chapter - 6 life processes · remember that a real image of 3 times the size of the object...

Class 10 Ch 6 Life Processes of 1

CHAPTER - 6

Life Processes Page No. 95

Q.1 Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms

like humans? (To be marked in textbook)

Ans. As in multicellular organisms, all the cells are not in direct contact with environment, simple

diffusion does not meet the requirement of all the body cells to get sufficient oxygen.

Q.2 What criteria do we use to decide whether something is alive?

Ans. Something can be considered alive if following features are present:

(i) Movements-

Visible movements like growth, breathing, etc.

Invisible movements - All the living organism must have movement at molecular level like

respiration.

(ii) Life processes- Occurrence of processes like nutrition, respiration, transportation and

excretion to be called alive.

(iii) Growth

(iv) Reproduction

(v) Response to stimuli

Q.3 What are outside raw materials used for by an organism? (To be discussed) Ans. Outside raw materials used by an organism include:

(a) Water (b) Oxygen (c) CO2 (d) sunlight and (e) Nutrients/Minerals

These raw materials are used by an organism for various purposes like respiration, growth and

maintenance.

Q.4. What processes would you consider essential for maintaining life? (To be discussed) Ans. The processes essential for maintaining life are

a. Nutrition

b. Respiration

c. Transportation

d. Excretion

DELHI PUBLIC SCHOOL, GANDHINAGAR ACADEMIC SESSION: 2020-21

Chapter 10 - Light: Reflection and Refraction

Exercise 168

Solution 1 Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.

Solution 2

Radius of curvature, R = 20 cm Radius of curvature of a spherical mirror = 2 × Focal length (f) R = 2 f

Hence, the focal length of the given spherical mirror is 10 cm. Concept Insight - Always remember that focal length of the spherical mirror is half its radius of curvature.

Solution 3

A concave mirror can give an erect and enlarged image of an object. Concept insight: When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect, and enlarged.

Solution 4

A convex mirror is preferred as a rear-view mirror in vehicles because it gives virtual, erect, and diminished image of the objects placed in front of it. Also, a convex mirror has a wider field of view, which allows the driver to see most of the traffic behind him. Concept insight: Recall the nature of images formed by convex mirror irrespective of the position of the objects.

Chapter 10 - Light: Reflection and Refraction Exercise 171

Solution 1

Radius of curvature, R = 32 cm Radius of curvature = 2 × Focal length (f) R = 2 f

Hence, the focal length of the given convex mirror is 16 cm. Concept Insight - Focal length of a spherical mirror is half the radius of curvature.

Solution 2 Magnification produced by a spherical mirror is given by the relation,

Let the height of the object, hO = h Then the height of the image, hi = - 3 h (Image formed is real)

Object distance, u = -10 cm v = 3 × (-10) = -30 cm Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror. Concept Insight - Following sign conventions in this type of question is very important. Remember that a real image of 3 times the size of the object corresponds to magnification of -3 (note the negative sign).


Solution 1

The light ray bends towards the normal. Reason: When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal. Concept insight: Air is rarer medium & water is denser medium. The direction of bending of light depends on whether the light is moving from rarer to denser medium or vice versa.

Solution 2

Refractive index of a medium nm is given by,

Given: Speed of light in vacuum, c = 3 × 108 ms-1

Refractive index of glass, ng = 1.50

Speed of light in the glass,

Concept insight: Remember this formula for refractive index of a medium. The refractive index of medium 2 with respect to medium 1 is given as,

The refractive index of medium 1 with respect to medium 2 is given as,

Solution 3 Highest optical density = Diamond Lowest optical density = Air Optical density of a medium is directly related with the refractive index of that medium. A

medium which has the highest refractive index will have the highest optical density and vice-versa.

It can be observed from the table that diamond and air respectively have the highest and lowest refractive indices. Therefore, diamond has the highest optical density and air has the lowest optical density.

Solution 4

Speed of light in a medium is given by the relation for refractive index (nm). The relation is given as

It can be inferred from the relation that light will travel the slowest in the material which has the highest refractive index and will travel the fastest in the material which has the lowest refractive index. It can be observed from the table that the refractive indices of kerosene, turpentine, and water are 1.44, 1.47, and 1.33 respectively. Therefore, light travels the fastest in water. Concept Insight: Higher is the refractive index of a medium, less is the speed of light in the medium.

Solution 5 Refractive index of a medium nm is related to the speed of light in that medium v by the relation:

Where, c is the speed of light in vacuum/air The refractive index of diamond is 2.42. This suggests that a light ray travelling in air and entering diamond gets slowed down and its speed becomes 1/2.42 times that in air. Chapter 10 - Light: Reflection and Refraction Exercise 184

Solution 1

1 diopter is defined as the power of a lens of focal length 1 metre.

Solution 2 When an object is placed at the centre of curvature, 2F1, of a convex lens, its image is formed at the centre of curvature, 2F2, on the other side of the lens. The image formed is real, inverted and of the same size as the object, as shown in the given figure.

It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Hence, the needle must be placed in front of the lens at a distance of 50 cm. Object distance, u = -50 cm Image distance, v = 50 cm Focal length = f According to the lens formula,

Hence, the power of the given lens is +4 D. Concept Insight - One should be extremely careful while substituting the values of u, v and f without forgetting to put the appropriate sign conventions.

Solution 3

Focal length of concave lens, f = -2 m

Here, negative sign arises due to the divergent nature of concave lens. Concept Insight: While using the formula of power, one should be careful to use the value of focal length expressed in meters only.


Solution 1 (d) Clay

Concept Insight: A lens allows light to pass through it. Since clay does not show such property, it cannot be used to make a lens.

Solution 2 (d) Between the pole of the mirror and its principal focus.

Concept Insight: When an object is placed between the pole and principal focus of a concave mirror, the image formed is virtual, erect and larger than the object.

Solution 3 (b) At twice the focal length

Concept Insight: When an object is placed at a distance equal to twice the focal length in front of a convex lens, its image is formed at a distance of twice the focal length on the other side of the lens. The image formed is real, inverted, and of the same size as the object.

Solution 4 (a) both concave

Concept Insight: By convention, the focal lengths of a concave mirror and a concave lens are taken as negative. Hence, both, the spherical mirror and the thin spherical lens are concave in nature.


Solution 1 (d) either plane or convex

Concept Insight: A convex mirror always gives a virtual and erect image of smaller size than the object placed in front of it. Similarly, a plane mirror always gives a virtual and erect image of the same size as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.

Solution 2 (c) A convex lens of focal length 5 cm. Concept Insight: A convex lens gives an erect and magnified image of an object when it is placed between the optical centre and focus of the lens. Also, magnification is more for convex lenses having shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.

Solution 3

Range of object distance = 0 cm to 15 cm A concave mirror gives an erect image when an object is placed between its pole (P) and

the principal focus (F). Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm,

the object must be placed anywhere between the pole and the focus (i.e. within 15 cm from the mirror). The image formed will be virtual, erect, and magnified in nature, as shown in the given figure.

Concept Insight: Recall that in case of a concave mirror, the erect & hence the virtual image can be formed only when the object is placed between the principal focus and pole.

Solution 4 (a) Concave (b) Convex (c) Concave Explanation:

(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce powerful parallel beams of light when the light source is placed at their principal focus.

(b) Convex mirror is used in the side/rear view mirror of a vehicle because convex mirrors give a virtual, erect, and diminished image of the objects placed in front of them and have a wide field of view. It enables the driver to see most of the traffic behind him/her.

(c) Concave mirrors are converging mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as principal focus. Hence, they can be used to produce a large amount of heat at that point.

Solution 5

The convex lens will form a complete image of an object, even if its one-half is covered with black paper. It can be understood by the following two cases.

Case I-When the upper half of the lens is covered: In this case, the rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.

Case II-When the lower half of the lens is covered:

In this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.

Concept Insight: In case of the half covered lens, the number of rays used up to make the image on the other side of the lens will be reduced to half.

Solution 6

Object distance, u = -25 cm Object height, ho = 5 cm Focal length, f = +10 cm According to the lens formula,

The positive value of v shows that the image is formed on the other side of the lens.

The negative sign shows that the image is real and formed behind the lens.

hi = m x ho = -0.66 x 5 = -3.3 cm The negative value of image height indicates that the image formed is inverted. The position, size, and nature of the image are shown in the following ray diagram.

Concept Insight: Remember to use appropriate sign conventions while substituting the values in lens formula.

Solution 7

Focal length (OF1) of the concave lens is f = -15 cm Image distance, v = -10 cm According to the lens formula,

The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.

Concept Insight: Remember to use appropriate sign conventions while substituting the values in lens formula.

Solution 8

Focal length of convex mirror, f = +15 cm Object distance, u = -10 cm According to the mirror formula,

The positive value of v indicates that the image is formed behind the mirror.

The positive value of magnification indicates that the image formed is virtual and erect. Concept insight - Correct interpretation of the signs are important. Remember:

Object distance u is conventionally always taken as negative. Magnification: Negative value corresponds to real & inverted images. Positive value

corresponds to virtual & erect images.

Solution 9

Magnification produced by a mirror is given by,

The magnification produced by a plane mirror is +1. It shows that the image formed by the plane mirror is of the same size as that of the object. The positive sign shows that the image formed is virtual and erect. Concept Insight - Positive magnification corresponds to virtual and erect image. The numeral value of magnification indicates that the size of the image is that many times the size of the object.

Solution 10

Given: Object distance, u = -20 cm Object height, H = 5 cm Radius of curvature, R = 30 cm Radius of curvature = 2 × Focal length i.e., R = 2 f

f = 15 cm

According to the mirror formula,

v = 8.57 cm The positive value of v indicates that the image is formed behind the mirror.

The positive value of magnification indicates that the image formed is virtual.

h' = m × h = 0.428 × 5 = 2.14 cm The positive value of the image height indicates that the image formed is erect. Therefore, the image formed is virtual, erect and smaller in size. Concept Insight - Write the given data in appropriate sign conventions. Then identify the unknown quantities that need to be evaluated. Then put the formulae that are most suitable.

Solution 11

Given: Object distance, u = -27 cm Object height, h = 7 cm Focal length, f = -18 cm According to the mirror formula,

The screen should be placed at a distance of 54 cm in front of the given mirror.

h' = h × m h' = 7 × (-2) = -14 cm The negative value of the image height indicates that the image formed is inverted. Concept Insight - The interpretation of sign convention is the key here.

Solution 12

P = -2 D

f = -0.5 m The focal length is negative. Hence, it is a concave lens. Concept Insight - The value of the focal length should be written in metre when substituting in the formula:

Solution 13

A convex lens has a positive focal length. Hence, it is a convex lens or a converging lens. Concept Insight: Positive focal length corresponds to convex lens and negative focal length corresponds to concave lens.

Class 10 Ch 6 Life Processes Page 1 of 4

CHAPTER - 6

Life Processes

Page No. 95

Q.1 Why is diffusion insufficient to meet the oxygen requirements of multicellular

organisms like humans? (To be marked in textbook)

Ans. As in multicellular organisms, all the cells are not in direct contact with environment,

simple diffusion does not meet the requirement of all the body cells to get sufficient

oxygen.

Q.2 What criteria do we use to decide whether something is alive?

Ans. Something can be considered alive if following features are present:

(i) Movements-

Visible movements like growth, breathing, etc.

Invisible movements - All the living organism must have movement at molecular level

like respiration.

(ii) Life process- Occurrence of processes like nutrition, respiration, transportation and

excretion to be called alive.

(iii) Growth

(iv) Reproduction

(v) Response to stimuli

Q.3 What are outside raw materials used for by an organism? (To be discussed) Ans. Outside raw materials used for by an organism includes:

(a) Water (b) Oxygen (c) CO2 (d) sunlight and (e) Nutrients/Minerals

These raw materials are used by an organism for various purposes like respiration,

growth and maintenance.

Q.4. What processes would you consider essential for maintaining life? (To be discussed) Ans. The processes essential for maintaining life are

a. Nutrition

b. Respiration

c. Transportation

d. Excretion

Page No. 101

Q.1 What are the differences between autotrophic and heterotrophic nutrition. (To be

discussed)

Autotrophic Nutrition Heterotrophic Nutrition

Organism prepares its own food using simple

inorganic molecules.

Organism obtain food from other organism

Energy required for synthesis of food No energy required

Digestion of food does not occur Generally digestion of food occur

Organisms act as producer Organisms act as consumer

Generally photosynthetic pigment (Chlorophyll)

present

Photosynthetic pigment (Chlorophyll) absent


Ex- Green plants ; Some bacteria and Some Protista Ex- Animals, Fungi, Some bacteria, Some

Protista and Some non green plants

Q.2 Where do plants get each of the raw materials required for photosynthesis? (To be

discussed) Ans. (a)Carbon dioxide from atmosphere.

(b) Light from Sun

(c) Water from Soil

(d) Minerals from soil

Q.3 What is the role of the acids in our stomach? Ans. HCl plays following role in our stomach:

(a) Make the medium acidic for action of enzyme pepsin.

(b) Kills the harmful bacteria present in food

(c) Prevents fermentation of food

(d) Softening of food

Q.4 What is the function of digestive enzymes? Give one example Ans. Digestive enzymes break-down the various complex components of food into simple and

soluble components so that they can be absorbed easily.

Eg., Starch salivary amylase

Maltose

Q.5 How is small intestine designed to absorb digested food? Ans. (i) The inner lining of small intestine has numerous finger-like projections called villi

which in turn have microvilli that increase the surface area for absorption.

(ii) The villi are richly supplied with blood vessels and Lacteals which transport the

absorbed food to each and every cells of the body.

Page No. 105

Q.1 What advantage over an aquatic organism does a terrestrial organism have

with regard to obtaining oxygen for respiration? (to be marked in the textbook)

Ans. The rate of breathing is slower in terrestrial organisms as compared to aquatic

organisms. This is due to the fact that in water, the amount of oxygen is less as

compared to air so, in aquatic organisms the rate of breathing is faster.

Q.2 What are different ways in which glucose is oxidized to provide energy in

various organisms? (To be discussed)

Ans. The pathways of break-down of glucose in various organisms are as below:


3. How is oxygen and carbon dioxide transported in human beings? (to be marked in

the textbook)

Ans. Transport of Oxygen:

(i) 97% carried in combined form with Haemoglobin.

(ii) 3% dissolved in plasma.

Transport of Carbon dioxide:

(i) 70% as Bicarbonate ion in plasma

(ii) 23% in combined state with Haemoglobin.

(iii) Upto 7% dissolved state in plasma.

Q.4 How are the lungs designed in human beings to maximize the area for exchange

of gases?

Ans. (i) In lungs, the bronchioles terminate in balloon-like structures called alveoli that

increase the surface area for exchange of gases.

(ii) The alveoli contain network of blood capillaries for easy exchange of gases.

TEXTBOOK EXERCISE

(Multiple choice questions to be done in the textbook)

Q.5 How are fats digested in our bodies? Where does this process take place?

Ans. Digestion of fats takes place in small intestine. Fats entering in intestine are in the form

of large globules. Bile juice breaks down these large globules into smaller globules

(Emulsification). After that fat digesting enzyme lipase present in pancreatic juice and

intestinal juice converts it into fatty acids and glycerol.

Q.6 What is the role of saliva in the digestion of food? (in the textbook)

Ans. (i) The saliva contains an enzyme called salivary amylase that breaks down starch

which is complex molecule into glucose.

(ii) Moistening & softening of food for easy chewing & mixing.

(iii) Helps in formation of slippery bolus for easy swallowing.

Q.7 What are the necessary conditions for autotrophic nutrition and what are its by-

products.

Ans. Conditions necessary for autotrophic nutrition are:

Light, Chlorophyll, Water, Carbon dioxide and Proper temperature.

By-products are: Oxygen and Water


Q.8 What are differences between aerobic and anaerobic respiration? Name

some organisms that use anaerobic mode of respiration.

Ans. Difference between aerobic and anaerobic respiration:

Anaerobic respiration takes place in yeast, some bacteria and some internal parasites like

tapeworm.

Q.9 How are the alveoli designed to maximize the exchange of gases? (To be discussed)

Ans. (i) The wall of the alveoli is folded and has large surface areas.

(ii) It contains an extensive network of blood vessels which provide a surface where the

exchange of gases can take place.

(iii) Wall of alveoli is thin/delicate & moist.

DELHI PUBLIC SCHOOL – GANDHINAGAR

ACADEMIC SESSION 2020 – 21

SCIENCE

CLASS X

Practice Questions: 1. What kind of image is formed on a cinema screen?

2. The depth of a bucket filled with water seems to be less than its actual depth. Name

the phenomenon responsible for this?

3. A light ray is incident on a rectangular glass slab at an angle of 60º. What is the angle

between the ray coming out of the slab and the normal to the face of rectangular slab

from which it comes out?

4. Calculate the angle of incidence of light ray incident on surface of a plastic slab of

refractive index , if the angle of refraction is 30º.

5. What type of lens is an air bubble inside the water?

6. You are provided with two lenses of focal length 20 cm and 30 cm, respectively.

Which lens will you use to obtain more convergent light?

7. Two lenses of power – 3.5 D and + 1D are placed in contact. Find the total power of

the combination of lens. Calculate the focal length of this combination.

8. Due to which property of light, sharp shadows of opaque objects are obtained?

9. What is the radius of curvature of plane mirror?

10. A ray is incident on a plane mirror as shown in figure.

What is the angle of reflection for the above incident ray?

11. Give the relationship of magnification with focal length of a spherical mirror.

12. What is the radius of curvature of the spherical mirror in the figure given, if PA = 10

cm?

a.

13. Is it possible that a convergent lens in one medium becomes divergent, when placed

in another medium?

14. What is the minimum distance between an object and its real image formed by a

convex lens?

15. When is the magnitude of the power of a lens equal to its focal length?

16. How can you distinguish between plane mirror, convex mirror and concave mirror by

merely looking at the image formed in each case? [CBSE 2011]

17. State two factors which determine lateral displacement of a ray of light passing

through a rectangular glass slab.

18. List out the factors on which refractive index of a medium depends.

19. Rohit while playing with an old lens discovers that, if he holds the lens 20 cm away

from a wall opposite to a window, he can see sharp but inverted images of outside

world on the wall. What is the power of the old lens holded by him?

20. A student wants to project the image of a candle flame on a screen 48 cm in front of a

mirror by keeping the flame at a distance of 12 cm from its pole. [CBSE 2014]

(i) Suggest the type of mirror he should use.

(ii) Find the linear magnification of the image produced.

(iii) How far is the image from its object?

(iv) Draw a ray diagram to show the image formation in this case.

21. (i) In refraction of light through a rectangular glass slab, the emergent ray is parallel

to the direction of the incident ray. Why?

22. (ii) What happens when a light ray is incident normally on one of the faces of a

rectangular glass slab? [CBSE 2012]

23. The image of an object formed by a lens is of magnification –1 . If the distance

between the object and its image is 60 cm, what is the focal length of the lens? If the

object is moved 20 cm towards the lens, where would the image be formed? State

reason and also draw a ray diagram in support of your answer. [CBSE 2016]

24. Ravi is given lenses with powers + 5 D, –5 D, +10 D, –10 D and –20 D. Considering

a pair of lenses at a time, which two lenses will he select to have a combination of

total focal length when two lenses are kept in contact in each case.

(i) –10 cm (ii) 20 cm (iii) –20 cm

25. It is desired to obtain an erect image of an object, using concave mirror of focal length

of 12 cm.

(i) What should be the range of distance of an object placed in front

of the mirror?

(ii) Will the image be smaller or larger than the object? Draw ray

diagram to show the formation of image in this case.

(iii) Where will the image of this object be, if it is placed 24 cm in

front of the mirror? Draw ray diagram for this situation also to

justify your answer.

26. Show the positions of pole, principal focus and the centre of curvature in the above

ray diagrams. [CBSE 2016]

27. What is meant by power of a lens? Define its SI unit.

28. You have two lenses A and B of focal lengths +10 cm and –10 cm, respectively. State

the nature and power of each lens. Which of the two lenses will form a virtual and

magnified image of an object placed 8 cm from the lens? Draw a ray diagram to

justify your answer. [CBSE 2015]

29. A dentist uses a mirror in front of a decayed tooth at a distance of 4 cm from the tooth

to get a four times magnified image in the mirror. Use mirror formula to find the focal

length and nature of the mirror used. [CBSE 2012]

30. A ray of light strikes the glass slab at an angle of 90º. What is the angle of incidence

and the angle of refraction? [CBSE 2012]

31. When two or more lenses are placed in contact, what will be their combined power?

[CBSE 2012] 32. A ray of light incident on one face of a rectangular glass slab emerges from the

opposite face of the slab parallel to the direction of the incident ray. Why does it

happen so? [CBSE 2012]

33. A ray passing through the centre of curvature of a spherical mirror after reflection,

returns along the same path. Why? [CBSE 2011]

34. Why does a ray of light bend from its path when it travels from one medium to other?

[CBSE 2011] 35. A ray of light travelling from a medium X enters obliquely into another medium Y. If

it bends away from the normal, then state which one of the two is relatively optically

denser. Why? [CBSE 2012]

36. Why is the refractive index of atmosphere different at different altitudes? [CBSE

2011] 37. For the same angle of incidence in medium P, Q and R, the angles of refraction are

45º, 35º and 15º, respectively. In which medium will the velocity of light be

minimum? Give reason for your answer. [CBSE 2012]

38. What is meant by refractive index? If the speed of light in a medium is 2/3 of the

speed of light in vacuum, find the refractive index of that medium. [CBSE 2011]

39. The radius of curvature of a concave mirror is 50 cm. Where an object should be

placed from the mirror, so as to form its image at infinity? Justify your answer.

[CBSE 2012

CLASS: 10 CHAPTER: 1 CHEMICAL REACTIONS AND EQUATIONS Page 1 of 4

Chapter: 1 Chemical Reactions and Equations

Pg. No. 6

Q.1 Why should a magnesium ribbon be cleaned before burning in air?

A.1 Magnesium is a reactive metal, so when kept in air for long it forms an inert layer of

magnesium oxide (MgO) on its surface. This inert layer of MgO prevents it from reacting

so this layer is removed before burning.

Q.2 Write the balance equation for the following reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

A.2 The chemical equations are as follows-

(i) →

(ii) →

(iii) →

Q.3 Write the balanced chemical equation with state symbols for the following reactions?

(i) Solutions of barium chloride and sodium sulphate in water react to give

insoluble barium sulphate and solution of sodium chloride.

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution

(in water) to produce sodium chloride solution and water.

A.3 Balance chemical reaction with state symbols are as follows:

(i) →

(ii) →

Pg. No. 10

Q.1 A solution of a substance ‘X’ is used for white washing

(i) Name the substance ‘X’ and writes its formula.

(ii) Write the reaction of the substance ‘X’ named in (i) above with water

(i)The substance whose solution is water is used for white washing is calcium oxide.

Its formula is CaO.

(ii)CaO(s) + H2O(l) Ca(OH)2(aq)

Delhi Public School, Gandhinagar

Class: X Subject: Science


Q.2

Why the amount of gas collected in one of the tubes in Activity 1.7 double of

the amount collected in the other? Name this gas.

The gas which is collected in double the amount in the electrolysis of water experiment

is hydrogen. This is because water contains two parts of hydrogen element as compared

to only one part of oxygen element.

Exercises

Q.4 What is balanced chemical equation? Why should chemical equation be balanced?

A.4 The reaction in which the number of atoms of each element is equal on the reactant side and

product side is called balanced equation.

Chemical reaction should be balanced because only a balanced equation tells us the relative

quantities of different reactants and products involved in the reaction.

Q.5

(a)

(b)

(c)

(d)

Translate the following statements into chemical equations and then balance them.

Hydrogen gas combines with nitrogen to form ammonia.

Hydrogen sulphide gas burns in air to give water and Sulphur dioxide.

Barium chloride reacts with aluminum sulphate to give aluminum chloride

and precipitate of barium sulphate.

Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

A.5

(a)

(b)

(c)

(d)

Q.6

(a)

(b)

(c)

(d)

Balance the following chemical equations:

→

→

→

→

A.6

(a)

(b)

(c)

(d)

Balanced chemical equations are:

→

2 →

→

→


Q.7 (a)

(b)

(c)

(d)

Write the balanced chemical equations for the following reactions.

Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water

Zinc + Silver nitrate → Zinc nitrate + Silver

Aluminum + Copper chloride → Aluminum chloride +Copper

Barium chloride + Potassium sulphate → Barium sulphate + potassium chloride

A.7

Balanced chemical equations are:

(a) →

(b) →

(c) →

(d) →

Q.8

(a)

(b)

(c)

(d)

Write the balanced chemical equation for the following and identify the type

of reaction in each case.

Potassium bromide(s)+Barium iodide(aq)→Potassium iodide(aq)+Barium bromide

Zinc carbonate (s) → Zinc oxide (s) + Carbon dioxide (g)

Hydrogen (g) + Chlorine (g) → Hydrogen chloride (g)

Magnesium (s)+ Hydrochloric acid (aq) → Magnesium chloride(aq)+Hydrogen (g)

A.8 (a)

(b)

(c)

(d)

→ (Double displacement reaction)

→ (Decomposition reaction)

→ (Combination reaction)

→ (Displacement reaction)


Chapter 1 Chemical reactions and equations Extra Questions 2020-21

Q.1 Decomposition reactions require energy either in the form of heat or light or

electricity for breaking down the reactants. Write one equation each for

decomposition reactions where energy is supplied in the form of heat, light and

electricity.

Q.2 Write chemical equation for the reactions taking place when:

(i) Iron reacts with stem

(ii) Magnesium reacts with dilute HCl

(iii) Copper is heated in air.

Q.3 Write the balanced chemical equations for the following reactions:

(a) Silver bromide on exposure to sunlight decomposes into silver and bromine.

(b) Sodium metal reacts with water to form sodium hydroxide and hydrogen gas.

Q.4 (a) A solution of a substance ‘X’ is used for testing carbon dioxide. Write the

chemical equation for the reaction of ‘X’ with carbon dioxide.

(b) How is ‘X’ obtained? Write the chemical reaction.

Q.5 Consider the chemical equations given below and answer the questions.

(a)

(b) HEAT

(c) + Cu

(d) +

(e) +

(f) +

Delhi Public School, Gandhinagar

Class: X Subject: Science

Find some changes

Hope you are familiar about these changes

Tell me if these changes are reversible or irreversible !!

Lets define a chemical change

You need to describe a lot

Clean a magnesium ribbon about 3-4 cm long by rubbing it with sandpaper.

Hold it with a pair of tongs. Burn it using a spirit lamp or burner and collect the ash so formed in a watch-glass as shown in Figure.

Burn the magnesium ribbon keeping it away as far as possible from your eyes.

What do you observe?

Changes are two types: Physical and Chemical

Take lead nitrate solution in a test tube.

Add potassium iodide solution to this.



Figure 1.2. Color change due to the mixing of lead nitrate and potassium iodide

Lets define a chemical reaction


Take a few zinc granules in a conical flask or a test tube.

Add dilute hydrochloric acid or sulphuric acid to this.

Do you observe anything happening around the zinc granules?

Touch the conical flask or test tube. Is there any change in its temperature?

Figure 1.3. Formation of hydrogen gas by the action of dilute sulphuric acid on zinc

Take a small amount of calcium oxide or quick lime in a beaker

Slowly add water to this. Touch the beaker. Is there any change in its

temperature?

Example: Generation of heat due to the reaction of calcium oxide with water


You need to describe a lot

Symbolic description is required

How will you understand that a chemical reaction is taking place !!

Changes in state: As in Activity 1.1

Changes in color: As in Activity 1.2

Evolution of gas: As in Activity 1.3

Change in temperature: As shown in Example

Chemical Equations

Reactant: A substance that takes part in the reaction and undergoes changes during reaction

Product: Substances that are produced due to the reaction

Word-equation shows change of reactants to products through an arrow placed between them. Reactants are written on LHS with a plus sign (+) between them. Similarly, products are written on RHS with a plus sign (+) between them.

Representation of change observed in figure into word equation

Understand this word equation

Word equation to Chemical equation

Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., are indicated above and/or below the arrow in the equation.

Writing Symbols of Physical States

Mg (s) + O2 (g) → MgO (s)


No. of atoms of Mg in LHS = 1 No. of atoms of Mg in RHS = 1

No. of atoms of O in LHS = 2 No. of atoms of O in RHS = 1

The equation is Not Balanced

The above equation is known as SKELETAL EQUATION

Example of balanced equation

No. of C atoms in LHS = 1 No. of C atoms in RHS = 1

No. of O atoms in LHS = 2 No. of O atoms in RHS = 2

This is Balanced Equation


Lead nitrate + Potassium iodide → Lead iodide + Potassium nitrate

Pb(NO3)2 + KI → KNO3 + PbI2

Elements No. of atoms in LHS No of atoms in RHS

Pb N O K I

1 2 6 1 1

1 1 3 1 2

Is this a Balanced Equation!!

Zinc + Sulphuric acid → Zinc sulphate + Hydrogen




Calcium oxide + Water → Calcium hydroxyde

CaO + H2O → Ca(OH)2

Elements No. of atoms in LHS No of atoms in RHS

Ca O H

1 2 2

1 2 2


We will practice more

Write the word equation followed by the chemical equations for the following chemical reactions

1. Calcium reacts with oxygen to form calcium oxide.

2. Nitrogen reacts with hydrogen to form ammonia.

3. Sodium reacts with water to form sodium hydroxide.

Calcium reacts with oxygen to form calcium oxide.

Calcium + Oxygen → Calcium oxide

Ca + O → CaO

Nitrogen reacts with hydrogen to form ammonia.

Nitrogen + Hydrogen → Ammonia

N2 + H2 → NH3

Sodium reacts with water to form sodium hydroxide.

Sodium + Water → Sodium hydroxide

Na + H2O → NaOH +H2

Can you tell me

What will be our next activity!!

Let us try to balance this chemical equation

Step I:

Step II

Let us try to balance this chemical equation

Step III

Step IV

Step V

Step VI

Step VII: Writing Symbols of Physical States

What is a balanced chemical equation? Why should chemical equations be balanced?

A chemical equation should be balanced to obey the law of conservation of mass i.e. the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.

1. Why should a magnesium ribbon be cleaned before burning air ?

The magnesium ribbon should be cleaned before burning in the air because the layer of magnesium oxide (which is formed due to the reaction of magnesium with air) can be removed in order to get the desired chemical reaction. So cleaning helps the burning process takes place easily.

2. Write the balanced equation for the following chemical reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride H2 + Cl2 → HCl

LHS: No. of H atom: 02; No. of Cl atom: 02

RHS: No. of H atom: 01, No. of Cl atom: 01

H2 + Cl2 = 2HCl

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

BaCl2 + Al2(SO4)3 → AlCl3 + BaSO4

Balanced

Not balanced

Not balanced

Make no. of SO42- same in both side

BaCl2 + Al2(SO4)3 → AlCl3 + 3 BaSO4

Make no. of Ba atom same in both side

3 BaCl2 + Al2(SO4)3 → AlCl3 + 3 BaSO4 Not balanced

3 BaCl2 + Al2(SO4)3 = 2 AlCl3 + 3 BaSO4

Make no. of Cl and Al atoms same in both side

(iii) Sodium + Water → Sodium hydroxide + Hydrogen Na + H2O → NaOH + H2

LHS: No. of Na atom: 01; No. of H atom: 01; No. of O atom: 01

RHS: No. of Na atom: 01, No. of O atom: 01; No. of H atom: 03

2Na + 2H2O = 2NaOH + H2

3. Write a balanced chemical equation with state symbols for the following reactions. i. Solutions of barium chloride and sodium sulphate in water react to give

insoluble barium sulphate and the solution of sodium chloride.

BaCl2 (aq) + Na2SO4 (aq) → BaSO4↓ + NaCl (aq)

BaCl2 (aq) + Na2SO4 (aq) = BaSO4↓ + 2NaCl (aq)

Not balanced

Balanced

3. Write a balanced chemical equation with state symbols for the following reactions. ii. Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)

Balanced

NaOH (aq) + HCl (aq) = NaCl (aq) + H2O (l)

Chemical Reactions

Atoms of one element do not change into those of another element.

Atoms do not disappear from the mixture or appear from

elsewhere. Chemical reactions involve the breaking and making of bonds

between atoms to produce new substances

Types of Chemical Reactions

1. Combination 2. Decomposition 3. Displacement 4. Double Displacement 5. Oxidation and Reduction

Figure 1.3. Formation of slaked lime by the reaction of calcium oxide with water

Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat.

CaO(s) + H2O(l) → Ca(OH)2(aq) + Heat (∆) (Quick lime) (Slaked lime)

Note: Single product is formed from two or more reactants

Combination Reaction

Do you know the use of slaked lime!!

Combination Reaction

Slaked lime (Ca(OH)2) is used for whitewashing walls.

Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) (Calcium hydroxide) (Calcium carbonate)

Calcium hydroxide reacts slowly with the carbon dioxide in air to form a thin layer of calcium carbonate on the walls.

Do you know that common name of Calcium carbonate is Marbel

More examples of combination reactions

(i) C(s) + O2(g) → CO2(g)

(ii) 2H2(g) + O2(g) → 2H2O(l)

CaO(s) + H2O(l) → Ca(OH)2(aq) + Heat

Reactions in which heat is released along with the formation of products are called Exothermic Reactions.

Can you give more examples of Exothermic Reactions?

(i) Burning of natural gas

CH4(g) + 2O2 (g) → CO2 (g) + 2H2O (g) + Heat

(ii) Do you know that respiration is an exothermic process?

Carbohydrates → Glucose O2 Energy

C6H12O6(aq) + 6O2(aq) → 6CO2(aq) + 6H2O(l) + Energy

(Glucose)

(iii) The decomposition of vegetable matter into compost

is also an example of an exothermic reaction

Decomposition Reaction

Take about 2 g ferrous sulphate crystals in a dry boiling tube. Note the colour of the ferrous sulphate crystals. Heat the boiling tube over the flame of a burner or spirit lamp as shown in Fig. Observe the colour of the crystals after heating.

Figure 1.4: Correct way of heating the boiling tube containing crystals of ferrous sulphate and of smelling the odour

2FeSO4(s) Fe2O3(s) + SO2(g) + SO3(g) (Ferrous sulphate) (Ferric oxide)

Heat

Green colour of the ferrous sulphate crystals will change. Characteristic odour of burning sulphur is obtained.

Note: Single reactant breaks down to give simpler products. This is a decomposition reaction.

Ferrous sulphate crystals (FeSO4, 7H2O) lose water when heated and the colour of the crystals changes. It then decomposes to ferric oxide (Fe2O3), sulphur dioxide (SO2) and sulphur trioxide (SO3).

Ferric oxide is a solid, while SO2 and SO3 are gases.

Can you give more examples of Decomposition Reactions?

FeSO4, 7H2O Heat FeSO4 (Ferrous sulphate crystals) -H2O

CaCO3(s) CaO(s) + CO2(g) (Limestone) (Quick lime)

Heat

It is an important decomposition reaction used in various industries including manufacture of cement.

When a decomposition reaction is carried out by heating, it is called

thermal decomposition.

Figure 1.5: Heating of lead nitrate and emission of nitrogen dioxide

Take about 2 g lead nitrate powder in a boiling tube.

Hold the boiling tube with a pair of tongs and heat it over a flame, as shown in Fig. 1.5.

What do you observe? Note down the change, if any.

2Pb(NO3)2(s) 2PbO(s) + 4NO2(g) + O2(g) (Lead nitrate) (Lead oxide) (Nitrogen dioxide) (Oxygen)

Heat

We will observe the emission of brown fumes. These fumes are of nitrogen dioxide (NO2).

Activity 1.6

www.youtu.be/O_1H2Cjb830

http://www.youtu.be/O_1H2Cjb830

Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in Fig. 1.6.

Connect these electrodes to a 6 volt battery. Fill the mug with water such that the electrodes

are immersed. Add a few drops of dilute sulphuric acid to the water.

Take two test tubes filled with water and invert them over the two carbon electrodes.

Switch on the current and leave the apparatus undisturbed for some time.

You will observe the formation of bubbles at both the electrodes. These bubbles displace water in the test tubes.

Is the volume of the gas collected the same in both the test tubes?

Once the test tubes are filled with the respective gases, remove them carefully.

Test these gases one by one by bringing a burning candle close to the mouth of test tubes.

Figure 1.6: Electrolysis of water

Activity 1.7

CAUTION: This step must be performed carefully. What happens in each case? Which gas is present in each

test tube?

(+) (-)

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http://www.youtu.be/R5msLljLJBI

Take about 2 g silver chloride in a china dish.

What is its colour?

Place this china dish in sunlight for some time (Fig. 1.7).

Observe the colour of the silver chloride after some time.

Activity 1.8

Figure 1.7. Silver chloride turns grey in sunlight to form silver metal

2AgCl(s) 2Ag(s) + Cl2(g) Sunlight

2AgBr(s) 2Ag(s) + Br2(g) Sunlight

The above reactions are used in black and white photography.

Decomposition reactions require energy either in the form of heat, light or electricity for breaking down the reactants.

Reactions in which energy is absorbed are known as

endothermic reactions.

Carry out the following Activity

Take about 2 g barium hydroxide in a test tube. Add 1 g of ammonium chloride and mix with the help of a glass rod. Touch the bottom of the test tube with your palm. What do you feel? Is this an exothermic or endothermic reaction?

Ba(OH)2 + 2NH4Cl BaCl2 +2NH3 + 2H2O

This is an endothermic reaction

Figure 1.8. (a) Iron nails dipped in copper sulphate solution

Take three iron nails and clean them by rubbing with sand paper. Take two test tubes marked as (A) and (B). In each test tube, take about 10 mLcopper sulphate

solution. Tie two iron nails with a thread and immerse them carefully in the copper sulphate solution in test

tube B for about 20 minutes [Fig. 1.8 (a)]. Keep one iron nail aside for comparison. After 20 minutes, take out the iron nails from the copper sulphate solution. Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B) [Fig.

1.8 (b)]. Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept

aside [Fig. 1.8 (b)].

Activity 1.9

Figure 1.8 (b) Iron nails and copper sulphate solutions compared before and after the experiment

Displacement Reaction

Why does the iron nail become brownish in colour and the blue colour of copper sulphate solution fades?

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) (Copper sulphate) (Iron sulphate)

Iron has displaced another element, copper, from copper sulphate solution. This reaction is known as displacement reaction

Other examples of displacement reactions

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) (Copper sulphate) (Zinc sulphate)

Pb(s) + CuCl2(aq) → PbCl2(aq) + Cu(s) (Copper chloride) (Lead chloride)

Zinc and lead are more reactive elements than copper. They displace copper from its compounds.

Double Displacement Reaction

Figure 1.9. Formation of barium sulphate and sodium chloride

Activity 1.10

Take about 3 mL of sodium sulphate solution in a test tube.

In another test tube, take about 3 mL of barium chloride solution.

Mix the two solutions (Fig. 1.9).


It will be observed that a white substance, which is insoluble in water, is formed. This insoluble substance formed is known as a precipitate.

Any reaction that produces a precipitate can be called a precipitation reaction.

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) ↓ + 2NaCl(aq) (Sodium sulphate) (Barium chloride) (Barium sulphate) (Sodiumchloride)

The white precipitate of BaSO4 is formed by the reaction of SO42– and Ba2+. The

other product formed is sodium chloride which remains in the solution. Such reactions in which there is an exchange of ions between the reactants are

called double displacement reactions.

Take lead nitrate solution in a test tube.

Add potassium iodide solution to this.


What do you observe? Figure 1.2. Color change due to the mixing of lead nitrate and potassium iodide

Recall this activity

Pb(NO3)2 + 2KI →2KNO3 + PbI2

Is it a double displacement reaction!!

Oxidation and Reduction

Figure 1.10: Oxidation of copper to copper oxide

Heat a china dish containing about 1 g copper powder (Fig. 1.10).


Activity 1.11

The surface of copper powder becomes coated with black copper(II) oxide. Why has this black substance formed?

2Cu + O2 2CuO Heat

If hydrogen gas is passed over this heated material (CuO), the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

CuO +H2 Cu+H2O Heat

2Cu + O2 2CuO Heat

CuO +H2 Cu+H2O Heat

If a substance gains oxygen during a reaction, it is said to be oxidised. If a substance loses oxygen during a reaction, it is said to be reduced.

Can you tell me which one is oxidised and reduced in aforementioned reactions!!

CuO +H2 Cu+H2O

Copper(II) oxide is losing oxygen and is being reduced. Hydrogen is gaining oxygen and is being oxidised.

Heat Such reactions are called oxidation-reduction

reactions or redox reactions.

Reduction

Oxidation

ZnO + C Zn +CO

MnO2 + 4HCl MnCl2 + 2H2O + Cl2

More examples of Redox Reactions

Carbon is oxidised to CO and ZnO is reduced to Zn.

HCl is oxidised to Cl2 whereas MnO2 is reduced to MnCl2.

If a substance gains oxygen or loses hydrogen during a reaction, it is oxidised.

If a substance loses oxygen or gains hydrogen during a reaction, it is reduced.

Reduction

Oxidation

Reduction

Oxidation

Recall Activity 1.1, where a magnesium ribbon burns with a dazzling flame in air (oxygen) and changes into a white substance, magnesium oxide. Is magnesium being oxidised or reduced in this reaction?

What is your answer !!

Effects of oxidation reactions in everyday life.

Iron articles are shiny when new, but get coated with a reddish brown powder when left for some time. This process is commonly known as rusting of iron. Some other metals also get tarnished in this manner.

Have you noticed the colour of the coating formed on copper and silver?

When a metal is attacked by substances around it such as moisture, acids, etc., it is said to corrode and this process is called corrosion. The black coating on silver and the green coating on copper are other examples of corrosion. Corrosion causes damage to car bodies, bridges, iron railings, ships and to all objects made of metals, specially those of iron.

Corrosion of iron is a serious problem. Every year an enormous amount of money is spent to replace damaged iron. You will learn more about corrosion in Chapter 3.

Corrosion

When fats and oils are oxidised, they become rancid and their smell and taste change.

Usually substances which prevent oxidation (antioxidants) are added to foods containing fats and oil.

Keeping food in air tight containers helps to slow down oxidation.

Do you know that chips manufacturers usually flush bags of chips with gas such as nitrogen to prevent the chips from getting oxidised?

Rancidity

Thank You

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