in liquids: This equilibrium is governed by Henry’s law, which states that the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent

General Characteristics of Equilibria involving Physical Processes
For the physical processes discussed above, following characteristics are common to the system at equilibrium:

Equilibrium is possible only in a closed system at a given temperature.
Both the opposing processes occur at the same rate and there is a dynamic but stable condition.
All measurable properties of the system remain constant.
When equilibrium is attained for a physical process, it is characterised by constant value of one of its parameters at a given temperature.
The magnitude of such quantities at any stage indicates the extent to which the reaction has proceeded before reaching equilibrium.

Equilibrium In Chemical Processes – Dynamic Equilibrium
Consider a general reversible reaction
A+B \(\rightleftharpoons \) C+D

Suppose the reaction is carried out in a closed container. In the beginning, the concentrations of A and B are maximum and the concentrations of C and D are minimum (equal to zero). As the reaction proceeds, the concentrations of A and B will decrease whereas the concentrations of C and D will increase. Hence the rate of the forward reaction will be high in the beginning and it will decrease gradually because of the fall in concentrations of A and B. On the other hand the velocity of the reverse reaction will be minimum at the beginning and it will increase gradually due to the increase in concentrations of C and D. Finally a stage will be reached when the rate of the forward reaction becomes equal to the rate of the reverse reaction. This state of the system is known as the state of chemical equilibrium. At this state the concentrations of the reactants and the products remain constant.
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 4
We can also start with C and D and make the reaction to proceed in the reverse direction. The concentration of C and D decreases and A and B increases. Finally, equilibrium is attained. One such example is given.
H_2(g) +l_2(g) \(\rightleftharpoons \) 2Hl_(g)

Law Of Chemical Equilibrium And Equilibrium Constant
The relation between rates of reaction and concentrations was given by Guldberg and Wage in 1864. This relation is known as law of mass action.
The relation is,
\(K_{c}=\frac{[C][D]}{[A][B]}\)
For a general reversible reaction of the type,
aA + bB \(\rightleftharpoons \) cC + dD
the equilibrium constant maybe represented as
\(K_{ c }=\frac { [c]^{ c }[D]^{ d } }{ [A]^{ a }{ \left[ B \right] }^{ b } } \)
The equation is known as the expression for the law of chemical equilibrium.

Chemical Equilibrium

The law of chemical equilibrium or equilibrium law may thus be stated as :
At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the prod-uct of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.
If equilibrium constant for the backward reaction is
K’_c then K’_c = \(\frac{1}{K_{e}}\)

Homogeneous Equilibria
In a homogeneous system, all the reactants and products are in the same phase. For example, in the gaseous reaction,
N_2(g) + 3H_2(g) \(\rightleftharpoons \) 2NH_3(g)

Heterogeneous Equilibria
Equilibrium in a system having more than one phase
is called heterogeneous equilibrium.
For example, H₂O_(l) \(\rightleftharpoons \) H₂O_(g)

Applications Of Equilibrium Constants

Predicting the Extent of a Reaction

If K_c >10³, products predominate over reactants, i.e., if K_c is very large, the reaction proceeds nearly to completion.
If K_c < 10^-3, reactants predominate over products, i.e., if K_c is very small, the reaction proceeds rarely.
If K_c is in the range of 10^-3 to 10³, appreciable concentrations of both reactants and products are present.

Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 5

Predicting The Direction Of The Reaction
The equilibrium constant is also used to find in which direction the reaction mixture of reactants and products will proceed. For this, we have to calculate the reaction quotient (Q_c) and compare with the equilibrium constant (K_c).

The concentrations of the species in Q_c are not necessarily equilibrium values.
For a general reaction aA + bB → cC + dD
\(Q_{ c }=\frac { [c]^{ c }[D]^{ d } }{ [A]^{ a }{ \left[ B \right] }^{ b } } \)
If Q_c > K_c, the reaction will proceed in the direction of the reactants (i.e., reverse reaction).
If Q_c < K_c, the reaction will proceed in the direction of the products (i.e., forward reaction).
If Q_c = K_c, the reaction mixture is already at equilibrium.
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 6

Calculating Equilibrium Concentrations
Step 1.
Write the balanced equation forthe reaction.

Step 2.
Under the balanced equation, make a table that lists foreach substance involved in the reaction:
a) the initial concentration,
b) the change in concentration on going to equilibrium, and
c) the equilibrium concentration.

In constructing the table, define x as the concentration ’ (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.

Step 3.
Substitute the equilibrium concentrations into the equilibrium equation forthe reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.

Step 4.
Calculate the equilibrium concentrations from the calculated value of x.

Step 5.
Check your results by substituting them into the equilibrium equation.

Problem
3.00 mol of PCl₅ kept in 1L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate composition of the mixture at equilibrium. K_c = 1.80

Solution
Let x mol of PCl₅ dissociated, At equilibrium:
(3 – x) x x
K_c = [PCl₃][Cl₂][PCl₅]
1.8 = x²/(3 – x)
x² + 1.8x – 5.4 = 0
x = [-1.8 ± √(1.8)² – 4(-5.4)]/2
x = [-1.8 ± √3.24 + 21.6]/2
x = [-1.8 ± 4.98]/2
x = [-1.8 + 4.98]/2
x = 1.59
[PCl₅] = 3.0 -x = 3 – 1.59 = 1.41 M
[PCl₃] = [Cl₂] = x = 1.59 M

Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G

∆G is negative, then the reaction is spontaneous and proceeds in the forward direction.
∆G is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ”G, the products of the forward reaction shall be converted to the reactants.
∆G is O, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:

∆G = ∆G° + RT InQ
where, G° is standard Gibbs energy.
At equilibrium, when ∆G = 0 and Q=K_c the equation becomes,
∆G = ∆G° +RTIn K = 0
∆G° = -RTInK
InK = -∆G° / RT
Therefore, K = e^{∆G^v/RT}

Factors Affecting Equilibria
In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’sprinciple. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to both physical and chemical equilibria.

Effect of Concentration Change
When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes.

Effect of Pressure Change
A pressure change obtained by changing the volume can affect the yield of products in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of gaseous products are different.

Effect of Inert Gas Addition
If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed. It is because the addition of an inert gas at constant volume does not change the partial pressures orthe molar concentrations of the substance involved in the reaction. So the reaction quotient does not change.

Effect of Temperature Change
Whenever an equilibrium is disturbed by a change in the concentration, pressure or volume, the composition of the equilibrium mixture changes because the reaction quotient, Q_c no longer equals the equilibrium constant, K_c However, when a change in temperature occurs, the value of equilibrium constant, K_c is changed. In general, the temperature dependence of the equilibrium constant depends on the sign of ∆H for the reaction.

The equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases.
The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases.

Temperature changes affect the equilibrium constant and rates of reactions.

Effect of a Catalyst
A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly ‘ the same amount.

Ionic Equilibrium In Solution
Michael Faraday classified the substances into two categories based on their ability to conduct electricity. One category of substances conduct electricity in their aqueous solutions and are called electrolytes while the other do not and are thus, referred to as non-electrolytes.

Faraday further classified electrolytes into strong and weak electrolytes.

The ionic strength calculator is a convenient tool to help you calculate the ionic strength of a solution.

Strong electrolytes on dissolution in water are ionized almost completely, while the weak electrolytes are only partially dissociated.

Acids. Bases And Salts

Arrhenius Concept of Acids and Bases
According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H⁺_(aq) and bases are substances that produce hydroxyl ions OH^–_(aq). The ionization of an acid HX _(aq) can be represented by the following equations:
HX_(aq) → H⁺(aq) + X^–(aq)
or
HX(aq) + H₂O(l) -> H₃O⁺(aq) + X^–(aq)

The Bronsted-Lowry Acids and Bases
The Danish chemist, Johannes Bronsted and the English chemist, Thomas M. Lowry gave a more general definition of acids and bases. According to Bronsted-Lowry theory, acid is a substance that is capable of donating a hydrogen ion l-T and bases are substances capable of accepting a hydrogen ion, H⁺. In short, acids are proton donors and bases are proton acceptors.

The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH^– is called the conjugate base of an acid H₂O and NH₄⁺ is called conjugate acid of the base NH₃. If Bronsted acid is a strong acid then its conjugate base is a weak base and vice versa.
Consider the example of ionization of hydrochloric acid in water.
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 7

Ionization Of Acids And Bases

The Ionization constant of water and its ionic product
Water undergoes self ionisation to a small extent as follows.
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 8
Since [H₂O] is constant, K[H₂O]² may be taken as a new constant K_w. Thus,
K_w= [H₃O⁺][OH^–]

Where K_w is called ionic product of water. Its value is 1 x10‘14 mol2 L2 at 298 K. In pure water, the concen-tration of hydronium ions and hydroxyl ions are equal. Therefore in pure water,
[H₃O⁺] = [OH^–] = 1 × 10^-7 mol L^-1

Since the ionisation of water increases with increase of temperature, K_w increases with rise of temperature.

Our buffer pH calculator will help you painlessly compute the pH of a buffer based on an acid or a base.

The pH Scale
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH scale.

The pH of a solution is defined as the negative logarithm to base 10 of the activity (a_H+) of hydrogen ion.
i.e., pH = – log aH^H+ = – log {[H+]/,mol L^-1}
Acidic solution has pH < 7 Basic solution has pH > 7
Neutral solution has pH = 7

Ionization Constants of Weak Acids
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 9
Here, c= initial concentration of the undissociated acid, HXat time, t = 0. α = extent up to which HX is ionized into ions.
K_a = c²a² / c(1 – α) = cα²/1 – A
K_a is called the dissociation or ionization constant.

Ionization of Weak Bases
The equilibrium constant for base ionization is called base ionization constant and is represented by K_b.

When equilibrium is reached, the equilibrium constant can be written as:
K_b = (cα)² / c(1 – α) = cα² / (1 – α)
considering the base-dissociation equilibrium reaction:
K_b = [BH⁺][OH^–]/[B]
Then multiplying and dividing the above expression by [H⁺], we get:
K_b = [BH⁺][OH^–][H⁺]/[B][H⁺]
= {[OH^–][H⁺]}{[BH⁺]/[B][H⁺]}
= K_w/K_a
Then we get the following relation;
pK_a + PK_b = pK_q = 14 (at 298 K)

Common ion effect in the ionization of Acids and Bases.
Common ion effect my be defined as the suppression of the dissociation of a weak electrolyte (weak acid or weak base) by the addition of some strong electrolyte containing a common ion.

Factors Affecting Acid Strength
Dissociation of an acid depends on the strength and polarity of the H-A bond.
Electronegativity of A increases CH₄ < NH₃ < H₂O < HF Acid strength increases

Common Ion Effect in the Ionization of Acids and Bases
Ka = [H⁺] [Ac^–] / [HAc] acetate ions to an acetic acid solution results in decreasing the concentration of hydrogen ions, [H⁺], Also, if H⁺ ions are added from an external source then the equilibrium moves in the direction of undissociated acetic acid. This phenomenon is an example of common ion effect.

Hydrolysis of Salts and the pH of their Solutions
Salts formed by the reactions between acids and bases in definite proportions, undergo ionization in water. The cations/anions formed on ionization of salts either exist as hydrated ions in aqueous solutions or interact with water to reform corresponding acids/bases depending upon the nature of salts. The later process of interaction between water and cations/anions or both of salts is called hydrolysis.

Buffer Solutions
The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions.

Solubilityequilibriaof Sparingly Soluble Salts

Solubility Product Constant
The equilibrium between the undisolved solid and the ions in a saturated solution can be represented by the equation:
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 10
We call K_sp the solubility product constant or simply solubility product.

Thus, solubility product of a salt is the product of concentration of ions in its saturated solution, raised to a power equal to the number of times the ions occur in the equation representing the dissociation of the salt.

The term K_sp in equation is given by Q_sp when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions K_sp = Q_sp but otherwise it gives the direction of the processes of precipitation or dissolution.

Common Ion Effect on Solubility of Ionic Salts
The solubility of salts of weak acids like phosphates increases at lower pH. This is because at lower pH the concentration of the anion decreases due to its proto-nation. This, in turn, increases the solubility of the salt so that K_sp = Q_sp.

Ncert Supplementary Syllabus

Designing Buffer Solution
Knowledge of pK_a, pK_b and equilibrium constant help us to prepare the buffer solution of known pH. Let us see how we can do this.

Preparation of Acidic Buffer
To prepare a buffer of acidic pH we use weak acid and its salt formed with strong base. We develop the equation relating the pH, the equilibrium constant, K_a of weak acid and ratio of concentration of weak acid and its conjugate
base. For the general case where the weak acid HA ionises in water,
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 11
ratio of concentration of conjugate base (anion) of the acid and the acid present in the mixture. Since acid is a weak acid, it ionises to a very little extent ‘and concentration of [HA] is negligibly different from concentration of acid taken to form buffer. Also, most of the conjugate base, [A^–], comes from the ionisation of salt of the acid. Therefore, the concentration of conjugate base will be negligibly different from the concentration of salt. Thus, equation (A-2) takes the form: pH-pK_a + log\(\frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\)

In the equation (A-1), if the concentration of [A^–] is equal to the concentration of [HA], then pH = pK_a because value of log 1 is zero. Thus if we take molar concentration of acid and salt (conjugate base) same, the pH of the buffer solution will be equal to the pK_a of the acid. So for preparing the buffer solution of the required pH we select that acid whose pK_a is close to the required pH. For acetic acid pK_a value is 4.76, therefore pH of the buffer solution formed by acetic acid and sodium acetate taken in equal molar concentration will be around 4.76.

A similar analysis of a buffer made with a weak base and its conjugate acid leads to the result,

pH of the buffer solution can be calculated by using the equation pH + pOH =14.

We know that pH + pOH = pK_w and pK_a + pK_b = pK_w On putting these values in equation (A-3) it takes the form as follows:
Plus One Chemistry Notes Chapter 7 Chemical Equilibrium 13
If molar concentration of base and its conjugate acid (cation) is same then pH of the buffer solution will be same as pK_a for the base. pK value for ammonia is 9.25; therefore a buffer of pH close to 9.25 can be obtained by taking ammonia solution and ammonium chloride solution of equal molar concentration. For a buffer solution formed by ammonium chloride and ammonium hydroxide, equation (A-4) becomes:

pH of the buffer solution is not affected by dilution because ratio under the logarithmic term remains unchanged.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 5 Basic Constituents of Matter

November 28, 2023November 23, 2023 by HSS Live

You can Download Basic Constituents of Matter Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 5 Basic Constituents of Matter

Basic Constituents of Matter Questions and Answers

The particles in sugar candy and potassium permanganate solutions cannot be seen as they are extremely small. If we take one or two pieces of camphor in a china dish and heat the china dish. Even though the camphor has disappeared completely, the smell of camphor remains in air. This is because the camphor has spread in air as extremely small particles.

Fill one quarter of a boiling tube with sugar and close it with cotton. Then heat the boiling tube strongly. The water in sugar is sticked on the sides of the boiling tube as water vapour. Carbon and water are the components of sugar. Water can be further divided. When electricity is passed through water it splits into hydrogen and oxygen. That is sugar, a pure substance, can be divided into carbon, hydrogen, and oxygen. At the same time, pure substances like carbon, hydrogen, oxygen, gold, and silver cannot be further divided into its components by chemical reactions.

Elements and Compounds

The pure substances which cannot be further decomposed through chemical processes are called elements.
Ex: hydrogen, oxygen, gold, silver Compounds are pure substances formed from two or more elements through chemical combination.
Ex : water, sugar, carbon dioxide

Origin of the names of elements

In the past, the elements were named after places, countries, continents, characteristic properties, scientists, planets, satellite etc.
Ex : Place – polonium, francium
Continent – Europium, Americium
Character, colour – rubedium, chromium
Scientist – bohrium, curium
Planet – neptunium

Symbols

The first letter of the English or latin letters are used as the symbols of the ele-ments.
Ex: Carbon – C
Oxygen – O
Sodium-Natrium – Na
Potassium-Kalium – K
But, for some elements, along with the first letter, the second or another prominent letter is also used in the symbol as a small letter.
Ex: Calcium – Ca
Chlorine – Cl

Atom and Molecule

The element carbon is made up of extremely small particles. Think of breaking a piece of carbon into smaller and smaller particles. When it is broken up in this manner, we get the smallest particle which retains all the characteristic properties of carbon at the end. This smallest particle is called an atom of carbon.

Atoms of some gaseous elements like helium, neon etc., can exist as free single atoms. But atoms of some other gaseous elements like hydrogen, oxygen etc., can exist only as a combination of two atoms. Atoms of some elements are seen to exist as a combination of more than two atoms. The smallest particles which can exist independently are called molecules.

Method of representing Atoms and Molecules

The symbol of helium is ‘He’. When we write ‘He’, it represents one atom of the element helium. 2He means two helium atoms.

Elements like helium, neon, argon, etc., are seen in nature as single atoms. But elements like hydrogen exist as diatomic molecules. Hydrogen atom is represented as H. But H₂ represents a hydrogen molecule.

Some elements exist as molecules of more than two atoms. Phosphorus (P₄), sulphur (S₈), etc., are examples of such
elements.

Molecules with only one atom are called monatomic molecules.
Ex: He, Na

Those with two atoms each are called diatomic molecules.
Eg: O₂, H₂, N₂

The molecules with more than two atoms are called polyatomic molecules.
Eg: P₄ S₈

Atoms and Molecules

In monatomic molecules, the number given on the left side of the symbol indicates the number of molecules and atoms. In polyatomic molecules, the subscript on the right side of the symbol indicates the number of atoms within one molecule. The total number of molecules is indicated on the left side.
O – One oxygen atom
O₂– One oxygen molecule
2O₂ – Two oxygen molecules
2He – Two helium atom/molecule
5Na – five sodium atoms

Compounds

We have seen how molecules are formed from the same type of atoms. Similarly, the compounds contain molecules formed from atoms of different elements.

Carbon dioxide molecule is formed when one carbon atom combines with one oxygen atom.

This is a compound. Water is a compound formed by the combination of hydrogen and oxygen.

Carbon dioxide is formed when carbon burns in oxygen. Carbon dioxide is also formed by the decomposition of calcium carbonate. Whatever be the sourcj of a compound, a fixed ratio is maintained between the atoms of the different elements present in it. In carbon dioxide molecule, the ratio of carbon atoms to oxygen atoms is always 1: 2. Thus carbon dioxide molecule can be represented as CO₂. Similarly in water molecule the ratio of hydrogen and oxygen atom is 2:1 This type of representation is known as the chemical formula of the compound.
H₂SO₄ ,CO₂ ,H₂O, HCl, NaCl

2CO₂ represents two carbondioxide molecules. We can understand that there are 6 atoms including 2 carbon atoms and 4 oxygen atoms in 2CO₂.

There are 7 atoms including 2 hydrogen atoms, 1 sulphur atom and 4 oxygen atoms in H₂ SO₄, Molecule. There will be 14 atoms in 2H₂ SO₄

Chemical Equations

The chemical equation of the reaction between zinc and hydrochloric acid forming zinc chloride and hydrogen is written as shown below.
Zn+HCl → ZnCl₂ + H₂

Here Zn and HCl are reactants. ZnCl₂ and H₂ are products.

The chemical reactions can be represented as equations containing the symbols and chemical formulae of the reacta¬nts and the products. When a chemical equation is written, the number of atoms of the same kind, on the side of reactants and that of products should be equal. That is the chemical equation given above is to be rewrite as
Zn + 2 HCl → ZnCl₂ + H₂

Now the number of atoms in both side of the arrow is equal. This type of chemical equations which is written such that the number of atoms of same type is equal in the side of reactants and products are called balanced chemical equations.

Basic Constituents of Matter Textbook Questions and Answers

Question 1.
Classify the following into elements and compounds. Ammonia, sugar, nitrogen, mercury, sodium chloride, water, copper sulphate, sodium, carbon.
Answe:

Elements	Compounds
Mercury	Ammonia
Sodium	Sugar
Carbon	Sodium chloride
Nitrogen	Water
	Copper sulphate

Question 2.
Analyse the chemical equation given for the reaction between nitrogen and hydrogen to give ammonia.
N₂+3H₂ → 2NH₃
a. What are the reactants and products in this reaction?
b. Find the total number of molecules and atoms of the reactants.
c. Find the total number of molecules and atoms of the products.
d. What is the relation between the number of atoms of the reactants and products?
Answer:
a. Reactants: nitrogen, hydrogen Product: ammonia (NH₃)
b. Total number of molecules in reactant – 4 total atoms- 8
c. Total number of molecules in product -2 total atoms- 8
d. Total number of atoms in reactants and products are equal.

Question 3.
N is the symbol of nitrogen.
a. What do N₂, 2N and 2N₂ indicate?
b. How many molecules and atoms are present in 5N₂?
Answer:
a. N₂ – One nitrogen molecule
2N – Two nitrogen atoms
2N₂ – Two nitrogen molecules
b. 5N₂ – 10 atoms, 5 molecules

Question 4.
Some chemical equations are given.
C + O₂ → CO₂
CH₄ + 2O₂ → CO₂ + 2H₂O
N₂ + O₂ → NO
CaCO₃ → CaO + CO₂
H₂ + I₂ → HI
Fe + HCl → FeCl₂ + H₂
CO₂ + C → CO
a. Which of these are balanced chemical equations?
b. Balance those equations which are not balanced.
Answer:
a. C + O₂ → CO₂
CH₄ + 2O₂ → CO₂ + 2H₂O
CaCO₃ → CaO + CO₂

b. N₂ + O₂ → 2NO
H₂ + I₂ → 2Hl
Fe + HCl → FeCl₂ + H₂
CO₂ + C → 2CO

Question 5.
Find whether the following statements are right or wrong.
a. All atoms of the same element show the same properties.
b. The atoms present in a compound are different.
c. Elements are pure substances.
d. Hydrogen is a monatomic molecule
Answer:
a. True
b. True
c. True
d. False

Question 6.
Take iron powder and sulphur powder in their mass ratio of 7:4 in a china dish. Heat it strongly for some time.
a. Try to separate the iron powder using a magnet. What can be observed? What is the reason?
b. Add a .little dilute hydrochloric acid to the above product; observe the changes taking place, and write them down.
c. Write down the inferences you have arrived at from these observations.
Answer:
Sulphur is reacted with iron powder to form ferrous sulphide therefore it cannot be separated.
There will be the smell of rotten egg because hydrogen sulphide is produced.

Question 7.
Make and exhibit the models of the following molecules using ball and sticks, different fruits and splints.
a. Water (H₂O)
b. Ammonia (NH₃)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
Answer:
Do your self

Question 8.
Haven’t you understood how symbols are assigned to elements? Given below are some elements that received symbols from their Latin names. Complete the table with the help of the Periodic Table.

Element	Latin name	Symbol
Silver	……………..	……………..
……………..	Hydrargium	……………..
Tin	……………..	……………..
……………..	……………..	Pb
Antimony	……………..	……………..
……………..	Aurum	……………..

Answer:

Element	Latin name	Svmbol
Silver	Arguntum	Ag
Mercury	Hydrargium	Hg
Tin	Stannum	Sn
Lead	Plumbum	Pb
Antimony	Stibium	Sb
Gold	Aurum	Au

Basic Constituents of Matter Additional Questions and Answers

Question 1.
Choose the correct symbol of the elements given below from the box.
(Calcium, carbon, sodium, potassium, Iron, silver)
(S, C, Na, Ca, Si, Ag, In, P, K, Fe)
Answer:
Calcium-Ca,
Carbon-C,
Sodium- Na,
Potassium- K,
Iron- Fe,
Silver- Ag

Question 2.
The chemical formula of calcium bisulphate is Ca (HSO₄)₂. Write down the number of atoms of each element present in this molecule.
Answer:
Ca=1, H=2, S=2, 0=8

Question 3.
Na. This is the symbol of an element. Which statement given below is wrong. Justify?

Symbol of Nitrogen
Indicates one atom.

Answer:
Symbol of Nitrogen.
The symbol that represent the element Nitrogen is ‘N’. ‘Na’ is the symbol of sodium.

Question 4.
We know about elements and compounds. Find out the odd one from the following. Give reason?
a. (A) Hydrogen (B) Oxygen (C)Nitrogen (D) Carbon dioxide
b. (A) Sodium (Na), (B) Pottasium (K) (C) Magnesium (Mg), (D) Iron (Fe)
c. Write the molecular formula of carbon dioxide?
d. What is the Latin name of sodium?
Answer:
a. Carbon dioxide – Compound, others are elements.
b. Magnesium – Others take symbols from their Latin name.
c. CO₂
d. Natrium

Question 5.
Arrange the following from smallest to largest.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 5 Basic Constituents of Matter 1

Answer:
Atom → Molecule → Small particle → Substances

Question 6.
Represent the basis of nomenclature of elements given in the following table.
a.

Element	Basis of nomenclature
Francium Rutherfordium

b. Which element symbol is Cl?
c. What are the constitute elements containing of CaF₂ molecule?
Answer:
a. Name of country
b. Name of scientist
c. Chlorine
d. Calcium, -Fluorine

Question 7.
Symbol of Nitrogen is ‘N’.
a. What represent symbols of 2N and N₂?
b. How many atoms are in 2NH₃?
c. How can represent 5 water molecules?
Answer:
a. 2N represent 2 atoms of Nftrogen. N₂ represent Nitrogen molecules can be formed by combining two atoms of Nitrogen.
b. 8 atoms
c. 5 H₂O

Question 8.
Evaluate the given chemical equation and answer the following questions.
SO₂ + O₂ – SO₃
a. Which are the reactants in the chemical reaction?
b. What is the product of the chemical reaction?
c. Write the balanced chemical equation?
Answer:
a. SO₂, O₂
b. SO₃
c. 2SO₂ + O₂ → 2SO₃

Question 9.
Find the monoatomic molecule from given elements.
a. (A) Hydrogen (B) Oxygen (C)Nitrogen (D) Helium
b. Write the symbols of Hydrogen and Helium?
c. Write chemical formula of Oxygen molecule?
d. What is the Latin name of sodium?
Answer:
a. (D) Helium
b. Hydrogen (H), Helium (He)
c. O
d. Natrium

Question 10.
Select the compounds from the following:
HCl, CO₂, O₂, NaCl, H₂, H₂O, Na, HCl
Answer:
NaCl, CO₂, H₂O, HCl

Question 11.
Form a table of elements which are related to the name of scientists from the following Americum, Mentelivium, Titanium, Rutherfordium, Rubedium, Einstenium, Europeum, Bohrium
Answer:
Mentelivium, Rutherfordium, Einstenium, Bohrium

Question 12.
Examine the following symbols and write the name of elements Ca, Cr, Cl, Be, Na, Fe, Cu
Answer:
Calcium, Chromium, Chlorine, Berelium, Sodium, Iron, Copper

Question 13.
Fill in Blanks suitably
Element – Atom,
Compound – …………
Answer:
Element – Atom,
Compound – Molecule

Question 14.
Complete the following table

Element	Basis of Naming
Americum
Francium
Rubedium
Plutonium
Titanium
Mentalivium
Rutherfordium

Answer:

Element	Basis of Naming
Americum	America- continent
Francium	France – country
Rubedium	Ruby – Colour
Plutonium	Pluto- planet
Titanium	Titan – God
Mentalivium	Mentelief – Scientist
Rutherfordium	Rutherford – Scientist

Question 15.
Examine the following molecules and tabulate the atoms contained in them.
1. Carbon dioxide
2. hydrogen chloride
3. Water
4. sugar
5. Mercuric oxide
Answer:
1. C, O;
2. H, Cl
3- H, O;
4. H, C, O;
5- Hg, 0

Question 16.
Complete the tables

Element	English name	Symbol
Carbon	Carbon	C
Oxygen	………….	………….
………….	Nitrogen	………….
………….	………….	H

Answer:

Element	English name	Symbol
Carbon	Carbon	C
Oxygen	Oxygen	0
Nitrogen	Nitrogen	N
Hydrogen	Hydrogen	H

Question 17.
Tabulate the number of atoms and molecules
5H₂, 2H₂ O, 3ZnCl₂, CO₂Answer:

Compound	Number of atoms	Number of molecules
5H₂	10	5
2H₂0	6	2
3ZnCl₂	9	3
co₂	3	1
co₂	3	1

Question 18.
Write the diatomic molecules from the following
H₂, Na, C, N₂, P₄, O₂
Answer:
H₂, N₂, O₂

Question 19.
Complete the table

Monatomic	Diatomic	Polyatomic
………..	H₂	P₄
Na	………..	………..
He	………..	………..

Answer:

Monatomic	Diatomic	Polyatomic
C	H₂	P₄
Na	O₂	O₃
Ne	N₂	S₈
He	Cl₂

Question 20.
How many hydrogen and sulphur atoms are there in H₂SO₄ molecule?
Answer:
Hydrogen-2 Sulphar-1

Question 21.
Complete the table.

Substance	No. Of molecules	No. of atoms
P₄
0₂
5Na
2Na
6Cl₂
2He

Answer:

Substance	No. Of molecules	No. of atoms
P₄	1	4
0₂	1	2
5Na	5	5
2Na	2	2
6Cl₂	6	12
2He	2	2

Question 22.
Find the total number of atoms in the following molecules
a. CO
b. 5 H₂O
c. 7 NH₃
d. 2 Cl₂ H₂₂ O₁₁
e. 3 NaCl
f. ZnCl₂
Answer:
a. CO – 2
b. 5 H₂O -15;
c. 7 NH₃ – 28;
d. 2C₂ H₂₂ O₁₁ – 90
e. 3 NaCl – 6
f. ZnCl₂ – 3

Question 23.
The chemical formula of water is H₂O. Write the ratio of hydrogen and oxygen atoms in water molecule.
Answer:
Ratio of hydrogen and oxygen is 2:1

Question 24.
Tabulate the reactants and products in the following chemical equation.
1. 2H₂ + O₂ → 2H₂O
2. Na + Cl₂ → 2NaCl
3. C + O₂ → CO₂
4. H₂ +Cl₂ → 2HCl
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 5 Basic Constituents of Matter 2

Question 25.
Balance the following equations which are not balanced.
C + O₂ → CO₂
H₂ + N₂ → NH₃
Mg + O₂ → Mgo
2H₂ + O₂ → 2H₂O
Answer:
3 H₂ + N₂ → 2NH₃
2Mg+O₂ → 2MgO

Question 26.
How did the symbol system introduced by Berzelius, help the study of chemistry?
Answer:

The study of chemistry became simple and easy.
Writing chemical equations representing chemical reactions became very easy.
The number of atoms in a molecule can be found out from the symbol of the molecule of the element.

Question 27.
Balance the equation
1. NaOH + H₂ SO₄ → Na₂ SO₄ + H₂O
2. H₂O₂ → H₂O + O₂
3. CH₄ + O₂ → CO₂ + H₂O
4. NaOH + HCl → NaCl + H₂O
5. H₂O + CO₂ → C₆H₁₂O₆ + O₂
Answer:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
2H₂ O₂ → 2HO₂ + O₂
CH₄ +O₂ → CO₂ + 2H₂O
2NaOH + 2HCl → 2NaCl + 2H₂O
6 H₂O + 6CO₂ → C₆ H₁₂ + 6 O₂

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

November 22, 2023November 21, 2023 by HSS Live

You can Download Force Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

When an object is pushed or pulled, a force is being applied on it. The unit of force is newton. It is indicated by the letter N.

Force is that which changes or tends to change the shape, size, volume, state of rest or state of motion of a body.

There are various types of forces such as muscular force, magnetic force, electric force, gravitational force, frictional force etc.

Contact force and non contact force

The force applied by the contact between objects is contact force. The force applied on an object without contact on it is non contact force.
Ex:pushing a trolley- contact force, Magnet attracting a nail-non contact force

Frictional force:

When a ball is rolling on the floor after some time it stops. This is because of the frictional force between the ball and floor. On the surface of the objects which are on contact with each other there are many mounts and pits get interlocked.

When we try to move a body by applying a force parallel to their surfaces an opposing force is experienced.
When a body moves or tends to move on the surface of another body, a force is experienced parallel to the surface which opposes the relative motion between them. This is friction.

Rolling friction and sliding friction

When a body rolls over the surface of another body, the force of friction which originates is rolling friction. When a body slides on the surface of another body the force of friction which originates is sliding friction.

Rolling friction is less than sliding friction. Striking a match stick on the side of a match box, wear and tear of machines, ability to hold objects, walking etc are advantages of the friction and treading of tyres of vehicles, wearing out of tyres, fuel loss are disadvantages of friction.

The materials like oil, grease which are used to reduce friction are the lubricants. Graphite is a solid lubricant.

Thrust and Pressure

The total normal force experienced by a ‘ surface is thrust. Thrust per unit area is pressure.
Pressure = \(\frac{Thrust}{area}\)

Unit of pressure is N/m2. This is known as pascal. The force acting on a larger area exerts a smaller pressure, while a force acting on a smaller area exerts a large pressure.

Liquid Pressure

The pressure is experienced in liquids also. As the height of liquid column increases the pressure exerted by it increases.

The pressure exerted by a liquid column increases with increase in height. The thrust acting per unit area by a liquid is liquid pressure.

Liquids exert force on all sides of the container in which they are taken.

If the height of the liquid column is Ti’, density of liquid ‘d’ and acceleration due to gravity ‘g’, then liquid pressure is P = hdg

Atmospheric pressure

Atmospheric air also can exert pressure. An envelope of air surrounds the earth. This is earth’s atmosphere. The density of atmospheric air near the surface of earth is greater and it decreases when we go up. The weight of air column over the unit area of earth’s surface is atmospheric pressure. One atmospheric pressure is the weight of mercury column of 0.76 m height and unit area (1m²). This is standard atmospheric pressure .

The unit of atmospheric pressure is* bar. The instrument used to measure atmospheric pressure is Barometre

Force Textbook Questions and Answers

Question 1.
Classify the following situations into contact and noncontact forces.
a. Applying break in a bicycle.
b. A mango falling from a mango tree.
c. The earth revolving around the sun.
d. The speed of a hall rolling on ground is reduced.
Answer:
Contact force
a. Applying break in a bicycle.
d. The speed of a ball rolling on ground is reduced
Non contact force
b. A mango falling from a mango tree.
c. The earth revolving around the sun.

Question 2.
State reason
a. We are able to walk on the ground without slipping.
b. It is easy to cut vegetables using a sharp knife.
c. The number of tyres is more for goods vehicles.
d. The moving parts of machines experience wear and tear.
Answer:
a. Because of the friction between ground and feet
b. When the area decreases pressure increases. As the area of the edge of the knife is veiy less, it is easy to cut
c. When the area of surface increases pressure decreases, so when more wheels are used it won’t depressed in soil.
d. Because of the friction between the area of contacts.

Question 3.
Match the following

A	B	C
Atmospheric pressure	pascal	coconut oil
Lubricant	attration	barometer
Coconut falling down	bar	thermometer
Magnet	friction	noncontact
	gravitation	repulsion

Answer:

Atmospheric pressure	bar	barometre
Lubricant	friction	Coconut oil
Coconut falling down	Gravitational force	Non contact
Magnet	attraction	repulsion

Question 4.
Bubbles rising from the bottom of the water filled in a bottle are depicted in the figure. Which is the correct figure? Justify your answer.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force 1

Answer:
Fig.c. There will be more pressure in the lower side of the liquid. The pressure decreases when it comes upward.

Question 5.
A toy car of about 50 g placed on a polished table with threads attached to it carrying two pans passed through pulleys is depicted.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force 2

(a) What do you observe if too g each is placed on both pans?
(b) What do you observe if too g is placed on pan A and 200 g in pan B
(c) Justify your answers.
Answer:
a. The car will be stationary because the force excreted from both sides are equal.
b. The pan of weight 200 will pull with more force and it moves that direction.
c. In the first phase same force is excerted. But in second phase force is exerted to one direction.

Force Additional Questions and Answers

Question 1.
Tabulate the following as contact force and non contact force.
1. pushing a trolley
2. falling coconut from coconut tree
3. magnet attract iron nail
4. Drowing water from the well
5. pushing the car
6. earth is moving around the sun
Answer:
contact force – 1,4,5
non contact force – 2,3,6

Question 2.
Ramu tried to push a round stone and rectangle stone of same weight. Which is easy to push.
Answer:
It is easy to push round stone because rolling friction is less than sliding friction.
3. Which are the following occation friction is beneficial and non beneficial
1. striking a match stick on the side of a match box
2. wear and tear of machines.
3. ability to hold objects
4. wearing out of tyres
5. walking 6. loss of fuel
Answer:
Beneficial : 1, 3, 5
non beneficial : 2, 4, 6

Question 4.
Why does the design of ship and aeroplane are streamlining?
Answer:
To reduce friction while moving through air and water.

Question 5.
Write one solid lubricant and liquid lubricant
Answer:
Solid-graphite
Liquid-oil

Question 6.
Fill up suitably.
Force – newton
Pressure – ………..
Answer:
Pressure – pascal

Question 7.
Write the reason
1. Constructing a knife with sharp edge
2. Constructing the building with wide basement
Answer:
1. When the area of contact decreases the pressure increases
2. When area increases the chance to depress the base in soil is very less.

Question 8.
Given figures, one can filled with water and other can filled with kerosine.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force 3
Which balloon is filled with water
Answer:
fig 2. When density of liquid increases the pressure increases, water is denser than kerosene.

Question 9.
What are the factors effecting liquid pressure? Write the mathematical equation.
Answer:
The height of the liquid column is ‘h’, density of liquid ‘d’ and acceleration due to gravity ‘g’ are the factors. The equation is liquid pressure P = hdg

Question 10.
As mountaineers climb higher altitude there is a possibility of nasal bleeding. Why?
Answer:
Going higher altitude the atmospheric pressure decreases. So blood veins are broken and blood is oozing out.

Question 11.
What is standard atmospheric pressure
Answer:
Atmospheric pressure at the sea level is equal to the pressure exerted by 0.76m of mercury column. This is standard atmospheric pressure.

Question 12.
When a hole is formed at the bottom of the water bottle, Anu closed the lid of the bottle tightly. The flow of water stops. Explain
Answer:
When the lid is closed tightly, the effect of atmospheric pressure is not experienced on the surface of the water in the bottle. Hence water cannot flow out from the bottle.

Question 13.
Mercury remains at a height of 76cm in a barometer. Raju made a hole at the top end of the tube. What will you observe? How can it be explained?
Answer:
Mercury will flow downwards when a hole is made at the top part of the tube. The atmospheric pressure exerted on the mercury in the vessels supports the mercury level in the tube. When a hole is made at the top, air enters the tube and the pressure is distributed throughout uniformly and hence the mercury level comes down.

Question 14.
A sharp nail pierces more easily through wood than a blunt nail. What is the reason?
Answer:
The surface area of the sharp tip of a nail is very small. Hence the pressure that it can exert on the wood is very great. That is why it pierces easily. For a blunt nail, the surface area of the tip is greater. Hence it can exert only a small amount of pressure.

Question 15.
What is meant by limiting friction.
Answer:
Limiting friction is the force of static friction acting just before the body starts moving.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life’s Mysteries in Little Chambers

November 21, 2023November 20, 2023 by HSS Live

You can Download life’s Mysteries in Little Chambers Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 1 life’s Mysteries in Little Chambers

life’s Mysteries in Little Chambers Questions and Answers

Microscope

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 1

The Microscope is a device used to magnify objects. The main parts of a microscope are Eye piece, Knobs, Objective Lens, Stage and clip, Diaphragm, Mirror etc.

The microscopes in which more than one lens is used are called compound microscopes: In a microscope there are two planes for the regulation of light. A plane mirror to reflect the sunlight and a concave mirror to reflect artificial light. The lens of the condenser that is fitted on the lower side of the stage focuses light on the material to be observed. Diaphragm regulates the intensity of light.

Indicators (Text Book Page No:9)

Question 1.
Why is a mirror fixed in a microscope?
Answer:
The Mirror is fixed to focus light on the objects to be observed.

Question 2.
Why do we use a slide made of glass to place the material to be observed?
Answer:
A transparent surface is required to pass light of the material to be observed. So glass slide is used to place it.

Magnification Power of Microscope

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 2

Magnification Power is marked as tox and 45X in the objective lens and eyepiece. The magnification power of a microscope is the product of the numbers seen in the objective lens and eyepiece.
Preparation of Observation Material

Following are the steps in the preparation of a slide to observe plant cells.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 3

Put the observable material in a petridish containing water to prevent dehydration.
Take thin cross sections of the material
Put the sections to a watch glass con-taining water.
Put the thinnest and complete sections to a watch glass containing stain.
Add one or two drops of glycerin to the slide to avoid drying up of the material.
Wash the stained section with clean water and place it in the glycerine on the slide.
Put a cover glass over the material.
Wipe out the excess glycerin using tissue paper.

Discovering of Cell

Robert Hooke discovered cells for the first time. The branch of science that deals with cells is called cell biology. The re markable findings that paved way for the development of cytology and the concerned scientists are indicated below.

Robert Brown – Discovered Nucleus (1831)
M. J. Schleiden – Discovered Plant Cells (1838)
Theodor Schwann- Discovered Animal Cells (1839)
Rudolf Virchow – Inferred that new cells arise only from existing cells (1858)

Cell Theory

M.J. Schleiden and Theodor Schwann are the founders of cell theory. Mainly two concepts are put forward by cell Theory.

All living body is made up of cells.
Cells are the structural and functional units of organisms.

Question 3.
What are the functions performed by cells?
Functions of Cells

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 11

Resist foreign particles,
Synthesis biomolecules.
Release energy from nutrients

Structure of cell is suitable to perform various biological functions. There are different parts inside the cell for this purpose.

Nucleus – Centre of the cell.
Ribosome – Helps in protein synthesis
Chloroplast – Helps in photo synthesis
Mitochondrion – Centre of energy production, storage and distribution.
Vacuole – Stores water, salts, excretory materials etc.
Golgi complex – Stores enzymes, hormones, mucous etc in small vesicles.

Question 4.
Complete the illustration given below related to different kinds of organelles.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 13

Nucleus

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 4

Innumerable life processes take place within the cell. The biomolecules that control and coordinate these processes are synthesized under the control of the genes in the chromatin network of the nucleus.

Major parts of the nucleus are nucleoplasm, nuclear membrane, chromatin reticulum, nuclear pore, nucleolus etc.

Nucleoplasm is the fluid part of the nucleus. Nuclear membrane is the double-layered membrane that surrounds the nucleus. The minute pores in the nuclear membrane are called nuclear pores. The spherical nucleolus seen in the nucleus play a major role in the synthesis of ribosome. The chromatin reticulum seen as a network in the nucleoplasm encloses genes.

Prokaryotes and Eukaryotes

There is no definite nucleus in the cell of bacteria, cyanobacteria and mycoplasma etc. There are called prokaryotes. But in the cells of amoeba, plants, and animals a well defined nucleus is present. These organisms are called eukaryotes.

Biological Membranes

Cells are covered by membranes. They regulate the exchange of materials between cells and between cell organelles and cytoplasm. These membranes are called selectively permeable as they pass only the essential materials across them.

Indicators (Text Book Page No: 17)

Question 5.
Plastids in the plant cells.
Answer:
Plastids are the cell organelles that are found only in plant cells. They are of 3 types – Chromoplast, Chloroplast and Leucoplast.

Question 6.
Pigments in the chromoplasts.
Answer:
The pigments in Chromoplast are Xanthophyll (yellow), Carotene (yellowish Orange), Anthocyanin (Red, Purple) etc.

Question 7.
Chromoplasts and sunlight.
Answer:
Chloroplast performs photosynthesis in the presence of sunlight. As the fruits ripen, the chloroplast in them changes to chromoplasts.

Question 8.
Change of colour and taste in fruits
Answer:
In the storage cells, leucoplasts are more. But as the fruit ripens the chlo- roplasts converted to chromoplasts. Starch changes to sugars. This is the reason behind the change in colour and taste when the fruit ripens.

life’s Mysteries in Little Chambers Let us complete the illustration (Text Book Page No: 18)

Question 1.
Compare the plant cell and the animal cell and complete the illustration given below.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 5

Answer:

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 14

life’s Mysteries in Little Chambers Let us assess (Text Book Page No: 19)

Question 1.
Complete the concept map related to the structure of animal cells.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 6

Answer:

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 15

Question 2.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 9

(i) Identify the parts A,B, C,D,E marked in the plant cell.
Answer:
A – Chloroplast
B – Mitochondrion
C – Ribosome
D – Vacuole
E – Nucleus

(ii) Write the function of the parts denoted by the letters given below.
Answer:
A – Photosynthesis
B – Production, Storage and distribution of energy
C – Protein synthesis

(iii) In which part of the cell are genes found?
Answer:
Genes are seen in the chromatin reticulum in the nucleus.

life’s Mysteries in Little Chambers Additional Questions and Answers

Question 1.
Find the odd one in each group. Also write the common characteristics of the others.
a. Bacteria, Cyanobacteria, Amoeba, Mycoplasma,
b. Golgi complex, Ribosome, Lysosome, Mitochondrion
c. Mitochondrion, Cell membrane, Ribosome, Golgi complex
d. Xanthophyll, Anthocyanin, Tonoplast, Carotene
Answer:
a. Amoeba; The others are prokary otes.
b. Lysosome; The others are present in both plant and animal cell.
c. Cell Membrane; Others are cell organelles

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. M.J. Schleiden- ………… ; Theodor Schwann- Animal cell
b. Power house – Mitochondrion; Skeleton of cell- …………
c. Bacteria – ………… ; Amoeba – Eukaryote
d. Robert Hooke – Cell; ………… – Nucleus
e. Yellowish orange colour – Ca-rotene; Yellow colour – …………
f. Powerhouse of the cell – Mitochondrion; Transport of materials – …………
Answer:
a. Plant Cell
b. Endoplasmic reticulum
c. Prokaryote
d. Robert Brown
e. Xanthophyll
f. Endoplasmic reticulum

Question 3.
Copy the given figure, then identify and mark those parts, their features are given.|
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 16
a. Cytoskeleton
b. Controll various process inside a cell
c. Power house of cell
Answer:
a. Endoplasmic reticulum – understand, marking
b. Nucleus – understand, marking
c. Mitochondrion – understand, marking

Question 4.
It was Rudolf Virchow who dis-covered one of the prominent characteristics of cell. Analyse the statement.
Answer:
He found the process cell division. New cells are formed only from existing cell.

Question 5.
Which components of plant cell are not found in animal cell?
Answer:
Centrosome, Lysosome.

Question 6.
Some components of Animal cell are also found in the plant cell. List out them.
Answer:
Cell wall, large vacuole, plastids.

Question 7.
What change occur when a mango ripens? What are the causes?
Answer:
Colour changes.
Chloroplast is changed into chromoplast.
The sour taste is changed sweet taste. Starch is changed into sugar.

Question 8.
Choose the parts of microscope from the following terms.

Diaphragm, test tube, objective lens, eye piece, mirror, hammer, watch glass, cover glass, knobs, stage, clip

Answer:
Diaphragm, objective lens, eye piece, mirror, Knobs, Stage, Clip

Question 9.
Make a table showing the different parts of microscope and their function.
Answer:
Eye piece – helps to magnify objects; helps to observe objects
Knobs – to adjust the position of lens
Objective lens – helps to magnify objects
Stage & Clip – object is placed on the stage and the slide is fixed using clips
Diaphragm – helps to reduce the intensity of light Mirror – reflects sunlight

Question 10.
Why does mirror is fixed in the microscope?
Answer:
To allow sunlight on to the material observed.

Question 11.
Why a glass slide is used to place the material to be observed?
Answer:
A transparent surface is needed to fall sunlight reflected by the mirror on the material to be observed.

Question 12.
Why plane mirror and concave mirror are used in the microscope?
Answer:
Plane mirror reflects sunlight.
Concave mirror reflects artificial light.

Question 13.
Ananthu noted the markings 10x and 45x in the objective lens of the microscope. He also noted the markings on the eyepiece too. But he couldn’t understand this? Can you help him?
Answer:
These markings (10x, 45x) denote their magnification power.

Question 14.
How can we find out the magni-fication power of microscope?
Answer:
The product of the numbers seen in the objective lens and eyepiece de-notes its magnification power.

Question 15.
What is the difference between simple microscope and compound microscope?
Answer:
In simple microscope there is only one lens. But in compound micro-scope there are more than one lens.

Question 16.
Why the sections of plantain roots are put in the water taken in the petridish?
Answer:
To avoid dehydration of roots.

Question 17.
Anagha and Priya placed the sections of plantain root on glass slides. After sometime the materials on Anagha’s slide dried up. But Priya’s slide was suitable for observation. What may he the reason for this?
Answer:
To avoid drying up it is needed to add a few drops of glycerine on slide. Priya has done it.

Question 18.
Write procedure for observation of the section of plantain root.
Aim – to observe the cells in the roots of plantain.
Materials required -Plantain root, Petridish, Blade, Brush, Glass slide, Cover slip, Stain, Glycerine, Water, Microscope.
Procedure: –
Answer:

Take thin sections of the roots of plantain. Put it in the water in a petridish.
Transfer the sections to the water taken in a watch glass using a brush.
Put the thinnest and complete sections in a watch glass containing stain.
Add a few drops of glycerine on slide to avoid dehydration.
Place the stained section on the slide after putting it in fresh water.
Cover it using a cover glass to avoid air bubble.
Wipe out the excess glycerine using a tissue paper.

Question 19.
Match the Scientists and their contributions.
Rudolf Virchow – a. Found out that plant body is made up of cells.
M. J. Schleiden – b. Observed cells for the first time.
Robert Hooke – c. Found out that animal body is formed of cells.
Robert Brown – d. Formulated the inference that new cells form only from the existing ones.
Theodor Schwann – e.Invented nucleus.
Answer:
Rudolf Virchow – b
M. J. Schleiden – a
Robert Hooke – b
Robert Brown – e
Theodor Schwann – c

Question 20.
State Cell theory
Answer:
All living body is formed of cells. Cells are the structural and functional units of organisms.

Question 21.
Who are the founders of Cell Theory?
Answer:
MJ. Schleiden and Theodor Schwann.

Question 22.
“Our body is composed of millions of cells. Why these much of cells?” – Ananthu
‘They have various functions to perform’ – Bhadra
List out the functions of cells in the body.
Answer:

Defend foreign particles.
Releases energy from the cells.
Synthesis biomolecules.

Question 23.
Classify the given cell organelles under suitable headings. Common terms can be included in both groups.
Ribosome, Nucleus, Cell membrane,Centrosome, Chloroplast, Vacuole, Golgi Complex, Mitochondrion, Lysosome
Answer:

Plant cell	Animal cell
Cell wall	Centrosome
Chloroplast	Lysosome
Vacuole	Ribosome
Ribosome	Nucleus
Nucleus	Golgi complex
Golgi Complex	Mitochondrion
Mitochondrion	Cell membrane
Cell membrane

Question 24.
Identify the cell organelles that perform the functions mentioned below.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 8

Encloses enzymes, hormones etc in small vesicles
Gives strength and shape to cells.
Power house of the cell.

Answer:

b. Golgi complex
c. Endoplasmic reticulum
a. Mitochondrion

Question 25.
Complete the labelling of the given diagram of plant cell suitably.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 18
Answer:
a. Nucleus
b. Centre of protein synthesis in the cell
c. Mitochondrion
d. Power house of the cell

Question 26.
Complete the illustration given below.
Vacuole, Mitochondrion, Endoplasmic Reticulum, Golgi complex
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 17

Answer:
Vacuole – Stores water, salts, and waste materials.

Mitochondrion – Power house of the cell, helps in the production and storage of energy.

Endoplasmic Reticulum – Pathway inside the cell. Provide rigidity and shape to cells.

Golgi Complex – Encloses enzymes, hormones & mucus in small vesicles.

Question 27.
While discussing about the cell Kunchu has a doubt” Are Protoplasm and Cytoplasm one and the same?” What is your response?
Answer:
Protoplasm constitutes all materials within the cell membrane. Cytoplasm is the part of protoplasm except the nucleus.

Question 28.
What is the difference between simple microscopes and electron microscopes?
Answer:
Simple Microscope

Magnifies an object 2000 times its actual size
Light source is used to see objects Electron Microscope
Magnifies objects million times
Electrons are utilized instead of light source.

Question 29.
Complete the table showing dif-ferent parts of nucleus and their functions.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 19

Question 30.
“Can organisms live without a definite nucleus”? I can’t believe it. Have you noticed Gopika’s doubt? How will you clarity this doubt?
Answer:
There are certain organisms without a definite nucleus. In bacteria, cyanobacteria and mycoplasma, there is no nucleus. They are called prokaryotes.

But in the cells of Amoeba, animals and plants there is a well defined nucleus covered with membrane. These are known as eukaryotes.

Question 31.
Complete the given flow chart.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 20
Answer:
a – Chloroplast
b – Leucoplast
c – Imparts colour to fruits and flowers.

Question 32.
Why do the cell membranes called selectively permeable membranes?
Answer:
Hence they allow only the essential materials to pass through.

Question 33.
How does the colour of fruits change when they ripen?
Answer:
As fruits ripens, the chloroplasts get converted to chromoplasts. Thus colour changes.

Question 34.
Describe the function of Centrosomes.
Answer:
Centrosomes is the cell organelle that is found only in animal cells. The centrioles that play a major role in cell division. It helps to the formation of spindle fibers.

Question 35.
Find out statements which are relate to animal cells from the following statements.
• seen cell wall.
• seen cell membrane.
• seen mitochondrion.
• seen vacuole and chloroplast.
Answer:
• seen cell membrane.
• seen mitochondrion.

Question 36.
How spindle fibres are formed in plant cells and fungus cells?
Answer:
In plant cells and fungus cells spindle fibres are formed through microtubule organising centre.

Question 37.
Why are biomembranes known as selectively permeable membrane?
Answer:
The exchange of materials between cells and between cell organelles and cytoplasm occurs through biomembranes. These membranes do not allow all but some materials to pass. Hence they are selectively permeable membrane.

Question 38.
The discovery of electron microscope enables the detailed study on the cells easy. How?
Answer:
The electron microscope can magnify an object into millions of time of the original size. So it helps to understand more about the micro structures of the cell.

Question 39.
What are unicellular organisms? Write examples for this?
Answer:
Some organisms have only a single cell in their body. Such organisms are called unicellular organisms,
eg: Amoeba, Paramoecium and Euglena.

Question 40.
What are multicellular organisms?
Answer:
There are a number of cells in the body of these creatures. They are therefore know as multicejlular organisms.

Question 41.
What is cell wall?
Answer:
Cell wall is the covering seen outer to the plasma membrane in plants. It is composed of cellulose. It has numerous pores. Movement of molecules is possible through these pores.

Question 42.
How do animal cells differ from the plant cells in structure?
Answer:
Plant cell and animal cell are often structurally same. But the plant cell is surrounded by a rigid protective cell wall formed of cellulose. Cell wall is the major factor by which we can identify the animal cells from plant cells. In plant cell large vacuole and chloroplasts are seen.

Question 43.
Draw the diagrammatic repres-entation of animal cells and plant cells and illustrate the parts?
Answer:
Animal cell

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 7

Plant cell

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 12

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

November 21, 2023November 20, 2023 by HSS Live

You can Download Measurements and Units Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

In olden days different scales were used for measuring length in different countries. Vaara, One forearm (muzham) etc are some of such units. These units were not accurate. For solving this problem a standard scale of a definite length has been recognised world widw for measuring length. This standard scale is the unut of length. The basic unit of length is metre (m).

Smaller units of length

Examine the metre scale. You can see small measurements on it. These are centimetre (cm) and millimetre(mm)
1m = 1oocm 1cm=1omm
The smallest length that can be accurately measured using a metre scale is its least count.

Thickness of paper

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 1

We cannot measure the thickness of the paper even with these small units. Take too sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 2

Similarly to find out the thickness of thin wire roll it on a pencil and mea-sure the length of ten rolls Then divide it by 10.

Length of a curved line

To measure the length of the curved line put a thread on it and measure its length using a meter scale

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 3

Distance of a sphere

A ball kept in between two wooden blocks, whose diameter is to be measured. Then we will get its diameter.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 4

Various measurements of length

The basic unit of length is metre. Smaller units like centimetre (cm), millimetre (mm), micrometre (micron – ™ nano-metre (nm) are also used in certain situa-tions for convenience. We use a bigger unit like kilometre (km) to measure the distance between two places. Units like astronomical unit (AU), light year (ly) are also used to measuring distance to planets or stars.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 7

Mass

Mass is the measure of quantity of matter contained in it. The basic unit of mass is kilogram(kg).

Except kilogram, milligram(mg), gram(g), quintal, tonne are also the units of mass.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 5

Time
The symbol is s.
60s = iminute; 6ominute = 1 hour One second is 1/ 86400 part of an average solar day.

Fundamental units

There are some quantities which are not related to one another and cannot be expressed using other quantities. Such quantities are fundamental quantities. The units of the fundamental quantities are the Fundamental Units. The system based on these fundamental units is the | International System of Units. Its short form is SI Units.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 6

Derived unit

Units which are expressed in terms of fundamental units or those units which are dependent on fundamental units are derived units.
Ex. Area = length x breadth

Unit of area = unit of length x unit of breadth = m xm = m².

Salient features of SI units

Unified units
Internationally accepted
Adequate to express all physical quantities.

Density and volume

The space occupied by a body is referred to as its volume.
The unit of volume is cubic metre (m³)

Mass per unit volume of substance is its density. Density = mass/volume Unit of density = unit of mass/unit of volume = kg/m³

The rules to be followed while writing units

We have to observe certain rules according to international standards while using these units and their symbols. They are given below:

The symbols of units are normally written using small letters in the English alphabet, eg. m (metre), s (second), kg (kilogram)
But there are certain occasions on which capital letters of the English alphabet are used as symbols. The units named after persons are written like this.
While writing the names of units never use capital letters, eg. kelvin (correct) Kelvin (wrong) newton (correct) Newton (wrong)
Never use the plural form for symbols, eg. 10 kg (correct) 10 kgs (wrong) 75 cm (correct) 75 cms (wrong)
Never use full stop or comma after a
symbol except at the end of a sentence. eg. 75 cm is the length of a table, (correct) 75 cm. is the length of a table, (wrong),
While writing derived units a slash (/) is used to denote division. But never use more than one slash in one derived unit. eg. m/s1 (correct) m/s/ s (wrong)
When a derived unit is expressed as the product of other units use a dot or a space between them. eg. N.m or Nm
8. Do not mix the name of a unit with the symbol.
While writing units along with a value, there must be single space between them.
Never use more than one unit to express a physical quantity.

Measurements and Units Textbook Questions and Answers

Question 1.
Write down the following units in the ascending order of their values
a) mm nm cm µ m.
b) 1 m 1 cm 1 km 1 mm
Answer:
a. nm → µ m → mm → cm
b. 1 mm → 1 cm → 1 m → 1 km

Question 2.
Which of the following units does not belong to the group?
a) kg mg g mm
b) m mm km mg
Answer:
a. mm
b. mg

Question 3.
Imagine that the distance from school to your friend’s house is 2250 m. State this in kilometres.
Answer:
2\(\frac{1}{4}\) km or 2.25 km

Question 4.
Convert the following units, into SI units without changing their values.
a. 3500 g
b. 2.5 km
c. 2 h
Answer:
a. 3.5 kg
b. 2500 m
c. 7200 s

Measurements and Units Additional Questions and Answers

Question 1.
Take a small quantity of solution A and solution B in a jar. Density of A is 2 kg/m³ and density of B is 5 kg/m³. Which of these flot above the jar ? Give reasion.
Answer:
Solution A
Density of A is less than the density of B. So solution with less density floats above the solution that has high density.

Question 2.
Find out the given word pair relation; and fill in the blanks.
a. Length : m ; Volume :………
b. Displacement : m ; Acceleration:…………….
Answer:
a. m³
b. m/s²

Question 3.
a. What is the basic unit of time?
b. What do you mean by one Solar day?
c. How much time earth will take to revolve once around the sun?
Answer:
a. Second
b. A day or a solar day is the time period from one noon to the next noon.
c. One year (365 days)

Question 4.
Find out odd one. Justify your answer?
a. meter, kilogram, kilogram/ meter³, second
b. Length, volume, mass, time
c. Volume, mass, density accelaration.
Ans:
a.kilogram/meter³ – meter, kilogram, second are basic units, kilogram/meter³ is unit of volume.
b. Volume – Volume is a derived quantity and others are basic quatities.
c. Mass – Mass is a basic quantity and others are derived quantity.

Question 5.
What is meant by least count of a metre scale?
Answer:
The smallest length that can be accurately measured using a metre scale is its least count. The least count is 1mm when metre scale is used.

Question 6.
Mention a method to measure the thickness of a paper sheet.
Answer:
Take 100 sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Question 7.
Using following objects how can you measure the diameter of a ball?
hall, two wooden blocks, metre scale
Answer:
The ball is kept in between two wooden blocks as shown in the figure whose diameter is to be measured.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 4

Question 8.
Fill in the blanks
a. 1 nm = ….. m
b. 1 km = …….cm
c. 1000g = …….kg
d. one light year = …… km
e. 60 sec = …… minute
Answer:
a. 1 nm = 10^-9 m
b. 1 km = 10⁵
c. 1000g = 1kg
d. one light year = 365 × 24 × 60 × 60 × 10⁵ km
e. 6o sec = 1 minute

Question 9. Calculate how many seconds in a solar day.
Answer:
24 ×60 × 60s

Question 10.
Explain the derived units with examples.
Answer:
Units which are expressed in terms of fundamental units or those units which are dependent on fundamen¬tal units are derived units.
Ex . Area = length × breadth=m × m = m²

Question 11.
Tabulate the fundamental quantities and their units.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 6

Question 12.
What are the salient features of SI units.?
Answer:

Unified units
Internationally accepted
Adequate to express all physical quantities.

Question 13.
Find the volume of a box having length 2m,breadth lm and height 0.5m
Answer:
volume = length × breadth × height =2 × 1 × 0.5 = 1 m³

Question 14.
Will the mass be same if you take one thermocol and a wooden block of same volume?
Answer:
No. Mass/Volume is density. The density of thermocol is lesser than that of wooden block. The mass will not be the same.

Question 15.
Take one jar with salt water and into both jars. In which jar the egg floated over the liquid. Justify your answer.
Answer:
In the jar having salt water because it has more density.

Question 16.
Tabulate following units and symbols
a. Length
b.Mass
c. Potential difference
d. Force
e.Pressure
f. Time
g.Density
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 8

Question 17.
Choose the derived quantities from following.
a. Length
b. volume
c. time
d. temperature
e. density
f. velocity
g. intensity of light
Answer:
volume, density, velocity

Question 18.
Tabulate two units connected to the name of personals
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units 9

Question 19.
Find the density of an object of mass 60 kg a having length 5m breadth 3m and height 2m.
Answer:
density =mass ÷ volume
\(=\frac{60}{5 \times 3 \times 2}=\frac{60}{30}=2 \mathrm{kg} / \mathrm{m}^{3}\)

Question 20.
Find the volume of a brick which has a length, breadth and height of 3m, 2m and 4m respectively.
Answer:
Volume = Length × breadth × height = 3 × 2 × 4 = 24m³

Question 21.
How to find the area of irregular surface? Explain with example.
Answer:
The area of irregular surface can be calculated by using graph paper.

Question 22.
How do you find out the diameter of a torch cell
Answer:
Place the cell between two wooden blocks tightly. Find out the distance between the two faces of the blocks, touching the cell using a metre scale. This reading gives the diameter of the cell.

Question 23.
The length of a room is 8m and the breadth ia 5m. Calculate the area of the room.
Answer:
Aiea = length × breadth = 8m × 5m = 40m²

Question 24.
What is meant by density of a substance?
Answer:
Density is the mass of a substance having unit volume. If the mass of the substance is M and V is its volume,
then \(\frac{M}{V}\)

Question 25.
Find out the fundamental units from the following table and mark them. Find out which unit is used to measure each of the physical quantity.

Ampere
Metre/second
Kilogram
Kilogram/metre
Mole
Kelvin
Metre³
Kilogram metre/second

Answer:
Fundamental units

Ampere
Kilogram
Mole
Kelvin

Physical quantity

Ampere – Electric current
Kilogram – Mass
Mole – Amount of substance.
Kelvin – Temperature

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

November 21, 2023November 20, 2023 by HSS Live

You can Download Properties of Matter Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

Properties of Matter Questions and Answers

Properties of matter

The matter commonly seen in three states, solid, liquid and gas. Matter has volume and mass. Matter is anything which occupies space and has volume. The solid has a definite shape. But liquid occupy the shape of the container. Gas has no definite shape.

Question 1.
explain the peculiarities of materials.
Answer:
Tiny particles of matter

The substances are made up of many tiny particles which cannot be seen by naked eye. Even though we cannot see the tiny particles of sugar in sugar solution, we can understand that tiny particles of sugar is dissolved in the solution while we taste it.

All the substances are made up of tiny particles which cannot be seen by naked eye and these tiny particles bear all the properties of the substances.

Question 2.
identify and picturise the arrangement of particles in different states of matter.
Answer:
Arrangements of particles in different states of matter

The particles of solids are very close to each other. The freedom of movement of particles is limited. The attractive force between particles is very high.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 1

The particles of liquid are relatively further apart and have more freedom for movement than in the solid state. The attractive force is less than that of solid state. The particles in gaseous state remains far away from one another and the attractive force is very low.

Change of State

When ice is heated, it changes into water and when water is further heated it boils and changes into steam.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 2

Question 3.
Which form of energy is responsible for the change of state here?
Answer:
Here the form of energy is responsible for the change of state is heat.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 4

When heat is absorbed energy of the particles, distance between the particles, movement of the particles are increased. The attractive force between particles is decreased.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 5

Some substances, when heated, change directly into gas without forming liquid. This phenomenon is sublimation.

Diffusion of substances in different states

The fragrance of incense stick fills the room when the stick is lit. The fragrance spreads because its particles are spread in air. If a drop of ink is added carefully into the water in a beaker, the ink spread in the water.

The spontaneous mixing of different particles having freedom of movement is Diffusion.

Question 4.
Distinguish pure substances and mixtures and tabulate them?
Answer:
Pure Substances and Mixtures
Depending on their nature, substances can be classified into two.
1. Pure Substances
2. Mixtures Materials made of particles of identical nature are called pure substances
Eg: Water, common salt, sugar

The substances made of particles of different nature are called mixtures.
Eg: Salt solution, sugar solution

Separating the Components of a Mixture.

Most of the substances found in nature are mixtures. We use different methods of separation of components in rice mixed with stones, tea mixed with tea dreg, mixture of methanol and ethanol.

Distillation

Distillation is used for separation of components in salt solution.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 6

When one component of the mixture is volatile and the others do not vapourise under the same condition, they can be separated by distillation. More over components of a mixture possess a large difference in their boiling points, they can be separated by distillation. Eg:Mixture of water (boiling point ioo°C)and acetone(56°C)

Fractional Distillation

If the boiling points of components have very small differences, fractional distillation is to be used to separate them.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 7

Eg: Mixture of Ethanol (boiling point 78°C) and methanol (boiling point 65°C).

When vapours of the mixture pass through the fractionating column, repeated. liquefaction and vapourisation take place. Subsequently, the vapours of low boiling methanol enter the condenser from the fractionating column, condense to liquid and get collected in the round bottomed flask first. Similarly, ethanol with higher boiling point can be collected later in another round bottomed flask.

Separating Funnel

Separating funnel is an apparatus used for separating immiscible liquids from their mixture.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 8
We can use separating funnel to separate the mixture of water and kerosene which are having difference in density.

Sublimation

This method can be used to separate the components which have the property of sublimation from the mixture. This method can be used for separating the components of a mixture of ammonium chloride and sand.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 9

Centrifugation

This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

In order to separate insoluble particles in a liquid mixture on the basis of their mass difference, centrifuge is used. The process is known as centrifugation.

Chromatography

Chromatography is the method used to separate more than one solute dissolved in the same solvent. This method was first employed for separating coloured substances and hence this process came to be known as Chromatography. Chromatography is employed to separate components from dyes and to separate poisonous substances mixed with blood.

Properties of Matter Textbook Questions and Answers

Question 1.
A few mixtures are given below. Tabulate the methods to separate their components and give the reasons for selecting the method.

Mixture	Method	Reason
Common salt and ammonium chloride
Sugar solution
Petrol and kerosene
Camphor and glass powder
Iron powder and sand

Answer:

Mixture	Method	Reason
Common salt and ammonium Chlo-	Sublimation	Components having the prope-
Sugar solution	Distillation	Components having the property of evaporation
Petrol and kerosene	Fractional distillation	Components have small differanee in boiling point
Camphor and glass powder	Sublimation	Components having the property of sublimation
Iron powder and sand	Magnetic separation	Components having magnetic properties

Question 2.
Given below are certain changes taking place to the particles during change of state. From this, find out and tabulate the changes in the particles when water boils to form steam and also when steam condenses to form water.

distance increases
attractive force decreases
energy increases
rate of movement increases
distance decreases
energy decreases
attractive force increases
rate of movement decreases

Answer:

Water changes to Steam	steam to water
Distance increases	Distance decreases
Energy increases	Energy decreases
Attractive force decreases	Attractive force increases
Movement of particles increases	Movement of particles decreases

Question 3.
Spirit kept open in a.watch glass disappears after some time. Which among the following phenomena are responsible for this?
a) sublimation
b) distillation
c) evaporation
d) diffusion
Answer:
a) sublimation

Question 4.
Which are the methods that can be used to separate the components of a mixture made of common salt, ammonium chloride and sand? Write the methods in the order in which they are applied.
Answer:
sublimation → filtration → distillation

Question 5.
Many minerals are present in ordinary water.
a. Which is the method that can used to remove the mineals and obtain pure water?
h.In which type of mixtures is this method employed?
c. Water purified by this method is distilled water .Write two instances of its use.
Answer:
a. distillation
b. When one component of the mixture is volatile and the others do not vapourise under the same condition, they canbe separated by distillation
c. For vaccination, In storage batteries

Question 6.
From the following statements, tick (V) those which apply to solid substances alone.
□ Particles have little freedom of movement.
□ Distance between particles is very high
□ Particles remain very close to each other
□ Energy of particles is very high
Answer:
✓ Particles have little freedom of movement.
✓ Particles remain very close to each other

Question 7.
Take a small wooden rectangular block and find its volume (volume = length x breadth x height).
Take a big measuring jar and fill three-fourth of it with water and mark the‘water level. Then dip the block in water in the jar. (To prevent floating, nails can be inserted in the block). Mark the difference in the water level.
a. Is there any relation between the difference in the water level and the volume of the block?
b. Which property of matter is revealed by this experiment?
Answer:
a.Yes. the volume of water rises is the volume of wooden block.
b. The substance that occupies a space.

Question 8.
Electronic balances are very popular now. On an electronic balance, find the weight of an empty balloon. Again, find its weight after filling air. Now, can you find the weight of the air in the balloon?
Repeat the experiment using balloons of different size by filling them with varying quantity of air.
Answer:
Reduce the weight of balloon from the weight of balloon with air

Question 9.
Take water mixed with chalk powder in a bottle. Tie a string to the bottle and swirl it at high speed along a circular path. Observe after sometime.
Repeat the activity using other mixtures which are suspensions. To which method of separation of components of a mixture can this be connected? Are there instances where this principle is made use of. Prepare a note.
Answer:
Centrifugation is the method used. This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

Question 10.
Take a long white chalk piece and put a mark with black ink slightly above the bottom. Keep the chalk piece dipped perpendicularly in water in a watch glass. After sometime observe the changes. Repeat the experiment using different chalk pieces marked with sketch pens of different colours. To which of the methods of separation you have studied is this related?
Answer:
Chromatography

Properties of Matter Additional Questions and Answers

Question 1.
Find the odd one out
a. Evaporation, Fermentation, Filtration, Chromatography.
b. Sea water, Muddy water, Table salt, Curd, Air
c. Distillation, Separating funnel, Filtration, Evaporation
Answer:
a. Fermentation
b. Table salt
c. Separating funnel

Question 2.
What is the use of chromatography?
Answer:
Solutes dissolved in the same solvent can be separated using chromatography.

Question 3.
What is the importance of evaporation? Where is this technique used on a large scale?
Answer:
Evaporation is used to separate solids dissolved in liquids, mainly water. It is used on a large scale to obtain common salt from sea water. Sea water is left in the open in large containers on sunny days. After a few days salt remains as residue and water evaporates. Common salt is then further purified.

Question 4. What is the advantage of distillation over evaporation?
Answer:
Distillation is more advantageous than evaporation, because liquid can be recover in distillation while in evaporation liquid is lost.

Question 5.
Which process is used to dry clothes in a washing machine?
Answer:
Centrifugation.

Question 6.
Method for separating drugs (medicine) which are miscible in blood Centrifugation, Evaporation, Filtration, Chromatography.
Answer:
Chromatography

Question 7.
From the below statements select correct one?
a. Every substance has a definite mass
b. Every substance has a definite shape
c. Every substance has a definite volume
Answer:
a. Every substance has a definite mass

Question 8.
Melting: Converting solid into liquid Sublimation:
a. Observe the first pair and fill up. the blanks in the second pair suitable.
b. Write two examples of sublimation
Answer:
a. Sublimation. A solid directly changes into gas
b. Camphor, iodine, naphthalene.

Question 9.
From Gopu’s hand the bottle containing coconut oil fall into a bucket containing water. Is it possible to separate oil and water?
Answer:
Using a separating funnel, they can be separated. The denser water will be at the bottom and the lighter oil will be at the top. Water is removed through the tap.

Question 10.
Classify the following substances according to its state and tabulate.
Water, pencil, book, air, kerosene, oxygen
Answer:

solid	liquid	gas
Pencil Book	Water kerosene	Air Oxygen

Question 11.
Will water fill in tumbler when a glass tumbler is immersed perpendicularly into the water in a turf ? Give reason.
Answer:
No because there is air in the tum-bler. Air need space to occupy so water cannot enter into the tumbler.

Question 12.
A balloon filled with air and an empty balloon weigled in a digital balance found difference in weight. What are the properties of matter understood by this experiment?
Answer:
Matter needs space to occupy and has mass.

Question 13.
Complete the table.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 10
Answer:

Question 14.
Which energy causes the change of state when ice is heated to water and water to water vapour? The three figures shows the arrangements of particles. Recoga- nise and write the state of matter.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 12
Answer:
Heat energy figure
(1) solid, figure
(2) liquid, figure
(3) gas

Question 15.
Mention three examples from daily life for diffusion.
Answer:
Spreads the smell of perfumes, smell of fruits and food item. Ink spreads in water.

Question 16.
Complete the table
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 13
Answer:

Question 17.
When naphthalene balls are kept somewhere after some days its size diminishes and disappears. Name the phenomenon caused for this process.
Answer:
Sublimation

Question 18.
Mention the reason behind the smell spreads everywhere in the room while in scent stick is lighted. Sublimation, Evaporation, Diffusion
Answer:
Diffusion

Question 19.
Tabulate following as pure substances and mixtures.
Gold, soda water, steam, sugar, soil, water, common salt, sugar solution, carbon dioxide, salt solution.
Answer:

Pure substances	Mixtures
Gold	Soda water
Sugar	Sugarsolution
Salt	Soil
Water	Salt solution
Steam
Carbon dioxide

Question 20.
Find the method of separation of components following mixt¬ures.
a. husk from paddy
b. butter from curd
c. tea dreg from tea
d. salt from sea water
e. blood cells from blood sample
f. mixture of petrol and kerosene
g. salt and ammonia
h. poison from blood
i. water and kerosene
j. water and acetone
Answer:
a. by picking
b. centrifugation
c. filtration
d. distillation
e. centrifugation
f. fractional distillation
g. sublimation
h. chromatography
i. using separating funnel
j. distillation

Question 21.
Write two examples of sublimation from daily life.
Answer:
1.camphor
2. naphthalene balls becomes smaller

Question 22.
Salt, rava and mustard are mixed while Suresh has brought it from shop. Say methods to separate these.
Answer:
Salt dissolves in water. Then filter it. Rava and mustard are separating by picking, salt is collected by distillation.

Question 23.
What will be seen when a filter paper is marked by sketch pen and one end of the paper is dipped in water? what is this process called?
Answer:
different colours are separated. Chromatography

Question 24.
Using given methods write one example for separating components of mixture.
Distillation, magnetic separation, centrifugation, fractional distillation, filtration, sublimation.
Answer:
• Separating common salt from sea water – Distillation
• Mixture if iron powder and sulphur – Magnetic separation
• Butter from curd – Centrifugation
• Separation of kerosene, petrol, die sel – Fractional distillation
• Separating tea dreg from tea – Fil-tration
• Ammonium chloride and common salt – Sublimation

Question 25.
Solid, Liquid, Gas are the three states of matter
a. Which of this having definite volume hut not definite shape?
b. Write any two changes occurs when we change a substance from liquid state to gaseous state?
c. How to convert a gasecfas substances into liquid?
Answer:
a. Liquid
b. Increase the distance between molecules.
• Decrease the attractive force between molecules.
• Increase the speed of molecules.
• Increase the energy of molecules.
c. to be cooled (or) Reduce temperature

Question 26.
Name of the mixtures in column A, properties of components are given in column B, method of separating of component are given in the column C also. Find out the match items from B & C to A.

Mixture	Character of components	Method of Separation
1. Alcohol and water	difference in boiling point	Centrifugation
2. Soil and mentor	difference in evaporating Character	Distillation
3. Curd and butter	difference in weight of insoluble precipitate	Chromatography
		Sublimation

Answer:

Mixture	Character of components	Method of Separation
1. Alcohol and water	difference in boiling point	Distillation
2. Soil and mentor	difference in evaporating character	Sublimation
3. Curd and butter	difference in weight of insoluble precipitate	Centrifugation

Question 27.
a. Which device given in the picture?
b. Why should this device doesn’t use for separating petrol and kerosene mixture?
c. Which method is used to separate the mixture containing petrol and kerosene?
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 8
Answer:
a. Separating funnel
b. Petrol and kerosene are miscible liquids.
c. Fractional distillation

Question 28.
Diffusion is a phenomenon which
causes the spread of smell.
a. What is the relation between the rate of diffusion and states of substance?
b. What is the relation between diffusion and temperature?
c. Given an example from everyday life which shows relation between diffusion and temperature.
Answer:
a.Rate of diffusion is more in gaseous state, less in solid state.
b. Rate of diffusion increases with increase in temperature.
c. The smell of hot food material spread quickly.

Question 29.
Find out the relation and fill in the blanks.
a. Separate salt from saline water :: Evaporation; Separation of components from black coloured ink :: ………..
b. Distillation :: Condenser; Fractional distillation :: ………..
c. Write down some situations for which the fractional distillation used?
d. When we use fractional distillation for separating components of mixture?
Answer:
a. Chromatography
b. Fractionating column
c. • Separating components to petroleum
• Separating components of air.
d. Used when slight differences in their boiling points.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

November 18, 2023November 17, 2023 by HSS Live

You can Download Cell Clusters Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

Cell Clusters Questions and Answers

Diversity among cells

Each part of human body is composed of different kinds of cells. These cells do not act independently. But they act in groups of similar cells. These cell clusters are called tissues.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 18

Tissues

The group of cells with common origin and perform a specific function are called tissues. The coordinated action of tissue, help to perform various physiological functions effectively.

Cell Differentiation

The single celled zygote divides continuously and forms the foetus that consists of cells differ in shape, size and contents. Foetal cells gradually attain change in structure and function. This process is called cell differentiation.

Indicators (Text Book Page No:24)

Question 1.
Formation of foetus.
Answer:
The single celled zygote formed by the fusion of sperm and ovum continuously divides and the foetus is formed.

Question 2.
Significance of cell differentiation.
Answer:
Foetal cells gradually attain change in structure and function due to cell differentiation. Thus, different kinds of cells and tissues are formed. It helps to make the life processes effective.

Stem cells

Stem cells are specialised cells that can be transformed into any kind of cells.

Stem cells can either transform into other cells through division or they can exist as such. When the cells in the tissues, get destructed, new cells emerge from stem cells.

It is expected that the research of stem cells may cause drastic change in the treatment of blood cancer, diabetes, Parkinson’s disease etc as desired cells are formed from stem cells.

Indicators (Text Book Page No:25)

Question 3.
What are the peculiarities of stem cells when compared to other cells?
Answer:
Stem cells are specialized cells that can be transformed into any kind of cell. They can either transform into other cells or can retain their existence. When the cells in the tissues get destroyed new cells originate from stem cells.

Question 4.
How is the destruction of cells in tissues compensated?
Answer:
When the cells in the tissues get destroyed, new cells originate from cells. Stem cells are found in bone marrow, skin, digestive fact etc.

Question 5.
Why is stem cell research gaining importance?
Answer:
Nowadays desired cells can be produced from stem cells in research laboratories. The research of stem cells can cause a drastic change in the treatment of blood cancer, diabetes, Parkinson’s diseases etc, and in the production of artificial organs.

Animal Tissues

Animal tissues are mainly classified into 4
(a) Epithelial tissue
(b) Nervous tissue
(c) Muscular tissue
(d) Connective tissue

1. Epithelial Tissue:- Covers and protects the body.
Covers the inner lining of the alimentary canal Performs the functions like protection, absorption, production of secretions, etc.
2. Nervous Tissue:- Controls and Co-ordinates physiological Activities Helps to respond to particular stimuli
3. Muscular Tissue:- Makes the movement of the body possible.
4. Connective Tissue:-

Connects different tissues.
Provide support and strength to the body
Performs the functions like material transport, defence, etc.

Meristematic tissues Meristematic cells are specialized cells seen at the tip of the stem and root in plants. They divide rapidly and helps in the growth of plants.

Table (Text Book Page No:27)

Compare the figures of a meristematic cell and a mature cell. Find out the differences and complete the table.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 3
Answer:

Characteristics	Meristematic Cells	Mature plant Cells
Relative size of Nucleus	(a) ……………….	(b) ……………
Thickness of cell wall	(c) ……………….	(d) ……………
Quantity of Cytoplasm	(e) ……………….	(f) ……………

Various Plant Tissues

Parenchyma, Collenchyma, and Sclerenchyma are the main plant tissues. They differ in structure and perform different functions.

The specialised tissues in plants that conduct water and salts that are absorbed by the roots to the leaves and the food prepared in the leaves to various parts, are called vascular tissues. These are called complex tissues as they are. formed by different types of cells. Xylem and Phloem are the main vascular tissues.

Work sheet (Text Book Page No:30)

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 5

Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 6

Organs like stomach and Intestine are mainly formed by nervous tissue, epithelial tissue, connective tissue, and muscular tissue.
Temporary storage of food materials, digestion and the secretion of digestive juices.
Complete digestion of food materials. Secretion of-digestive juices, reabsorption of water, absorption of nutrients.
Organ performs the co-ordinated functions of various tissues. The function of an organ is not the same as that of the individual tissues which constitute the organ.
Various organs combine to form organ system. As a result physiological functions can be performed very effectively.
eg: teeth, tongue, glands, oesophagus, stomach, small intestine, large intestine etc constitute digestive system. It enables the complete digestion of food.

Table (Text Book Page No:31)
Complete the following table by finding out the systems to which the organs listed in the table belong to:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 4
Answer:

Organs	Organ System
Heart, Blood vessels	Circulatory System
Nose, Trachea, Lungs.	Respiratory System
Kidney, Ureter, Urinary bladder	Excretory System
Brain, Nerves	Nervous System

An organism is composed of various organ systems. Therefore the structure of organisms is highly complicated.

Cells are the basic unit of life. Cell parts are composed of different substances.

In higher organisms diverse tissues act complementary to each other and perform various life processes.

Cell Clusters Let us assess (Text Book Page No:33)

Given below in the illustration are various tissues related to the structure of hand.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 19
Answer:
1. • A
• C
2. Formed from various cells
3. Only the corners of the cells are thicked

Cell Clusters Text Book Questions and Answers

Question 1.
Arrange an exhibition showing pictures and descriptions of different types of cells.

Answer:
Cell
Cell is the structural and functional unit of life.
• All organisms are formed of one or more cells.
• All vital activities of life take place inside the cells.
• Here dietary information that controls the functions of the cells are contained in the cell.
• There are three types of cells in the blood. Red blood cells, white blood cells and platelets.

Red Blood cells

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 7

Red blood cells are the most abundant cells in the blood. They are also known as erythrocytes. Invertebrates, RBC carries oxygen to the tissues. In RBC, the cytoplasm contains an iron containing biomolecule, the haemoglobin. If in parts red colour to the blood. In human, RBCs has biconcave disc shape.

White blood cells

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 8
WBC’S are also called leucocytes, they protect the body from pathogens and other antigens as a part of the immune system. They are formed from the stem cells of the bone marrow. The number of WBCs is an indicator of health. They are classified into neutrophil, eosinophil, basophil, monocyte and lymphocyte according to the diversity of cytoplasm, nucleus etc.

Platelets

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 9
They are also known as thrombocytes. They help in the coagulation of blood. They are formed from the cells called megakaryocytes.

Question 2.
Prepare a magazine specifying the importance, relevance, and scope of stem cells.
Answer:
Stem cell

Stem cell are specified cells that can transform to any kind of cell. They are also known as root cells. They transform to other cells by a long process of differentiation.

Stem cells can either transform to other cells or exist as such. When the cell in the tissues get destructed, new cells originate from stem cells. Stem cells are seen in bone marrow, skin, digestive tract etc.

Today it is possible to develop desired cells from stem cells under specific conditions of laboratories. It is expected to make tremendous change in the treatment of blood cancer, diabetes, parkinson’s diseases etc. and in the development of artificial organs through the research works of in stem cells.

Cell Clusters Additional Questions and Answers

Question 1.
Findi the odd one in each group.
Also write the common characterestic of the others.
a. Collenchyma, Sclerenchyma Xylem, Nervous tissue.
b. Blood, Muscle, ligament, Bone
c. Basophil, Platelet, Neutrophil, Lymphocytes
d. Collenchyma, Aerenchyma, Parenchyma, Chlorenchyma
e. Intercalary meristem, Primary meristem, Apical meristem, Lateral meristem
f. Xylem, Collenchyma, Scler-enchyma, Parenchyma.
g. Man, Pigeon, Duck, Amoeba
h. Large intestine, Digestive tract, Heart, Stomach
i. Bony tissue, Muscular tissue, Cartilaginous tissue, Blood
j. Parenchyma, Collenchyma, Epithelial tissue, Sclerenchyma.
Answer:
a.Nervous tissue; Others are plant tissues.
b. Muscle, Others are connective tissues.
c. Platelets, Others are white blood cells.
d. Collenchyma, Others are the different forms of parenchyma.
e. Primary meristem : Others are divide and new tissues. Primary meristem divide and from apical meristem and lateral meristem.
f. Xylem : It is conduction tissue and others are basic tissues.
g. Amoeba – Cell level organism and
others are organ system level organisms.
h. Heart : It is a circulatory organ. Others are digestive organ.
i. Muscular tissue : The others are connective tissues
j. Epithelial tissue : The others are plant tissues.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Water and salts : xylem :: ………… : Phloem
b. Photosynthesis : Parenchyma : Support and Strength : ………..
c. Between bones and muscles : tendons ; Between bones and bone: ………..
d. Blood : Connective tissue : : Skin: ……….. ;
e. Control and Co-ordination : Nervous tissue : : Body Move ment:: ………..
f. Fibrous tissue : Connectother tissue; ……….. : Transport of materials
g. Storage of food : ……….. ; Flexibility to plant parts : Col-lenchyma
h. White blood cells : ……….. ; Redblood cell; O₂ transportation
i. Xylem – ………..; Phloem: Transports synthsised food
j. Intercalary meristem : Inter node lengthening; Lateral meristem: ………..
k. Aerenchyma : Air cavities ; Chlorenchyma : ………..
l. Cell :Tissue; system : ………..
Answer:
a. Food
b. Sclerenchyma
c. Ligaments
d. Epithelial Tissue
e. Muscular Tissue
f. Blood
g. Parenchyma
h. Immunity
i. Transports water and minerals
j. Thickening of stem
k. Chlorophyll
l. Organ

Question 3.
Complete the flowchart which shows levels of organisation of human being.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 10
Answer:
a. cell organelle
b. Organ

Question 4.
Observe the following figure and answer the given questions.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 20
a. Find out correct one from the bracket shows the plant tissue given in the figure.
A. parenchyma
B. sclerenchyma
C. xylem
D. collenchyma
b. Write down the characteristic features of the plant cell in the given figure?
c. Give scientific explanation for the features of plant part are given below.
i. Desterity of shaft of colacasia.
ii. Hardness of shell
Answer:
a. B. sclerenchyma
b. composed of cells that are uniformly thick all over the cell wall.
c. i. presence of collenchyma cells, ii. presence of sclerenchyma cells.

Question 5.
Find out wrong items from the following statements, and also correct their underlined words.
a. Sclerenchyma can provides strength and support to plant parts.
b. Fibrous tissue can enables to respond by identifying the changes inside and outside the body.
c. Production of secretion is one of the function of Epithelial tissue.
d. muscular tissue enables the movement of the body.
Answer:

b, d are wrong statements.
b. Nervous tissue can enables to respond identifying the changes inside and outside the body.
d. Muscular tissue enables the movement of the body.

Question 6.
What is the difference between blood and blood circulatory system?
Answer:
Blood is a tissue in the blood circulatory system. But blood vessels, heart, nerves, etc. are the other factors of the blood circulatory system. The combined work of all these factors form the blood circulatory system.

Question 7.
All parts of the plant do not grow. Why?
Answer:
In plants, only the meristematic cells have the ability to divide and grow. Meristematic cells are mainly found in the root apex and stem apex. So the plant growth is concentrated at the root apex and stem apex.

Question 8.
What are tissues? Give examples.
Answer:
A group of similar cells is called tissue
eg : Parenchyma

Question 9.
“Body is composed of millions of cells. Are they perform the same function? How do they act independently?”
This is question raised in the class by the teacher for a discussion. What will be your answer to this question?
Answer:
Cells act in groups of similar kind. Group of similar cells that are originated from a single cell and perform a specific function are called tissues. They perform different functions.
Eg: muscular tissue, Nervous tissue, Epithelial tissue.

Question 10.
Our body is developed from a single-celled zygote. How do this much organs and organ system sare formed in our body?
Answer:
Zygote divides continuously and forms the foetus that consists of cells differ in shape, size, and content. Foetal cells gradually attain diversity in structure and function. Thus different organs and organ systems are formed.

Question 11.

Stem cells for Cancer treatment

Tvm:- Scientists say that the production of tissues from stem cells may cause drastic change in the treatment of Cancer…

Did you read the news report? What are stem cells? How do they become useful in the treatment of Cancer?
Answer:
Stem cells are specialized cells that can be transformed to any kind of cell. Stemcells modify to other cells through prolonged differentiation. When cells destroyed, new cells originates from stem cells. In research laboratories desired cells are formed from stem cells. New cells, instead of cancer cells, can be produced from stem cells.

Question 12.
Complete the given flow chart
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 11
Answer:
a. Nervous Tissue
b. Connective Tissue
c. • Covers and protects the body
• The inner lining of alimentary ‘ canal
• Perform the function of protection, absorption, production of secretion, etc.
d. helps to respond by recognizing the changes within and out of the body
e. Makes the movement of body possible.

Question 13.
Categorise the given statements under suitable headings.
1. Provide shape to body.
2. Defence
3. Covers and protects internal organs.
4. Conduction of respiratory gases.
Answer:

Blood	Bone, Cartilage (Connective Tissue)
2. Defence	1. Provide shape to body.
4. Conduction of respiratory gases.	3. Covers and protects internal organs.

Question 14.
Identify the figures given below.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 12
a. Nervous tissue
b. Muscular tissue

Question 15.
What are the differences between the process of growth in plants and animals?
Answer:
Plant growth

Plants grow throughout their life.
In plants, growth is mainly restricted at the tips of stem and root(meristem)

Animal growth

Animals grow upto a particular period
In animals, growth is not confined to any specific regions.

Question 16.
Why plant growth is confined to specific parts?
Answer:
In plants, specific cells called meristematic cells are concentrated the tips of roots and stems. They have the ability to divide rapidly.

Question 17.
Complete the table properly

Characteristics	Meristematic Cells	Mature Cells
Relative size of Nucleus	(a) ……………….	(b) ……………
Thickness of cell wall	(c) ……………….	(d) ……………
Quantity of Cytoplasm	(e) ……………….	(f) ……………

Answer:
a. Comparatively large nucleus
b. Small nucleus
c. Thin cell wall
d. Thick cell wall
e. More
f. Comparatively less

Question 18.
How do the water and salts absorbed by the roots reach the leaves?
Answer:
In plants, the water and salts absorbed by the roots are transported to the leaves through specialised tissues called vascular tissues.

Question 19.
Vascular tissues are called complex tissues. Why?
Answer:
Vascular tissues are formed by the union of different types of cells. Hence they are called complex tissues.

Question 20.
Picture of section of stem is given. Label the parts.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 13
Answer:
a. Collenchyma
b. Parenchyma
c. Sclerenchyma
d. Phloem
e. Xylem

Question 21.
Pair the given cells with their specific character

Parenchyma, Collenchyma, Sclerenchyma.

a. formed of cells whose cell wall is uniformly thickened in all parts.
b. seen in tender parts of plant.
c. formed of cells whose cell wall has thickenings in the comers only
Answer:
parenchyma – b
Collenchyma –
c sclerenchyma – a

Question 22.
Write examples for vascular tissues?
Answer:
Xylem and Phloem

Question 23.
Arrange the given statements in the right column.
1. Formed of inter related cells seen as ducts.
2. Carries water and salts absorbed by the roots to the leaves.
3. Carries the food prepared in the leaves to various plant parts.
4. Provide support and strength to plant
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 15
Answer:
Xylem 2, 4 Phloem -1, 3

Question 24.
Complete the worksheet based on the hints given.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 16
Hints
A. Carries the food prepared in the leaves to various plant parts.
B. Carries water and salts absorbed by the roots to the leaves.
C. Seen in the tender parts of the leaves.
D. Thickenings seen only in the corners of the cell wall
E. Cells have uniform thickenings in the cell wall.
Answer:
A – Phloem
B – Xylem
C – Parenchyma
D – Collenchyma
E – Sclerenchyma

Question 25.
Which are the tissues that constitute the organs like stomach and intestine?
Answer:
Nervous tissue, Epithelial tissue, Connective tissue, Muscular tissue

Question 26.
Identify the organ system to which the given organs are associated?
a. Heart, Blood Vessels
b. Nose, trachea, lungs
c. Kidney, Ureter, Urinary Bladder.
d. Brain, Nerves
Answer:
a – Circulatory system
b – Respiratory system
c – Excretory system
d – Nervous system

Question 27.
Which of the statements is not related to tissues?
a. Different types of cells are seen.
b. Similar kind of cells are seen.
c. Performs specific function.
d. Formed from different cells
Answer:
d. Formed from different cells

Question 28.
Which indicator helps to identify collenchyma when tissues are observed through a microscope.
a. No thickenings in the cell wall.
b. all parts of the cell wall is thick,
c. No nucleus in the cell
d. Thickenings are seen only in the comers of the cell.
Answer:
d. Thickenings are seen only in the comers of the cell.

Question 29.
Complete the flow chart showing the levels of organization in organisms
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 17
Answer:
a. tissue
b. Organ system
c. Organism
d. Species

Question 30.
What is meant by voluntary muscles and involuntary muscles? Give examples:
Answer:
Tissues that work according to our wish are called voluntary muscles.
eg: muscles of limbs.
Muscles which works independently and are out of our control are called involunatary muscles.
eg : Muscles of the alimentary canal, muscles of the eyelids, etc.

Question 31.
Xylem vessels are thicker than phloem vessels. Why?
Answer:
The cell wall of the xylem vessels is comparatively thicker than that, of the phloem vessels. Lignin is used to make the cell wall strong. So the xylem vessels are thicker.