CBSE Class 10

NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds

In this chapter, students will learn about bonding in carbon –The Covalent Bond, versatile nature of Carbon, saturated and unsaturated carbon compounds, chemical properties of carbon compound, soaps and detergents.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 4 – Carbon and Its Compounds solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 4 – Carbon and Its Compounds Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂ ?
Answer:
The atomic number (Z) for carbon is six and its electronic configuration is 2, 4. Carbon has four valence
electrons. Each oxygen atom (Z = 8) has six valence electrons (2, 6). In order to complete its octet, the carbon atom shares its four valence electrons with the four electrons of the two oxygen atoms as follows :

Thus, in carbon dioxide molecule, the carbon atom is linked to two oxygen atoms on both sides by two shared pairs of electrons resulting in double bonds on either sides. Both carbon and oxygen atoms complete their octet as a result of electron sharing.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur ?
Answer:
The atomic number (Z) of sulphur is sixteen and its electronic configuration is 2, 8, 6. The sulphur atom has six valence electrons. The chemical formula of sulphur molecule is S₈. Each sulphur atom is linked to similar atoms on either sides by single covalent bonds and thus, completes its octet. The molecule is in the form of a ring also represented by crown shape.
Ring structure of S₈ molecule Crown shape of S₈ molecule

Question 3.
How many structural isomers you can you draw for pentane ?
Answer:
Pentane (C₅H₁₂) has a skeleton of five carbon atoms. It can exist as a straight chain as well as two branched chains. There are three structural isomers for the hydrocarbon which is an alkane.

Question 4.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us ?
Answer:

1. Catenation: Carbon has the unique property of self linking which is known as catenation. In fact, any number of carbon atoms can be linked to one another by covalent bonds. This is on account of the stability of C—C bonds since the size of the carbon atom is quite small.
2. Linking of carbon with other atoms. Carbon is tetravalent in nature and can readily unite with atoms like hydrogen, oxygen, nitrogen, sulphur etc. by electron sharing.

Question 5.
What will be the formula and electron dot structure of cyclopentane ? (CBSE 2013)
Answer:
Cyclopentane is a cyclic compound with formula C₅H₁₂ The structure of the compound may be represented as :

Question 6.
Draw the structures of the following compounds :
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone Are structural isomers possible for bromopentane ?
Answer:

Bromopentane has a chain of five carbon atoms. It can exist in a number of forms which are structural isomers.

The structural isomers (i), (ii) and (iii) which differ in the position of the Br atom are known as position isomers.
The structural isomers (iv), (v) and (vi) which differ in the arrangement of carbon atoms in the chain are called chain isomers.
In writing the IUPAC name, the name of prefix bromo is written before that of prefix methyl. In fact, alphabetical order is followed while naming the different prefixes.

Question 7.
How would you name the following compounds ?

Answer:
(i) Bromoethane
(ii) Hex-l-yne
(iii) Methanal

Question 8.
Why is the conversion of ethanol into ethanoic acid an oxidation reaction ?
Answer:
Ethanoic acid (CH₃COOH) has one oxygen atom more and two hydrogen atoms less than ethanol (C₂H₅OH). In general,

• Loss of hydrogen is known as oxidation.
• Gain of oxygen is known as oxidation.

Therefore, it is an oxidation reaction.

Question 9.
A mixture of ethyne and oxygen is used for welding. Can you tell why a mixture of ethyne and air is not used ?
Answer:
When ethyne is burnt in oxygen, large quantity of heat alongwith light is produced. The heat evolved can be used for gas welding which is usually carried to weld small broken pieces of articles made up of iron.

Air mainly contains a mixture of nitrogen (4 parts) and oxygen (1 part). As we know, nitrogen gas does not support combustion. This means that in air, only oxygen will help in the combustion of ethyne. Therefore, it is always better to use oxygen for the combustion of ethyne.

Question 10.
How would you distinguish experimentally between an alcohol and a carboxylic acid ?
Answer:
The distinction can be made by the following tests :

1. Dip a strip of blue litmus separately in both alcohol and carboxylic add taken in two glass tubes. The colour will change to red in the tube containing carboxylic acid and not in the tube which contains alcohol.
2. Add a small amount of solid sodium hydrogen carbonate (NaHCO₃) in both the tubes. A brisk effervescence accompanied by bubbles will be noticed in the tube containing carboxylic acid and not in the tube containing alcohol.

Question 11.
What are oxidising agents ?
Answer:
Oxidising agents are the substances which either their own or on reacting with another substance release
oxygen in order to carry oxidation reactions. The commonly used oxidising agents are : Ozone, bromine
water, a mixture of potassium dichromate and sulphuric acid or a mixture of potassium permanganate and sulphuric acid etc.

Question 12.
Would you be able to check if water is hard by using a detergent ?
Answer:
No, it is not possible. Actually detergents produce foam in any type of water ; whether hard or soft.
Therefore, a distinction between the two cannot be made. However, soaps can be used for this purpose.

Question 13.
People use a variety of methods to wash clothes. Usually after adding soap, they beat the clothes on a stone or beat them with a paddle, scrub with a brush or the mixture is agitated in a washing machine Why is this agitation necessary to get clean clothes ?
Answer:
The purpose of soap or detergent in washing is to reduce friction between oil drops carrying dirt particles and water so that they may mix with each other. All the methods that have been suggested loosen the bonds between the dust or oil particles and fabrics of clothes. The agitation helps in washing the clothes.

NCERT End Exercise

Question 1.
Ethane, with the molecular formula C₂H₆ has :
(a) G covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Answer:

(b). The molecule has seven covalent bonds.

Question 2.
Butanone is a four carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol.
Answer:
(c). The functional group is ketone (>C=0) also known as one.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely.
Answer:
(b). The fuel is not burning completely. The unburnt particles present in smoke blacken the vessel from outside.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The molecule of chloromethane (CH₃Cl) consists of three elements i.e., carbon (Z = 6) hydrogen (Z = 1) and chlorine (Z = 17). Carbon atom has four valence electrons (2, 4) ; hydrogen has one (1) while chlorine has seven electrons in the valence shell (2, 8, 7). In order to complete its octet, carbon shares three valence electrons with three hydrogen atoms while one is shared with the electron of chlorine atom. The structure of covalent molecule may be written as follows :

Question 5.
Draw the electron dot structures for
(i) ethanoic acid
(ii) H₂S
(iii) propanone
(iv) F₂.
Answer:

Question 6.
What is homologous series ? Explain with an example.
Answer:
A series of similarly constituted compounds in which the members present have the same functional group, same chemical properties and any two successive members in a particular series differ in their molecular formula by —CH₂ group and molecular mass by 14 u.
For example, the boiling points of members in the family of alkanes follow the order :

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties ?
(CBSE Delhi 2011)

Answer:
Distinction based on physical properties :

1. Smell. Ethanol has a characteristic smell known as alcoholic smell which is pleasant. Ethanoic acid has vinegar like smell.
2. Boiling points. Boiling point of ethanol (351 K) is less than that of ethanoic acid (391 K).
3. Litmus test. Ethanol is neutral in nature and does not bring any change in the colour of litmus whether blue or red. Ethanoic acid is acidic and changes the colour of a blue litmus strip to red when dipped in it.

Distinction based on chemical properties :

1. Action with sodium hydrogen carbonate. On adding a small amount of sodium hydrogen carbonate to ethanoic acid, carbon dioxide gas is evolved with brisk effervescence. However, no such reaction is noticed in case of ethanol.
   CH₃COOH + NaHCO₃ ———> CH₃COONa + CO₂ + H₂O
2. Action with caustic alkalies. Ethanoic reacts with both sodium hydroxide (NaOH) and potassium hydroxide (KOH) to form corresponding salt and water. Ethanol fails to react with either of these.
   

Question 8.
Why does micelle formation take place when soap is added to water ? Will a micelle be formed in other solvents such as ethanol also ? (CBSE Delhi 2011)
Answer:
Soap may be represented by the formula RCOO^–Na⁺ where R is an alkyl group which represents long chain of carbon with fifteen or more atoms. Now, oil drops containing dirt particles and water donot mix. Soap helps in their mixing by reducing interfacial tension or friction. Actually it forms a sort of bridge between oil drops and water in which the alkyl portion (hydrophobic end) points towards oil drop while other portion COONa (hydrophilic end) is directed towards water. This is known as micelle formation. Thus, soap helps in the formation of a stable emulsion between oil and water. Ethanol and other similar solvents which are of organic nature, donot help in micelle formation because soap is soluble in them.

Question 9.
Why are carbon and its compounds used as fuels in most cases ?
Answer:
Carbon burns in oxygen or air to form carbon dioxide gas. The reaction is highly exothermic. That is why different forms of coal are used as fuels. The most important compounds of carbon are hydrocarbons. Just like carbon, hydrogen also readily burns in oxygen or air to form water producing heat. The hydrocarbon methane (CH₄) is a constituent of natural gas. Propane (C₃H₈) and butane (C₄H₁₀) are present in liquid petroleum gas (L.P.G.). Petrol and kerosene also contain different hydrocarbons. Therefore, these are used as fuels.

Question 10.
Explain the formation of scum when hard water is treated with soap. (CBSE Delhi 2011)
Answer:
Soap is basically sodium or potassium salt of higher fatty acids. Hard water contains in it Ca²⁺ and Mg²⁺ ions as their salts. When soap is added to hard water, the corresponding calcium and magnesium salts are formed. These are in the form of precipitates, also called ‘scum’.

Question 11.
What change will you observe by testing soap with litmus paper (blue or red) ?
Answer:
When soap is dissolved in water, the solution is alkaline in nature due to the formation of alkali NaOH or KOH. The solution changes the colour of red litmus to blue. However, the solution does not change the colour of blue litmus.

Question 12.
What is hydrogenation ? What is its industrial application ?
Answer:
The addition of hydrogen to unsaturated hydrocarbon in the presence of a metal catalyst is also known as catalytic hydrogenation.
The reaction is extremely useful in the hydrogenation of vegetable oils also called edible oils e.g. ground nut oil, cotton seed oil etc. These are also called cooking oils and are unsaturated in the sense that their molecules contain atleast one C=C bond in their structures. Upon passing hydrogen gas through oil in the presence of nickel catalyst, the double bond changes to single bond. As a result, the unsaturated oil changes to solid fat which is of saturated nature. Vegetable ghees such as Dalda, are of saturated nature and are formed by catalytic hydrogenation reaction.

Question 13.
Which of the listed hydrocarbons undergo addition reactions : C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄ ?
Answer:
In order that a hydrocarbon may undergo addition reaction, it must be unsaturated in nature. It must be either an alkene (C=C) with general formula C_nH_2n or an alkyne (C = C) with general formula C_nH_2n-2. Out of the list of the hydrocarbons given :

• C₃H₆ (Propene) is an alkene with C=C bond. It corresponds to general formula C_nH_2n (n = 3)
• C₂H₂ (Ethyne) is an alkyne with C = C bond. It corresponds to general formula C_nH_2n-2 (n = 2).

Both these hydrocarbons take part in addition reactions. For example, they react with hydrogen upon heating to 473 K in the presence of Nickel catalyst to form corresponding alkanes.

Question 14.
Give a test that can be used to differentiate between butter and cooking oil.
Answer:
Butter is saturated in nature while cooking oil is unsaturated. This means that cooking oil has atleast one C=C bond present in the constituting compounds while butter does not have any such bond. The distinction between them can be made by reacting with bromine water or bromine dissolved in carbon tetrachloride. Cooking oil will discharge the yellow colour of bromine while butter will not.

Question 15.
Explain the mechanism of cleansing action of soap.
Answer:
Cleansing action of Soaps and Detergents
Both soaps and detergents resemble in their cleansing action. They have two main parts. These are non—polar hydrocarbon chain which is water repellent or hydrophobic (phobic or phobia stands for repulsion or hatred) and the polar carboxyl group as its salt which is attracted towards water or is hydrophilic (philic stands for love or attraction). The former is called the tail of the molecule and the latter is regarded as the head.

In order to understand the cleansing action of both soaps and detergents, let us try to analyse how the clothes become dirty. They first become oily because of the perspiration coming out of the skin and also from the organic matter dispersed in the atmosphere. Dust particles stick to oil drops and the clothes become dirty. In order to wash these, they are dipped in water and soap or detergent is applied. In solution, it dissociates to give carboxylate ions (RCOO ) or sulphonate ions (RSO₃^–) and the cations (Na⁺). In the carboxylate ion, the alkyl portion which contains a long chain of hydrocarbons is a tail pointing towards the oil drops while the COO^– portion is the head directed towards water. In a detergent it is the SO₃^– portion which points towards water. This is quite evident from the figure where the solid circles (•) represent the polar groups and the wavy lines (^^^) represent the alkyl portions. The molecules of soap or detergent have a unique orientation in water. They actually form a cluster of molecules in which the hydrophobic or alkyl portion is in the interior while the ionic or polar portion is on the surface of the cluster as shown in the figure. This formation is known as micellear formation or simply micelle. Soaps or detergent thus help in forming a stable emulsion of oil and water by acting as a bridge between the two. The oil droplets alongwith the particles of the dirt get detached from the fibres of the clothes and pass into the emulsion. In this manner, the clothes become free from dust or dirt.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Hope given NCERT Solutions for Class 10 Science Chapter 4 helpful to you.

NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

In this chapter, students will learn about the physical properties of metals, and non-metals, chemical properties of metals, how metals react with air, water, acid, other solutions, and metal salts, reactivity series.

Further students will come to know how metals and non-metals react, properties of ionic compounds, the occurrence of metals, extraction of metals, refining of metals, corrosion, prevention of corrosion. The chapter contains questions and answers exercise along with multiple choice questions.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 3 – Metals and Non-metals solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 3 – Metals and Non-metals Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 3 NCERT Questions

NCERT Solutions for Class 10 Science Chapter 3 In Text Book Questions

Question 1.
Give example of a metal which
(a) is a liquid at room temperature
(b) can be easily cut with a knife
(c) is the best conductor of heat
(d) is the poorest conductor of heat.
Answer:
(a) Mercury
(b) Sodium
(c) Silver
(d) Lead.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Explain the meaning of malleable and ductile.
Answer:
Malleable: The property due to which a substance can be beaten into sheets is known as malleability. Metals are malleable in nature.
Ductile. The property due to which a substance can be drawn into wires is known as ductility. Metals are ductile in nature.

Question 3.
Why is sodium kept immersed in kerosene oil ? (CBSE 2011)
Answer:
Sodium reacts with air and water both. It is a highly reactive metal. When kept in open, it readily combines with oxygen present in air to form its oxide. Similarly, it reacts with water vapours or moisture to form sodium hydroxide.

In order to preserve sodium metal, we generally keep it under kerosene so that neither air nor moisture may come in its contact.

Question 4.
Write the equations for the reactions of
(a) iron with steam
(b) calcium with water
(c) potassium with water.
Answer:
(a) 3Fe(s) + 4H₂O (steam) ———–> Fe₃O₄(s) + 4H₂(g)
(b) Ca(s) + 2H₂O(aq) ———–> Ca(OH)₂(s) +H₂(g)
(c) 2K(s) + 2H₂O(aq) ———–> 2KOH(aq) + H₂(g)

Question 5.
Samples of four metals A, B, C and D were taken and were added to the following solutions one by one. The results obtained have been tabulated as follows :

Use the table given above to answer the following questions :
(a) Which is the most reactive metal ?
(b) What would you observe when B is added to solution of copper(II) sulphate ?
(c) Arrange the metals A, B, C and D in order of increasing reactivity. (CBSE 2011)
Answer:
Based on the activity series, the relative position of the metals in which involved in solutions is : Zn > Fe > Cu > Ag. On the basis of the results given in the table .

• Metal A is more reactive than copper and less reactive than iron.
• Metal B is more reactive than iron and less reactive than zinc. –
• Metal C is only more reactive than silver and less reactive than other metals.
• Metal D is the least reactive in nature.

In the light of above information, we can conclude that
(a) Metal B is the most reactive.
(b) Since B is more reactive than iron, it is also more reactive than copper. This means that it would displace copper from copper(II) sulphate solution. The blue colour of solution will slowly fade.
(c) The decreasing order of reactivity of metals is: B>A>C>D.

Question 6.
Which gas is produced when a reactive metal reacts with dilute hydrochloric acid ? Write the chemical reaction when iron reacts with dilute H₂SO₄. (CBSE 2010)
Answer:
Hydrogen gas (H₂) is produced when a reactive metal reacts with dilute hydrochloric acid. Iron and dilute H₂SO₄ react as follows :
Fe(s) + H₂SO₄(aq) ————> FeSO₄(aq) + H₂(g)
Hydrogen gas is evolved in this reaction also.

Question 7.
What would you observe when zinc is added to a solution of iron (II) sulphate ? Write the chemical reaction that takes place. (CBSE 2010)
Answer:
The green colour of the solution would slowly disappear. Zinc would gradually dissolve and iron would get precipitated at the bottom of the beaker.

Question 8.
(i) Write electron-dot structures for sodium, magnesium and oxygen.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds ?
Answer:
(i)

(ii) Formation of sodium oxide (Na₂O)

Formation of magnesium oxide (MgO)

(iii) For answer, consult structures given above.

Question 9.
Why do ionic compounds have high melting points ? (CBSE 2014)
Answer:
In the formation of ionic compounds, positive ions (cations) and negative ions (anions) participate. These are closely packed and the ionic compounds exist as crystalline solids. They have strong inter ionic forces of attraction and have high melting and boiling points.

Question 10.
Define the following terms :

1. Minerals
2. Ores
3. Gangue.

Answer:

1. Minerals : These are the combined states of metals and non-metals present in earth’s curst.
2. Ores : The minerals from which metals can be conveniently and profitably extracted, are called ores.
3. Gangue : It represents the earthy impurities such as mud, sand and clay associated with the ore.

Question 11.
Name two metals which are formed in nature in free state.
Answer:
The metals are gold (Au) and platinum (Pt).

Question 12.
Which chemical process is used for obtaining a metal from its oxide ?
Answer:
The chemical process is known as reduction.

Question 13.
Metallic oxides of zinc, magnesium and copper were heated with the following metals. In which cases, will you find displacement reactions taking place ?

Relative positions of these metals in the activity series are : Mg, Zn, Cu : In the light of this :
Answer:
Magnesium (Mg) will displace both zinc (Zn) and copper (Cu) from their oxides
Mg + ZnO ———-> MgO + Zn
Mg + CuO ———-> MgO + Cu
Zinc will displace copper from copper oxide.
Zn + CuO ———–> ZnO + Cu
Copper is least reactive and will not initiate displacement reaction.

Question 14.
Which metals do not corrode easily ?
Answer:
Metals such as gold (Au) and platinum (Pt) present at the bottom of the activity series do not corrode easily.

Question 15.
What are alloys ? (CBSE 2011)
Answer:
Alloys are the homogeneous mixture of two or more metals or even metals and non-metals.

NCERT Solutions for Class 10 Science Chapter 3 NCERT End Exercise

Question 1.
Which of the following will give displacement reactions ?
(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal.
Answer:
(d). Only AgNO₃ solution will give displacement reaction with copper (Cu) because copper is placed above silver in the activity series.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting ?
(a) applying grease
(b) applying paint
(c) applying a coating of zinc
(d) all the above.
Answer:
Theoretically all the three methods are helpful for preventing an iron frying pan from rusting. However, the constituents of both grease and paint are mostly organic compounds. They cannot withstand the heat and do not last. Therefore, applying a coating of zinc (galvanisation) is the best method. Option (c) is correct.

Question 3.
An element reacts with oxygen to give a compound with high melting point. This compound is also water soluble. The element is likely to be :
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron
Answer:
(a). Calcium (Ca) combines with oxygen to form calcium oxide (CaO) with very high melting point. CaO dissolves in water to form calcium hydroxide

Question 4.
Food cans are coated with tin and not with zinc because
(a) Zinc is costlier than tin
(b) Zinc has higher melting point than tin
(c) Zinc is more reactive than tin
(d) Zinc is less reactive than tin.
Answer:
(c). Zinc is more reactive than tin and reacts with organic acids present in food to form poisonous compounds. Since tin is placed below zinc in the activity series, it is less reactive and does not react with the organic acids. Therefore, (c) is the correct option.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch :
(a) Flow could you use them to distinguish between samples of metals and non metals ?
(b) Assess the usefulness of these tests to distinguish between metals and non metals.
Answer:
(a) With the help of hammer, convert both the metal and non-metal (solid) in the form of plates or rods. Metal will readily form these since they are malleable. Non-metals being brittle will break when struck with hammer. They will form plates with difficulty Now construct a cell in both the cases using these plates as electrodes and switch on the current. If the bulb glows, this means that the electrodes are of metals. In case, this does not glow, this means that the electrodes are of non-metals.
(b) From these tests, we may conclude that

1. Metals are malleable while non-metals are not.
2. Metals are good conductors of electricity while non-meals are not (graphite is an exception).

Question 6.
What are amphoteric oxides ? Give examples of two amphoteric oxides.
Answer:
These are the oxides which can act both as acids and bases. For example, aluminium oxide (Al₂O₃) and zinc oxide (ZnO). The amphoteric character of the two oxides are shown by the following reactions.

Question 7.
Name two metals which can displace hydrogen from dilute acids and two metals which can not do so.
Answer:
Sodium and calcium can displace hydrogen from dilute acids Copper and silver can not displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining of metal M, name anode, cathode and electrolyte.
Answer:
Anode : Rod of the impure metal
Cathode : Rod of pure metal
Electrolyte : Aqueous solution of soluble salt of metal M.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it as shown in the figure.

What will be the action of gas on

1. dry litmus paper ?
2. moist litmus paper ?

Write a balanced chemical equation for the reaction taking place.
(CBSE 2011)
Answer:
The gas evolved upon heating sulphur powder on a spatula is sulphur dioxide

1. SO₂(g) has no action of dry litmus paper.
2. SO₂(g) dissolves in moisture (water) present in moist litmus paper to form sulphurous acid. In acidic solution, moist litmus paper will change to red.
   

Question 10.
State two ways to prevent the rusting of iron.
Answer:

1. By applying a coating of grease or paint on the surface of iron.
2. By depositing a layer of zinc on the surface of iron. The process is called galvanisation.

Question 11.
What types of oxides are formed when non-metals combine with oxygen ?
Answer:
The oxides are generally acidic in nature which means that when dissolved in water, their solutions change blue litmus to red. For example,

Question 12.
Give reasons for the following :
(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal but still used for making cooking utensils.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction. (CBSE 2013, 2014)
Answer:
(a) These metals placed at the bottom of the activity series are very little reactive in nature. Gold and
platinum are known as noble metals. They are not affected by air, water and by chemicals. Since they have bright lustre, jewellery can be made from these metals.
(b) There are reactive metals placed high in the activity series. In air, their surface gets tarnished due to presence oxygen, water vapours and carbon dioxide in air. With water, these react violently to evolve so much heat that is not possible to handle them. These metals are generally kept under kerosene which does not contain air and water.
(c) When exposed to air, the metal changes its oxide called aluminium oxide (Al₂CO₃). It gets deposited over the surface of the metal and forms a protective coating on the surface. Due to the presence of this layer, the metal becomes unreactive and can be used in making cooking utensils.
(d) Both carbonate and sulphide ores of metals cannot be directly reduced to metallic state. Flowever, metal oxides can be easily reduced with coke or other reducing agents. Both are therefore, converted into their respective oxides by calcination process (for carbonate ores) and by roasting process (for sulphide ores).
Metal oxides can be easily reduced to metallic form with coke (C) or any other suitable reducing agent. Therefore, carbonates and sulphides are converted to the oxide form by processes of calcination and roasting and are not directly reduced.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels. (CBSE 2014)
Answer:
Copper metal slowly reacts with water, carbon dioxide and oxygen present in air to form basic copper carbonate which is green in colour. Its layer slowly gets deposited on the surface of the metal. Now lemon

juice contains citric acid while tartaric acid is present is tamrind. Both these acids react with basic copper carbonate to form soluble salts such as copper acetate (with citric acid) and copper tartarate (with tartaric acid). The equations for the reactions are complicated and are not given. These salts are removed from the surface of the copper metal and the surface of the metal shines.

Question 14.
A man went door to door posing as a goldsmith. He promised to bring back the glitter on dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparked like new but their weight was reduced drastically. The lady got upset and after a futile argument, the man beat a hasty retreat. Can you predict the nature of the solution used by the impositer ?
Answer:
The man had actually used the solution of aqua regia (mixture of cone. HCl and cone. HNO₃ in the ratio of 3 : 1 by volume) which has dissolved gold forming soluble auric chloride (AuCl₃). Since gold actually reacted, there was a loss in weight of the gold bangles. With the removal of the dull layer of gold from the surface, there was original shine on the bangles. The chemical reactions have been

Question 15.
Give reason as to why copper is used to make hot water tanks and not steel (an alloy of iron).
Answer:
Copper is a better conductor of heat than steel which is an alloy of iron. Though copper is costlier than steel, it is used to make hot water tanks for storing hot water in preference to steel.

Question 16.
Differentiate between metals and non-metals on the basis of chemical properties.
For the distinction in the chemical characteristics,

Hope given NCERT Solutions for Class 10 Science Chapter 3 are helpful to complete your science homework.

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NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

NCERT Solutions for Class 10 Science Chapter 2 Acids Bases and Salts

In this chapter, students will understand the chemical properties of acids, and bases, how acids and bases react with metals, how metal carbonates, and metal hydrogen carbonates react with acids, how acids and bases react with each other, the reaction of metallic oxides with acids. Students will also learn about what all acids and bases have in common, what happens to acids and bases in water solution, the importance of pH in everyday life, pH of salts.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 2 Acids Bases and Salts. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 2 – Acids, Bases and Salts solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 2 – Acids, Bases and Salts Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 2 In Text Book Questions

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube ?
Answer:
Take a small volume of all the three liquids in three test tubes. Dip red litmus paper strips separately in all the three. The tube in which red litmus strip turns blue, contains basic solution. Now remove the blue litmus paper and dip it one of the remaining test tubes. If the colour of the blue litmus paper changes to red, the tube contains acidic solution. In case, it remains blue then the tube contains distilled water.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Why should not curd and sour substances be kept in containers made up of brass or copper ?
Answer:
Both curd and sour substances contain some acids in them. They react with copper or brass vessels to form certain salts which are of poisonous nature. Therefore, it is not advisable to keep them in these containers.

Question 3.
Which gas is usually liberated when an acid reacts with a metal ? Illustrate with an example. How will you test the presence of this gas ?
Answer:
Metals are mostly reactive in nature. They react with dilute acids (HCl and H₂SO₄) to evolve hydrogen gas. For example,

The gas burns with a pop sound when a burning candle is brought near it.

Question 4.
A metallic compound ‘A’ reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
Since the gas is evolved with effervescence and extinguishes fire, it is expected to be CO₂ gas. As calcium chloride is formed as one of the products, this means that the substance A’ can be calcium carbonate. It reacts with dilute hydrochloric acid as :

Question 5.
Aqueous solutions of HCl, HNO₃ and H₂SO₄ etc. show acidic character while those of the compounds like ethyl alcohol (C₂H₅OH) and glucose (C₆H₁₂O₆) fail to do so. Explain.
Answer:
All the acids that are listed, have replaceable hydrogen atoms which they release in aqueous solution as H⁺ ions. Therefore, they show acidic character. However, both ethyl alcohol (C₂H₅OH) and glucose (C₆H₁₂O₆) do not have replaceable hydrogen atoms. They fail to evolve hydrogen gas and do not show any acidic character.

Question 6.
Why does aqueous solution of an acid (HA) conduct electricity ?
Answer:
Aqueous solution of an acid (HA) releases H+ ions or H₃O⁺ ions and anions (A^–) in solution. Since ions are the carrier of charge, the aqueous solution of an acid conducts electricity.

Question 7.
Why does not dry HCl gas change the colour of the dry litmus paper ? (CBSE 2013)
Answer:
Dry HCl gas fails to release any H⁺ ions which means that it is not acidic. It fails to change the colour of the dry litmus paper which has also no moisture present.

Question 8.
While diluting an acid, why is it not recommended that acid should be added to water and not water to the acid ? (CBSE 2011)
Answer:
Acids particularly the mineral acids like H₂SO₄, HNO₃ and HCl etc., have strong affinity for water. The dilution process is highly exothermic in nature. The heat evolved may crack or break the container and may also convert the acid into fog which is likely to pollute the atmosphere. In order to control the heat evolved, it is advisable to add acid drop by drop to water. In case water is added to acid, then the entire acid will get itself involved in the exothermic process. It may not be possible to control the heat evolved.

Question 9.
How is concentration of hydronium ions (H₃O⁺) affected when solution of an acid is diluted with water ?
Answer:
An acid dissociates into hydronium ions (H₃O⁺) and anions when dissolved in water. Upon dilution, the volume of the solution increases and the number of ions per unit volume decreases. Therefore, the concentration of H₃O⁺ ions per unit volume decreases.

Question 10.
How is concentration of hydroxyl (OH^–) ions affected when excess of base is dissolved in solution of sodium hydroxide ?
Answer:
Sodium hydroxide (NaOH) is a strong base. It immediately dissociates in solution to give OH^– ions and cations. Upon dissolving more of base in the solution, the concentration of OH^– ions further increases.

Question 11.
You have two solutions A and B. The pH of solution A is 6 and that of solution B is 8. Which solution has more hydrogen ion concentration ? Which of these is acidic and which one is basic ?
Answer:
The pH of a solution is inversely proportional to the concentration of H⁺ ions in solution. Lesser the pH of the solution, more will be the H⁺ ion concentration. The solution A with pH 6 has more H⁺ ion concentration than the solution with pH equal to 8. The solution A is acidic because its pH is less than 7 and the solution B is basic because its pH is more than 7.

Question 12.
What effect does concentration of H⁺(aq) ions have on acidic nature of absolution ?
Answer:
The acidic nature of a solution is directly related to the concentration of H⁺ ions. As the concentration of H⁺ ions increases, the acidic nature of solution also increases.

Question 13.
Do basic solutions also have H⁺(aq) ions ? If yes, then why are these basic ?
Answer:
Yes, basic solutions have also H⁺(aq) ions present in them. Actually, these solutions are prepared in water. Being a weak electrolyte, it dissociates to give H⁺ and OH^– ions. However, the number of H⁺ ions is very small as compared to the number of OH^– ions which are released by the base and also by water. Therefore, the solutions as a whole are of basic nature.

Question 14.
Under what soil conditions, do you think a farmer would spread or treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate) ?
Answer:
A soil usually becomes acidic when there is either a high peat content, iron minerals or there is some rotting vegetable. In order to reduce the acidic strength, ‘liming of soil’ is usually done. For this, any of the substances that have been mentioned are added to the soil since these are of basic nature.

Question 15.
Name the substance which upon treating with chlorine gives bleaching powder. Write the chemical equation for the reaction. (CBSE 2011)
Answer:
Slaked lime is the substance which reacts with chlorine to give bleaching powder

Question 16.
Name the sodium compound used for softening hard water.
Answer:
Washing soda or sodium carbonate. It is chemically sodium carbonate decahydrate (Na₂CO₃.10H₂O). What will happen if the solution of sodium hydrogen carbonate is heated ?

Question 17.
Write the chemical equation involved.
Answer:
Carbon dioxide gas will evolve and sodium carbonate will be left.

Question 18.
Write the chemical equation for the reaction between Plaster of Paris and water.
Answer:

NCERT Solutions for Class 10 Science Chapter 2 NCERT End Exercise

Question 1.
A solution turns red litmus blue. Its pH is likely to be
(a) 2
(b) 4
(c) 7
(d) 10.
Answer:
The solution is basic. Its pH is likely to be 10.
Therefore, (d) is the correct answer.

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime water milky. The solution contains
(a) NaCl
(b) HCl
(c) LiCl
(d) KCl.
Answer:
The crushed egg-shells consist of layer of calcium carbonate which reacts with dilute HCl to evolve CO₂(g). The gas turns lime water milky.
Therefore (b) is the correct choice.

Question 3.
10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the volume of HCl solution (the same solution as before) required to neutralise will be
(a) 4 mL
(b) 8 mL
(c) 12 mL
(d) 16 mL.
Answer:
10 mL of NaOH will require HCl = 8 mL and 20 mL of NaOH will require HCl=16 mL.
Therefore, (d) is the correct answer.

Question 4.
Which of the following types of medicines is used for treating indigestion ?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic.
Answer:
Antacid is used for treating indigestion.
The correct answer is (c).

Question 5.
Write the word equations and the balanced equations for the reactions when :
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron fillings.
(d) dilute hydrochloric acid reacts with iron fillings.
Answer:

Question 6.
Compounds like alcohol and glucose also contain hydrogen but are not characterised as acids. Describe an activity to prove it.
The chemical formula of ethyl alcohol is C₂H₅OH which is an alcohol and of glucose is C₆H₁₂O₆. Both are organic compounds and contain hydrogen atoms. However, they do not behave as acids. (CBSE 2011, 2013)
Answer:
This can be shown by the following activity :
In a glass beaker, take a dilute solution of glucose (C₆H₁₂O₆). Fix two small nails of iron in a rubber cork and place the cork in the beaker as shown in the figure. Connect the nails to the terminals of a 6 volt battery through a bulb. Switch on the current. The bulb will not glow. This shows that the electric current has not passed through the glucose solution. As the current is carried by the movement of ions, this shows that in solution, glucose has not given any H⁺ ions.

Now repeat the same experiment with ethyl alcohol (C₂H₅OH). The bulb will not glow in this case also. This shows that both of them do not behave as acids although they contain hydrogen atoms in their molecules.

Question 7.
Why does not distilled water conduct electricity whereas rain water does ?
Answer:
Pure water (or distilled water) is a very weak electrolyte and does not dissociate into ions. Therefore, it does not conduct electricity. However, rain water contains some dissolved acids like carbonic acid (H₂CO₃) and sulphurous acid (H₂SO₃). Actually air contains traces of both CO₂ and SO₂ gases which dissolve in rain water to produce corresponding acids. As a result, water becomes acidulated and gets ionised easily. Therefore, rain water conducts electricity.

Question 8.
Why does not an acid show any acidic behaviour in the absence of water ?
Answer:
An acid gets ionized only in aqueous solution i.e. in the presence of water. In other words, an acid releases H+ ions or shows acidic behaviour only in the presence of water.

Question 9.
Five solutions A, B, C, D and E when tested with universal indicator show pH as 4, 2, 12, 7 and 9 respectively. Which solution is :
(a) neutral
(b) strongly alkaline
(c) strongly acidic
(d) weakly alkaline
(e) weakly acidic
(f) Arrange the pH in increasing order of H⁺ ion concentration.
Answer:
(a) Neutral : D with pH = 7
(b) Strongly alkaline : C with pH = 12
(c) Strongly acidic : B with pH =2
(d) Weakly alkaline : E with pH = 9
(e) Weakly acidic : A with pH = 4
(f) Increasing order of H+ ions concentration :C<E<D<A<B

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A while acetic acid (CH₃COOH) is added to test tube B. In which case, fizzing occurs more vigorously and why ?
Answer:
Fizzing in the reaction is due to the evolution of hydrogen gas by the action of metal on the acid

Since hydrochloric acid is a stronger acid than acetic acid, fizzing occurs more readily in tube A than in tube B. Actually hydrogen gas will evolve at more brisk speed in test tube A.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd ? Explain your answer. (CBSE 2011)
Answer:
When milk changes into curd, the pH decreases. Actually, lactose (carbohydrate) present in milk gets converted into lactic acid. As more of acid is formed, pH of the medium decreases.

Question 12.
A milkman adds a very small amount of baking soda to fresh milk.
(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?
(b) Why does this milk take a long time to set as curd ? (CBSE 2011)
Answer:
(a) We know that fesh milk is slightly acidic due to presence of lactic acid and its pH is 6. Upon
standing, its pH slowly decreases and it becomes sour since more of acid to released. The purpose of adding baking soda or sodium hydrogen carbonate (NaHCO₃) is to make medium slighly alkaline. The base released will neutralise the effect of lactic acid present in milk. This will check the milk from getting sour.
(b) When milk sets as curd, it becomes more acidic and pH decreases. In the alkaline medium, it takes longer time to achieve acidic medium back so that milk may set as curd.

Question 13.
Why should Plaster of Paris be stored in a moisture-proof container ?
Answer:
In the presence of moisture, Plaster of Paris gets hydrated and changes to Gypsum which is a hard mass.

It can be no longer be used for making moulds and statues. Therefore, Plaster of Paris is kept in moisture proof containers or bags.

Question 14.
What is neutralisation reaction ? Give two examples.
Answer:
Neutralisation reaction is the reaction between acid and base dissolved in aqueous solution to form salt and water.

Question 15.
Both NaCl and KNO₃ are neutral in nature. They neither change blue litmus red nor red litmus blue. That is why the reaction is called neutralisation reaction.
Give two important uses of washing soda and baking soda. (CBSE 2011)
Answer:
Uses of washing soda:

1. In the manufacture of glass, soap, paper and chemicals like caustic soda (NaOH) and borax (Na₂B₄O₇) etc.
2. As a cleansing agent for domestic purposes.

Uses of baking soda:

1. In baking powder used for preparing cakes.
2. In antacids to reduce acidity in the stomach.

NCERT Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts

NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations

In this chapter students will learn about writing chemical equations, balancing chemical equations, different types of chemical equations, decomposition reaction, displacement reaction, double displacement reaction, oxidation and reaction, corrosion, rancidity.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 1 – Chemical Reactions and Equations solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 1 – Chemical Reactions and Equations Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 1 In Text Book Questions

Question 1.
Why should magnesium ribbon be cleaned before burning in air ?
Answer:
Magnesium ribbon has generally a coating of basic magnesium carbonate on its surface. It is a mixture of magnesium hydroxide and magnesium carbonate and is slowly deposited on the surface of the metal by the action of moist air. The coating or layer prevents the metal from burning when flame is brought in contact with the metal. The surface should be properly cleaned preferably with a sand paper before burning the ribbon in air in order to remove the layer of magnesium oxide.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Write the balanced equations for the following chemical reactions :
(i) Hydrogen + Chlorine ———> Hydrogen chloride
(ii) Sodium + Water ———> Sodium hydroxide + Hydrogen
(iii) Barium chloride + Aluminium sulphate ———> Barium sulphate + Aluminium chloride.
Answer:
The balanced equations are written for the symbol equations and not for word equations.

Question 3.
Write the balanced equations 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 (in water) and water.
Answer:
The symbol equations in balanced form for the reactions are :

Question 4.
A solution of the substance ‘X’ is used for white washing.
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ with water.
Answer:
(i) The substance lX’ is calcium oxide (also called quick lime). Its formula is CaO

Question 5.
With the help of an experiment show that in the electrolysis of acidulated water, the volume of one gas is twice the volume of the other gas. Name the gas.
Answer:
The gas is hydrogen and its volume is twice that of oxygen

Question 6.
When you mix solutions of lead (II) nitrate and potassium iodide,
(i) What is the colour of the precipitate formed ? Name the compound involved.
(ii) Write a balanced chemical equation for the reaction
(iii) Is this a double displacement reaction ?
Answer:
(i) The precipitate is yellow in colour. The compound is lead (II) iodide with chemical formula Pbl₂.
(ii) pb(NO₃)₂ (aq) + 2KI (aq) ———-> Pbl₂ (s) + 2KNO₃ (aq)
(iii) Yes, it is a double displacement reaction.

Question 7.
Why does the colour of copper sulphate change when an iron nail is dipped in it ?
Answer:
Iron nail acquires a brown coating of copper as a result of the displacement reaction in which iron has displaced copper from copper sulphate solution. The solution becomes light green.

Question 8.
Give one example of the double displacement reaction.
Answer:

Question 9.
Identify the substances that are oxidised and the substances that are reduced in the following reactions :

Answer:
(i) In this reaction, sodium (Na) is oxidised to sodium oxide (Na₂O). This means that oxygen (O₂) has been reduced.
(ii) In this reaction, hydrogen (H₂) is oxidised to water (H₂O) while copper (II) oxide (CuO) is reduced to copper (Cu).

Question 10.
Magnesium ribbon burns with a dazzling flame in air (or oxygen) and changes to white substance magnesium oxide. Is magnesium being oxidised or reduced in this reaction ?
Answer:
The balanced chemical equation for the reaction is :

Magnesium (Mg) is oxidised to magnesium oxide (MgO) in this reaction.

NCERT Solutions for Class 10 Science Chapter 1 NCERT End Exercises

Question 1.
WTiich of the following statements about the reaction given below are incorrect ?

(a) Lead is getting reduced
(b) Carbon dioxide is getting oxidised
(c) Carbon is getting oxidised
(d) Lead oxide is getting reduced
(i) a and b
(ii) a and c
(iii) a, b and c
(iv) all are incorrect
Answer:
(i) a and b are both incorrect
Pb is getting oxidised to PbO in backward reaction.
CO₂ is getting reduced to C in backward reaction.

Question 2.
The above reaction is an example of :
(a) combination reaction
(b) double displacement reaction
(c) decomposition reaction
(d) displacement reaction
Answer:
(d). It is an example of displacement reaction. The aluminium (Al) metal has displaced iron (Fe) from Fe₂O₃ when the reaction is carried in aqueous solution. Please note that Al lies above Fe in the reactivity series.

Question 3.
What happens when dilute hydrochloric acid is added to iron fillings ? Tick the correct answer :
(a) Hydrogen gas and iron chloride are produced
(b) Chlorine gas and iron hydroxide are produced
(c) No reaction takes place
(d) Iron salt and water are produced.
Answer:
(a). Iron chloride and hydrogen gas are produced according to the reaction

Question 4.
What is a balanced chemical equation ? Why should chemical equations be balanced ? (CBSE 2013)
Answer:
Balanced chemical equation. A chemical equation is said to be balanced if :

1. the atoms of different elements on both sides of the equation are equal.
2. the equation is molecular i.e.. the gases if involved in the equation must be in the molecular form (eg., H₂, O₂, N₂, Cl₂ etc.)

Necessity to balance chemical equations. The chemical equations have to be balanced to fulfill the requirement of law of conservation of mass. According to the law,
In a chemical reaction, the mass of reactants must be the same as the mass of products and this is possible only if the atoms of different elements of the reactants and products are equal.

Question 5.
Transfer the following into chemical equations and balance them :
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer:
(a) The symbol equation for the reaction is :
H₂ + N₂ ———> NH₃
The balancing of equation is done in the following steps :
Step I: Let us count the number of atoms of all the elements of the reactants and the products on both sides of the equation.

A simple look at the equation reveals that neither the number of H nor of N atoms are equal on both sides of the equation.
Step I: In order to equate the number of H atoms on both sides, put coefficient 3 before H₂ on the reactant side and coefficient 2 before NH₃ on the product side.
3H₂ + N₂ ———–> 2NH₃
Step III: On counting, the number of N atoms on both sides of the equation are also the same (2). This means that the equation is balanced.
(b) The symbol equation for the reaction is :
H₂S + O₂ ———–> H₂O + SO₂
The balancing of equation is done in the following steps :
Step I: Let us count the number of atoms of all the elements on both sides on the equation.

A simple look at the equation reveals that the number of H and S atoms are equal on both sides. At the same time, the number of O atoms are not equal.
Step II: In order to equate the number of O atoms, put coefficient 3 before O₂ on the reactant side and coefficient 2 before SO₂ on the product side.
H₂S + 3O₂ ———–> H₂O + 2SO₂
Step III: O atoms are still not balanced. To achieve this, put coefficient 2 before H₂O on the product side.
H₂S + 3O₂ ————> 2H₂O + 2SO₂
Step IV: To balance S atoms, put coefficient 2 before H₂S on the reactant side.
2H₂S + 3O₂ ————> 2H₂O + 2SO₂
On inspection, the number of atoms of all the elements in both sides of the equation are equal. Therefore, the equation is balanced.
(c) The symbol equation for the reaction is :
BaCl₂ + Al₂(SO₄)₃ ———-> A1Cl₃ + BaSO₄
The balancing of equation is done in the following steps :
Step I: Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that only Ba atoms are equal on both sides. The rest of the atoms are to be balanced. It is done as follows :
Step II: In order to equate the number of Al atoms, put coefficient 2 before AlCl₃ on the product side.
BaCl₂ + Al₂(SO₄)₃ ———-> 2AICl₃ + BaSO₄
Step III: In order to balance Cl atoms, put coefficient 3 before BaCl₂ on the reactant side.
3BaCl₂ + Al₂(SO₄)₃ ———–> 2AlCl₃ + BaSO₄
Step IV: To balance Ba atoms, put coefficient 3 before BaSO₄ on the product side.
3BaCl₂ + Al₂(SIO₄)₃ ———-> 2AlCl₃ + 3BaSO₄
Step V: On inspection, the number of S and O atoms on both sides of the equation are also found to be equal. Thus, the equation is in balanced form.
(d) The symbol equation for the reaction is :
K + H₂O ———–> KOH + H₂
The balancing of the equation is done in the following steps :
Step I. Let us count the number of atoms of all the elements on both sides.

A simple look at the equation reveals that the number of K and O atoms on both sides of the equation are equal. At the same time, the number of H atoms are not equal.
Step II: To balance the number of H atoms, put coefficient 2 before KOH on the product side and 2 before H₂O on the reactant side.
K + 2H₂O ———> 2KOH + H₂
Step III: To balance the number of K atoms in the above equation, put coefficient 2 before K atom on the reactant side.
2K + 2H₂O ———> 2KOH + H₂
Step IV: On inspection, the number of atoms of all the elements are found to be equal on both sides of the equation. It is balanced.

Question 6.
Balance the following chemical equations :

Answer:
(a) The symbol equation as given for the reaction is :
HNO₃ + Ca(OH)₂ ——–> Ca(NO₃)₂ + H₂O
The balancing of the equation is done in the following steps :
Step I: Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of Ca atoms are equal on both sides.
Step II: In order to equate the number of N atoms, put coefficient 2 before HNO₃ on the reactant side.
2HNO₃ + Ca(OH)₂ ———-> Ca(NO₃)₂ + H₂O
Step III: In order to equate the number of H atoms, put coefficient 2 before H₂O on the product side.
2HNO₃ + Ca(OH)₂ ———-> Ca(NO₃)₂ + 2H₂O
Step IV: On inspection the number of O atoms on both sides of the equation is the same i.e., 8. Therefore, the equation is balanced.
(b) The symbol equation as given for the reaction is :
NaOH + H₂SO₄ ———-> Na₂SO₄ + H₂O
Step I: Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of O and S atoms are equal on both sides.
Step II: In order to equate the number of Na atoms, put coefficient 2 before NaOH on the reactant side.
2NaOH + H₂SO₄ ———-> Na₂SO₄ + H₂O
Step III: In order to equate the number of H atoms, put coefficient 2 before H₂O on the product side.
2NaOH + H₂SO₄ ———> Na₂SO₄ + 2H₂O
Step IV: On inspection, the number of O atoms on both sides of the equation is the same i.e., 6. Therefore, the equation is balanced.
(c) The symbol equation as given for the reaction is already balanced.
NaCl + AgNO₃ ———-> AgCl + NaNO₃
On inspection, the number of atoms of all the elements are found to be equal on both sides of the equation. It is in the balanced form.
(d) The symbol equation as given for the reaction is :
BaCl₂ + H₂SO₄ ———-> BaSO₄ + HCl
Step I: Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of Ba, S and O atoms are equal on both sides.
Step II: In order to equate the number of Cl atoms, put coefficient 2 before HCl on the product side.
BaCl₂ + H₂SO₄ ———> BaSO₄ + 2HCl
Step III: On inspection, the number of H atoms on both sides of the equation is the same i. e., 2. Therefore, the equation is balanced.

Question 7.
Write the balanced equations for the following reactions :
(a) Calcium hydroxide + Carbon dioxide ———-> Calcium carbonate + Water
(b) Lead + Copper chloride ———-> Lead chloride + Copper
(c) Barium chloride + Sodium sulphate ———-> Barium sulphate + Sodium chloride
(d) Zinc + Silver nitrate ———–> Zinc nitrate + Silver.
Answer:
All these equations are in word form whereas the balanced equations are written in symbol form.
(a) Ca(OH)₂ + CO₂ ———-> CaCO₃ + H₂O
(b) Pb + CuCl₂ ———-> PbCl2 + Cu
(c) BaCl₂ + Na₂SO₄ ———> BaSO₄ + 2NaCl
(d) Zn + 2AgNO₃ ———–> Zn(NO₃)₂ + 2Ag

Question 8.
Write the balanced chemical equations for the following reactions and identify the type of reaction :
(a) Potassium brormide (aq) + Barium iodide (aq) ————> Potassium iodide(aq) + Barium bromide (aq)
(b) Zinc carbonate (aq) ————> Zinc oxide (aq) + Carbon dioxide (aq)
(c) Hydrogen (aq) + Chlorine (aq) ———> Hydrogen chloride (aq)
(d) Magnesium (aq) + Hydrochloric acid (aq) ———–> Magnesium chloride (aq) + Hydrogen(aq).
Answer:
(a)          2KBr (aq) + BaI₂ (aq) ——–> 2Kl (aq) + BaBr₂ (aq)
The reaction is known as double displacement reaction.
(b)          ZnCO₃(s) ———-> ZnO(s) + CO₂(g)
The reaction is known as decomposition reaction. ‘
(c)          H₂(g) + Cl₂(g) ———–> 2HCl (g)
The reaction is known as combination reaction.
(d)          Mg (aq) + 2HCl (aq) ———–> MgCl₂ (aq) + H₂(g)
The reaction is known as displacement reaction.

Question 9.
What do you mean by exothermic and endothermic reactions ? Give examples.
Answer:
A chemical reaction is said to be exothermic in which a certain amount of heat energy is released. The container in which reaction is carried gets heated up. For example,
N₂ (g) + 3H₂(g) ———> 2NH₃ (g) + 92 kj
CH₄(g) + 2O₂(g) ———> CO₂ (g) + 2H₂O (l) + 890 kj
NaOH (aq) + HCl (aq) ———> NaCl (aq)+ H₂O (aq) + 57.5 kj
A chemical reaction is said to be endothermic in which a certain amount of heat energy is absorbed. The container in which the reaction is carried becomes cold. For example,
N₂(g) + O₂ (g) ———-> 2NO (g) – 180 kj
C(s) + H₂O (aq) ———-> CO (g) + H₂(g) – 130 kj

Question 10.
Why is respiration considered an exothermic reaction ? (CBSE 2011)
Answer:
Respiration is the most important biochemical reaction which releases energy in the cells. When we breathe in air, oxygen enters our lungs. It binds itself to haemoglobin present in red cells and is carried to millions of cells present on the body. Respiration occurs in these cells and is accompanied by the combustion of glucose producing carbon dioxide water and heat energy.

Question 11.
Why are decomposition reactions called opposite of combination reactions ? Write equations for these reactions.
Answer:
Combination reaction may be defined as the reaction in which two or more substances combine under suitable conditions to form a new substance. For example,

A decomposition reaction may be defined as the reaction in which a single substance decomposes or splits into two or more substances under suitable conditions.
For example,

It may be concluded that a certain substance is formed or synthesised in combination reaction and it breaks or splits in decomposition reaction. Therefore, the two reactions oppose each other.

Question 12.
Write equations for each decomposition reaction, where energy is supplied in the form of heat, light and electricity.
Answer:

Question 13.
What is the difference between displacement and double displacement reactions ? Write equations for these reactions. (CBSE 2011)
Answer:
In a displacement reaction, one element takes the place of another in a compound dissolved in a solution. For example,
Fe(s) + CuSO₄ (aq) ———> FeSO₄ (aq) + Cu(s)
In a double displacement reaction, one component each of both the reacting molecules get exchanged to form the products. For example,
HCl (aq) + NaOH(aq) ———> NaCl (aq) + H₂O (aq)

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write chemical equation involved.
Answer:
The chemical equation for the displacement reaction is :
Cu(s) + 2AgNO₃ (aq) ———-> Cu(NO₃)₂ (aq) + 2Ag (s).

Question 15.
What do you mean by precipitation reaction ? Explain giving examples.
Answer:
In a precipitation reaction, one of the products formed during the reaction does not dissolve in solution and gets settled at the surface of the container (beaker or tube). It is known as a precipitate. For example,

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples of each,
(a) oxidation
(b) reduction.
Answer:
(a) Oxidation involves the gain of oxygen by a substance in a chemical reaction. For example,

(b) Reduction involves the loss of oxygen from a substance in a chemical reaction. For example,

Question 17.
A shining brown coloured element ‘X’ on heating in’ air becomes black in colour. Name the element ‘X’ and the black coloured compound formed. (CBSE 2013)
Answer:
The element ‘X’ is copper and the black coloured compound is copper (II) oxide also known as cupric oxide.

Question 18.
Why do you apply paint on iron articles ?
Answer:
Paint forms a protective coating on the surface of iron. Both oxygen and moisture (present in air) cannot have a direct contact with the surface of the iron metal. Therefore, the surface gets protected against rusting.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why ?
Answer:
Oil and fat containing food items or etables get rancid due to oxidation by air or oxygen. In case the container or bag is flushed with nitrogen, then oxidation or rancidity will be checked.

Question 20.
Explain the following terms wih one example of each.
(a) Corrosion
(b) Rusting.
Answer:
Corrosion may be defined as the chemical process of slow eating up of the surfaces of certain metals when kept in open for a long time.
Quite often, when we open the bonet of a car after a long time, we find a deposit around the terminals of the battery. This is because of corrosion of the terminals. Black coating on the surface of silver and green layer on the surface of copper are the examples of corrosion. In case of iron, corrosion is called rusting. Rust is a chemical substance which is soft and also porous. It is brown in colour and is formed by the chemical action of moist air (containing CO₂ and H₂O) on iron. It is basically an oxidation reaction and formula of rust is Fe₂O₃.xH₂O. It is very slow in nature and once started keeps on.

Both corrosion and rusting are very harmful and cause damage to the buildings, railway tracks, automobiles and other objects/materials where metals are used. We quite often hear that an old building has collapsed of its own causing loss of both lives and property. This is on account of the rusting of iron which is used in making the structures particularly the roof. Corrosion also causes constant damage to the statues made up of marble which is chemically calcium carbonate (CaCO₃). Both sulphuric acid and nitric acid present in the rain water dissolve calcium carbonate to form calcium sulphate and calcium nitrate respectively.
CaCO₃ + H₂SO₄ ———— > CaSO₄ + CO₂ + H₂O

CaCO₃ + 2HNO₃ ———— > Ca(NO₃)₂ + CO₂ + H₂O

In addition to this, traces of hydrogen sulphide gas (H₂S) present in atmosphere form black stains on these statues due to calcium sulphide which is black in colour.
CaCO₃ + H₂S  ———–> CaS + H₂O + CO₂
Corrosion has caused huge damage to our historical monuments including ‘Taj Mahal’ which is regarded as the eighth wonder. Marble is chemically CaCO₃. Polluted air contains both H₂SO₄ and HNO₃ along with traces of H₂S gas. They react chemically with CaCO₃ as shown above. As a result, this historical monument which is the pride of India is being constancy corroded. Every effort must be made to protect ‘Taj Mahal’ and other similar monuments which are our treasure. Corrosion (or rusting) occurs in metals only. We shall study it detail and the measures to check corrosion in chapter 3 on Metals and Non-Metals.

NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations

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