These NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.
Chemical Reactions and Equations NCERT Solutions for Class 10 Science Chapter 1
Class 10 Science Chapter 1 Chemical Reactions and Equations InText Questions and Answers
In-text Questions (Page 6)
Question 1.
Why should magnesium ribbon be cleaned before burning air ?
Answer:
A layer of magnesium oxide is already present on the surface of magnesium ribbon which does not allow burning of magnesium ribbon in air. So magnesium ribbon should be cleaned before burning in air.
Question 2.
Write the balanced equation for the following chemical reactions:
(i) Hydrogen + Chlorine → Hydrogen chloride
(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
(iii) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
(i) H2(g) + Cl2(g) → 2HCl (g)
(ii) 3 BaCl2 + Al2(SO4)3 → 3 BaSO4 + 2AlCl3
(iii) Na (s) + H2O (l) → NaOH (aq) + H2(g)
Question 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.
(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride and water.
Answer:
(i) BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (↓) + 2NaCl (aq)
(ii) NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
In-text Questions (Page 10)
Question 1.
A solution of a substance ‘X is used for white-washing:
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
Answer:
(i) The substance ‘X’ is quick lime or calcium oxide.
Question 2.
Why is the amount of gas collected in one of the test tubes in activity 1.7 double of the amount collected in the other ? Name this gas.
Answer:
When electricity is passed through the water in presence of an acid., water is electroysed and produces hydrogen and oxygen gas.
\(2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\)
It is clear from the above equation that 2 moles .water undergoes electrolysis and produce 2 moles of hydrogen and one mole of oxygen in the given set of conditions of temperature and pressure because the volume of a gas is directly proportional to its no. of moles so the amount of one gas is doubled than the other.
The double amount gas is hydrogen.
In-text Questions (Page 13)
Question 1.
Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
Answer:
Iron is more reactive that copper so it displaces copper metal from copper sulphate solution and produces sulphate of iron.
Displacement calculator that determines the size of an engine based on Bore, Stroke and number of cylinders.
Question 2.
Give an example of a double displacement reaction other than the one given in Activity 1.10.
Answer:
Question 3.
Identify the substances that are oxidised and the substances that are reduced in the
following reactions.
(i) Na(s) + O2(g) → 2Na2O(s)
(ii) CuO(s) + H2 (g) → CU(s) + H2O(l)
Answer:
(i) 2 Na (s) + O2 (g) → 2 NO2O (s)
Na (s) → Oxidised
O2(g) → Reduced
(ii) CuO (s) + H2(g) → Cu (S) + H2O (l)
CuO (s) → Reduced
H2 (g) → Oxidised
Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Questions and Answers
Page No. 14
Question 1.
Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) → 2Pb(s) + CO2(g)
(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) (d), (b) and (c)
(iv) all
Answer:
(i) a and b
Question 2.
Fe2O3 + 2Al → Al2O3 + 2Fe
The above reaction is an example of a
(a) combination reaction.
(b) double displacement reaction.
(c) decomposition reaction.
(d) displacement reaction.
Answer:
(d) displacement reaction.
Question 3.
What happens when dilute hydrochloric add 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) Hydrogen gas and iron chloride are produced.
Question 4.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
A chemical equation is said to be balanced when it has the same no. of atoms of different elements in both the sides, i. e., reactant and products sides.
e.g. H2 + O2 → H2O
The above equation is not a balanced equation because it does not have the equal no. of atoms of oxygen in both sides.
But 2H2 + O2 → 2 H2O is a balanced equation because it does not have the equal no. atoms of different elements in both the sides.
A chemical equation should be balanced because during a chemical change the no. of atoms of each element remain same. In other words the law of concervation of mass i.ethe total mass of all the products of a chemical reaction has to equal to the total mass of all reactants. And it is only possible when a chemical equation is balanced.
Question 5.
Translate the following statements into chemical equations and then balance them.
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas bums in air to give water and sulpur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer:
(a) H2 (g) + N2(g) → NH3 (g)
Step I. Look more closely the number of atoms of different elements present in the unbalanced chemical equation.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| H | 2 | 3 |
| N | 2 | 1 |
Step II. It is often convenient to start with the compound that contains maximum number of atoms, whether a reactant or a product. So we select NH3 and the element hydrogen in it. There are three hydrogen atoms on the right and only two hydrogen atoms on the left.
| Atoms of Hydrogen | In reactants | In Products |
| Initial | 2 (in H2) | 3 (in NH3) |
| To balance | 2 × 3 | 3 × 2 |
Now the partly balanced equation becomes as follows :
3H2 + N2 → 2NH3
Step III. But nitrogen in the above equation is automatically balanced. So the balanced equation will be
3 H2 + N2 → NH3
The above equation balanced because it contains the equal no. of atoms of hydrogen and nitrogen on both sides.
(b) H2S + O2 → H2O + H2O
The above equation can be balanced by algebraic sum method.
Let ‘a’, ‘b’, ‘c’ and ‘d’ are the no. of molecules of ‘H2S’, ‘O2‘, ‘H2O’ and ‘SO2‘ respectively in the balanced chemical equation. So the above equation can be written as
a H2S + b O2 → c H2O + d SO2
Now in a balanced chemical equation.
L.H.S. R.H.S.
Hydrogen 2 a= 2c …(i)
Sulphur a = d …(ii)
Oxygen 2b = c + 2d ….(iii)
Let a = 1 …..(iv)
So from equations (i) and (iv)
2 × 1 = 2 c
⇒ c = 1
From equation (iii)
2b = c + 2d
2b = 1 + 2 × 1
⇒ 2b = 3
⇒ b = \(\frac {3}{2}\)
Now by putting the value of a, b, c and d in the given equation, we get
1 × H2S + \(\frac {3}{2}\)O2 → 1 × H2O + 1 × SO2
OR 2 H2S + 3O2 → 2 H2O + 2SO2
(c)
Step I:
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Ba | 1 | 1 |
| Cl | 2 | 3 |
| Al | 2 | 1 |
| S | 3 | 1 |
| O | 12 | 4 |
Step II: Now start from Al2 (SO4)3 and balance oxygen first. The simple ratio between the atoms of oxygen in reactants to the product will be 12 : 4 or 3 : 1. So multiplying Al2 (SO4)3 by ‘1’ ‘BaCl2‘ by ‘3’, we get
BaCl2 + Al2 (SO4)3 → AlCl3 + 3BaSO4
Step III: Now balance ‘Al’ be multiplying ‘2’ in the product side (i.e. in AlCl3)
BaCl2 + Al2 (SO4)3 → 2 AlCl3 + 3 BaSO4
Step IV : Balance barium by multiplying ‘3’ in the reactant to BaCl2.
3 BaCl2 + Al2 (SO4)3 → 2 AlCl3 + 3 BaSO4
‘Cl’ is automatically balanced in the above equation.
(d) K + H2O → KOH + H2
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| K | 1 | 1 |
| H | 2 | 3 |
| O | 1 | 1 |
Step II: Now select ‘KOH’ in the reactant side because it contains the maximum no. of elements.
K + 2H2O → 2 KOH + H2
Step III: Balance potassium metal in the above partially balanced equation by multiplying ‘2’ in the potassium i.e. reactant side.
2K + 2H2O → 2KOH + H2
Question 6.
Balance the following chemical equations.
(a) HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O
(b) NaOH + H2SO4 → Na2SO4 + H2O
(c) NaCl + AgNO3 → AgCl + NaNO3
(d) BaCl2 + H2SO4 → BaSO4 + HCl
Answer:
(a)
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| H | 3 | 2 |
| N | 1 | 2 |
| O | 5 | 7 |
| Ca | 1 | 1 |
Step II. Select Ca (NO3)2 because it contains the maximum no. of elements. To balance oxygen multiply Ca (OH)2 by two, we get
HNO3+ 2Ca (OH)2 → Ca(NO3)2 + H2O
Step III. Now balance calcium atoms by multiplying two in reactants sides. We get
HNO3 + 2 Ca (OH)2 → Ca (NO3)2 + H2O
Step IV : Balance nitrogen in the above partially balanced equation, multiply HNO3 by ‘4’. We get
4 HNO3 + 2 Ca (OH)2 → 2 Ca (NO3)2 + H2O
Step V : Balance hydrogen by multiplying ‘4’ in the product side. We get,
HNO3 + 2 Ca (OH)2 → 2 Ca (NO3)2 + 4H2O
This is a balanced chemical equation.
(b) NaOH + H2SO4 → Na2SO4 + H2O
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Na | 1 | 2 |
| O | 5 | 5 |
| H | 3 | 2 |
| S | 1 | 1 |
Step II: Balance sodium in the above equation by multiplying ‘2’ in NaOH.
2 NaOH + H2SO4 → Na2SO4 + H2O
Step III: Balance hydrogen by multiplying ‘2’ in H2O.
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
This is a balanced equation.
(c) NaCl + AgNO3 → AgCl + NaNO3
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Na | 1 | 1 |
| Cl | 1 | 1 |
| Ag | 1 | 1 |
| N | 1 | 1 |
| O | 3 | 3 |
The above equation contains equal no. of atoms of different elements in reactants and products sides.
(d) BaCl2 + H2SO4 → BaSO4 + HCl
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Ba | 1 | 1 |
| Cl | 2 | 1 |
| H | 2 | 1 |
| S | 1 | 1 |
| O | 4 | 4 |
Step II : Now balance chlorine atoms in the above equation by multiplying ‘2’ in the product side to HCl
BaCl2 + H2SO4 → BaSO4 + 2HCl
This is a balanced chemical equation.
Question 7.
Write the balanced chemical equations for the following reactions.
(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride.
Answer:
(a) Ca (OH2) + CO2 → CaCO3 + H2O
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Ca | 1 | 1 |
| O | 4 | 4 |
| H | 2 | 2 |
| C | 1 | 1 |
The above equation is balanced.
(b) Zn + AgNO3 → Zn (NO3)2 + Ag
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Zn | 1 | 1 |
| Ag | 1 | 1 |
| N | 1 | 2 |
| O | 3 | 6 |
Step II: Balance oxygen by multiplying ‘2’ in AgNO3, we get
Zn + 2 AgNO3 → Zn (NO3)2 + Ag
Step III ; Now Balance ‘Ag’ metal by multiplying ‘2’ in product side.
Zn + 2 AgNO3 → Zn (NO3)2 + 2Ag
(c) Al + CuCl2 AlCl3 + Au
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Al | 1 | 1 |
| Cu | 1 | 1 |
| Cl | 2 | 3 |
By multiplying ‘AlCl3‘ with ‘2’ and ‘CuCl2‘ with ‘3’, ‘Al’ with ‘2′ and ‘Cu’ with ‘3’ we get
2Al + 3 CuCl2 → 2AlCl3 + 3 Cu
This is a balanced equation
(d) BaCl2 + K2SO4 → BaSO4 + KCl
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| Ba | 1 | 1 |
| Cl | 2 | 1 |
| K | 2 | 1 |
| S | 1 | 1 |
| O | 4 | 4 |
Balance chlorine by multiplying ‘2’ in NaCl, we get
BaCl2 + K2SO4 → BaSO4+ 2KCl
Question 8.
Write the balanced chemical equation for the following reactions and identify the type of reaction in each case.
(a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
(b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
(c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
(d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)
Answer:
(a) KBr (aq) + Bal2 (aq) → KI (aq) BaBr2 (aq)
Step I.
| Element | No. of atoms in Reactants | No. of atoms in Products |
| K | 1 | 1 |
| Br | 1 | 2 |
| Ba | 1 | 1 |
| I | 2 | 1 |
Balance bromine and iodine in the above equation.
2 KBr (aq) + Bal2 (aq) → 2 Kl (aq) + BaBr2 (aq)
It is a double displacement reaction.
(b) ZnCO3 (s) → ZnO (s) + CO2 (g)
It is decomposition reaction.
(c) H2(g) + Cl2(g) → 2 HCl (g)
It is combination reaction.
(d) Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
It is a displacement reaction.
Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:
Exothermic Reaction: A reaction in which heat energy is given out along with the product is called exothermic reaction.
C (s) O2 (g) → CO2 (g) + Heat energy (394 kJ)
N2 (g) + 3H2 → (g) 2 NH3 (g) + Heat energy (92.4 kJ)
Endothermic Reaction : A reaction in which heat energy is absorbed along with the product is called endothermic reaction.
CaCO3 (s) → CaO (s) CO2 (g)-Heat energy
2HgO (s) → 2 Hg (l) + O2 (g)-Heat energy
Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:
It is a well known fact that animals and human being require energy to stay alive. We and animals get this energy from food. During oxidation or digestion, food is broken down into simple substances. Rice, potatoes and bread contains carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and liberates energy. This reaction is known as respiration.
C6H12O6 (aq) + 6O2 (aq) → 6 CO2 (aq) 6H2O (l) + energy
Because energy is released during respiration so it is an exothermic reaction.
Question 11.
Why are decomposition reactions called opposite of combination reactions? Write equations for these reactions.
Answer:
In decomposition reactions a single reactant breaks down to give two or more simpler products.
In combination reaction two or more than two reactants combine with each other and to give simple product.
So it is dear from above discussion that decomposition reactions are opposite of the combination reactions,
Question 12.
Write equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Answer:
In a decomposition reaction a single reactant breaks down to give two or more simpler products. These reactions need energy in different forms to proceed.
The decomposition of CaCO3 (s) takes place by supplying energy in the form of heat.
\(\mathrm{CaCO}_{3}(s) \stackrel{\Delta}{\rightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(g)\)
In the presence of sunlight white silver chloride decomposes and turns grey because silver metal is formed.
Water decomposes into H, and O, by passing electricity in an electrolytic cell.
\(2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Electricity }}{\rightarrow} 2 \mathrm{H}_{2}(g) \mathrm{O}_{2}(g)\)
Question 13.
What is the difference between displacement and double displacement reactions ? Write equations for these reactions.
Answer:
Displacement Reaction : In a displacement reaction, more active element displaces or removes another element from its compound.
e.g. Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Double displacement reactions: In a double displacement reaction two different atoms or groups of atoms are displaced by other atoms or group of atoms.
e.g
Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
Answer:
Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
Precipitation reaction : A reaction in which precipitates are formed is called precipitation reaction.
e.g.
Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each.
(a) Oxidation
(b) Reduction
Answer:
(a) Oxidations A process which involves the gain of oxygen is called oxidation. In oxidation reaction oxygen is added to the reactant.
e.g. 2 Mg (s) + O2 (g) → 2 MgO (s)
2 H2S (g) + O2 (g) → 2 S (s) + 2 H2O (l)
(b) Reduction: A process which involves the loss of oxygen is called reduction.
2 Pb (NO3)2 (S) → 2 PbO (s) + 2NO2 (g) + \(\frac {1}{2}\)O2 (g)
2 KNO3 → 2 KNO2 + O2
Question 17.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:
The shiny brown coloured element ‘X’ is copper.
When copper is heated in air it forms copper oxide which is black in colour.
Question 18.
Why do we apply paint on iron articles?
Answer:
When metals are exposed by air, moisture or other atmospheric gases, it results in the formation of compounds like sulphides, oxides, carbonates etc. on the surface of metals. This phenomenon is called corrosion.
To Protect iron articles form corrosion they are painted i.e., a thin layer of paint is applied on the iron articles so they do not interact within the atmospheric gases and no corrosion of iron articles is occured.
Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
When fats and oils are oxidised, they become rancid and their smell and taste change. Usually special type of substances i.e., antioxidants are added to fatty foods to prevent oxidation. It is known as rancidity. To protect food from oxidation or to slow down oxidation process of food it is kept in the refrigerator on we flush food with nitrogen.
Question 20.
Explain the following terms with one example each.
(a) Corrosion (b) Rancidity
Answer:
(a) Corrosion : The process of slowly eating away of the metals when they are exposed by air, moisture or other atmospheric gases, resulting into the formation of compounds such as oxides, sulphides, carbonates etc. is called corrosion.
Iron is corroded when exposed by moisture or air. Corrosion of iron is also known as rusting. Rust is mainy hydrated iron (III) oxide i.e. Fe2O3. xH2O.
Rusting weakness the structure of car bodies, bridges, iron railings and other iron articles. Rusting of iron is a serious problem. Every year a large amount of money is spent to replace damaged iron.
(b) Rancidity: When fats and oils are oxidised they become rancid and their smell and taste change. Generally special type of substances i.e., antioxidants are added to fatty foods to prevent oxidation. At home oxidation of food can be slowed down by keeping it in the refrigerator. Keeping and food in air-tight containers can also help.
Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Activities
Activity 1.1 (Page 1)
Caution : This activity needs teacher’s assitsance. It would be better if students wear eye
Fig : Burning of a magnesium ribbon in air and collection of magnesium oxide in watch glass
- Clean a magnesium ribbon about 2 cm long by rubbing with sand paper.
- Hold it with a pair tongs. Born it using a spirit lamp or burner and collect the ash so formed in a watch glass as shown in Fig. Burn a magnesium ribbon keeping it as far as possible from your eyes.
Question 1.
What do you observe ?
Answer:
Observation : What observe that magnesium ribbon burns with a dazzling white flame and changes into a white powder. This powedered material is magnesium oxide. It is formed due to the reaction between magnesium and oxygen present in air.
Activity 1.2 (Page 2)
- Take lead nitrate solution in a test tube.
- Add potassium iodide solution in this.
Question 1.
What do you observe ?
Answer:
Pb (NO3)2 (aq) + 2KI (aq) → PbI2(ppt.) + 2KNO3
When the solution of potassium iodide is added to the lead nitrate solution, yellow colour precipitate of lead iodide is formed. It represents a chemical change.
Activity 1.3 (Page 2)
- Take few zinc granules in a conical flask or test tube.
- Add dilute hydrochloric acid or sulphuric add to this (Figure).
Caution : Handle the acid with care.
Question 1.
What do you observe immediately above the zinc granules in the conical flask or test tube ?
Touch the conical flask or test tube. Is there any change in temperature.
Answer:
Zinc granules react with dil. Hydrochloric acid or sulphuric acid liberates hydrogen gas in the form of bubbles along with zinc chloride or zinc sulphate.
Yes, the test tube becomes hot, because it is an exothermic reaction.
Activity 1.4 (Page 6)
- Take a small amount of calcium oxide or quick lime in a beaker.
- Slowly add water to this.
- Touch the beaker as shown in Fig.
Question 1.
Do you feel any change in temperature.
Answer:
Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat.
The temperature of the beaker is raised because this is an exothermic reaction.
Activity 1.5 (Page 8)
- Take 2 gm 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 fig.
Question 1.
Observe the colour of the crystals after heating.
Answer:
The colour of ferrous sulphate crystals is green. After heating for sometime the green colour of ferrous sulphate has changed into reddish brown.
Activity 1.6 (Page 8)
- Take about 2 gm 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.
Question 1.
What do you observe ? Note down the change, if any.
Answer:
Observation : When lead nitrate powder is heated in a boiling tube, we observe brown fumes of a nitrogen dioxide (NO.) release. The following decomposition reaction takes place :
Activity 1.7 (Page 9)
- Take a plastic mug. Drill two holes at its base and fit rubber stoppers in there holes, Insert carbon electrodes in these rubber stoppers as shown in figure.
- Connect these electrodes to a 6 volt battery.
- Fill in the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
- Take two graduated test tubes filled with water and invert them over the two carbon electrodes.
- Switch on the current and leave the appratus undisturbed for some time.
- You will observe the formation of bubbles at both the electrodes.
- These bubbles displace water in the graduted tubes.
Question 1.
Is the volume of the gas collected the same in both the test tubes ?
Answer:
- Once the test tube 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 the test tubes. Caution :
This step must be performed carefully by the teacher.
Question 2.
What happens in each case ?
Which gas is presen in each test tube ?
Answer:
Observation : No, the volume of the gases in both test tubes are not same.
The two test tubes contain ‘H2‘ and ‘O2‘ respectively.
Hydrogen burns with a pale blue flame but oxygen is not combustible.
Activity 1.8 (Page 9)
- Take about 2 gm silver chloride on a china dish.
- What is its colour ?
- Place this china dish under sunlight for sometime. (Fig.)
Question 1.
Observe the colour of the silver chloride after some time.
Answer:
The colour of the silver is white. We observe that the white colour of silver chloride turns grey in sunlight. This is due to the decomposition of silver chloride and chlorine by light.
\(2 \mathrm{AgCl}(\mathrm{s}) \stackrel{\text { Sunlight }}{\longrightarrow} 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g})\)
Activity 1.9 (Page 10)
- 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 ml copper 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. (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. (b)].
- Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside [Fig. (b)].
Question 1.
Why does the iron nail become brownish in colour and the blue colour of copper sulphate solution fade?
Answer:
Observatons: The following chemical reaction takes place in this Activity :
In this reaction, iron has displaced or removed another element, copper, from copper sulphate solution. This reaction is known as displacement reaction.
Activity 1.10 (Page 11)
- 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.).
Question 1.
What do you observe?
Answer:
Observations : We observe that white precipitate of barium sulphate is formed. Any reaction that produces a precipitate can be called a precipation reaction.
This is a double displacement reaction,
Recall Activity, 1.2, where you. have mixed the solutions of lead (II) nitrate and potassium iodide.
(i) What was the colour of the precipitate formed? Can you name the compound precipitated?
(ii) Write the balanced chemical equation for this reaction.
(iii) is this also a double displacement reaction ?
Answer:
(i) Yellow colour precipitates of lead iodide are formed.
(iii) Yes, this is a double displacement reaction.
Activity 1.11 (Page 12)
- Heat a china dish containing about 1g copper powder (Fig.).
Question 1.
What do you observe?
Answer:
Observation : The surface of copper powder becomes coated with black copper(II) oxide. This is because oxygen is added to copper and copper oxide is formed.
\(2 \mathrm{Cu}+\mathrm{O}_{2} \stackrel{\text { Heat }}{\longrightarrow} 2 \mathrm{CuO}\)
Recall Activity 1.1, where a magnesium ribbon bums with a dazzling flame in air (oxygen) and changes into a white substance, magnesium oxide. Is magnesium being oxidised or reduced in this reaction?
Answer:
Magnesium is being oxidised.
Class 10 Science Chapter 1 Chemical Reactions and Equations Additional Important Questions and Answers
Very Short Answer Type Questions
Question 1.
What are the state symbols of solid, liquid and gas ?
Answer:
Solid (s), liquid (l) and gas (g).
Question 2.
What is the state symbol of aqueous state?
Answer:
Aqueous (aq).
Question 3.
What are the symbols used to represent gaseous product and precipitate product ?
Answer:
Gaseous Product (↑); Precipitate Product (↓)
Question 4.
What is the formula of quick lime ?
Answer:
CaO.
Question 5.
What is the formula of slaked lime ?
Answer:
Ca(OH)2.
Short Answer Type Questions
Question 1.
Name the component oxidised, component reduced, oxidising agent and reducing agent in the following displacement reactions.
(i) Mg + CuSO4 MgSO4 + Cu
(ii) Cu + 2 AgNO3 Cu (NO3) → 2 + 2 Ag.
Answer:
(i) Mg : Oxidised, recuding agent.
CU2 : Reduced, oxidising agent.
(ii) CU : Oxisided, redusing agent.
Ag+ : Reduced, oxidising agent.
Question 2.
What type of chemical reactions take place when:
(a) A magnesium wire is brunt in air ?
(b) Limestone is heated ?
(c) Proteins are converted into amino acids.
(d) Electricity is passed through water ?
(e) Ammonia and hydrogen chloride are mixed ?
Answer:
(a) Combination reaction.
(b) Decomposition reaction.
(c) Decomposition reaction.
(d) Decomposition reaction.
(e) Combination reaction.
Question 3.
In the reaction represented by the equation:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
(a) Name the substance oxidised.
(b) Name the oxidising agent.
(c) Name the substance reduced.
(d) Name the reducing asgent.
Answer:
(a) HCl, (b) MnO2, (iii) MnO2, (iv) HCl
Long Answer Type Question
Question 1.
What is corrosion ? What are the factors which promote corrosion ? How it is prevented ?
Answer:
The process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, sulphates, etc. is called corrosion.
Factors which promote corrosion :
- Reactivity of the metal: More active metals are corroded readily e.g. Iron corroded more than zinc.
- Impurities : Impure metal are corroded more than pure metals e.g. pure iron does not rust.
- Presence of air and moisture : Air and moisture increase the rate of corrosion. Gases like ‘SO2‘ and ‘CO2‘ in air accelerate corrosion. Iron does not rust in vacuum.
- Electrolytes: Presence of electrolytes like NaCl, HCl etc. accelerate the corrosion. For example, iron rusts faster in saline water than in pure water.
- Strains in metals: Bends, scratches, nicks and cuts in a metal increase the rate of corrosion.
Prevention : When a metal surface is not allowed to come in contact with moisture, oxygen and carbon dioxide, it stops corrosion.
The can be achieved by the following methods:
(i) The metal surface is coated with paint which keeps it out of contact with air, moisture etc. till the paint layer develops cracks.
(ii) By applying film of oil and grease on the surface of the iron tools and machinery, the rusting of iron can be prevented since it keeps the iron surface away from moisture, oxygen and carbondioxide.
Multiple Choice Questions
Question 1.
The colour of lead iodide is:
(a) Red
(b) Blue
(c) White
(d) Yellow
Answer:
(b) Blue
Question 2.
Magnesium bums in air to form:
(a) Magnesium oxide
(b) Magnesium trioxide
(c) Magnesium dioxide
(d) A mixture of magnesium trioxide
Answer:
(a) Magnesium oxide
Question 3.
The following observations (s) help (s) us to determine that a chemical reaction has taken place:
(a) Change in colour
(b) Evolution of gas
(c) Change in temperature
(d) All the above
Answer:
(d) All the above
Question 4.
Zinc reacts will dil. H2SO4 to evolve:
(a) Oxygen
(b) Nitrogen
(c) Sulphur dioxide
(d) Hydrogen
Answer:
(d) Hydrogen
Question 5.
The molecular formula of rust is:
(a) Fe2O3
(b) FeO. xH2O
(c) Fe2O3. xH2O
(d) FeO.
Answer:
(c) Fe2O3. xH2O