{"id":26768,"date":"2022-05-24T11:00:37","date_gmt":"2022-05-24T05:30:37","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26768"},"modified":"2022-05-24T11:53:57","modified_gmt":"2022-05-24T06:23:57","slug":"ncert-solutions-for-class-10-science-chapter-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-science-chapter-1\/","title":{"rendered":"NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations"},"content":{"rendered":"

These NCERT Solutions for Class 10 Science<\/a> 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.<\/p>\n

Chemical Reactions and Equations NCERT Solutions for Class 10 Science Chapter 1<\/h2>\n

Class 10 Science Chapter 1 Chemical Reactions and Equations InText Questions and Answers<\/h3>\n

In-text Questions (Page 6)<\/span><\/p>\n

Question 1.
\nWhy should magnesium ribbon be cleaned before burning air ?
\nAnswer:
\nA 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.<\/p>\n

Question 2.
\nWrite the balanced equation for the following chemical reactions:
\n(i) Hydrogen + Chlorine \u2192 Hydrogen chloride
\n(ii) Barium chloride + Aluminium sulphate \u2192 Barium sulphate + Aluminium chloride
\n(iii) Sodium + Water \u2192 Sodium hydroxide + Hydrogen
\nAnswer:
\n(i) H2<\/sub>(g) + Cl2<\/sub>(g) \u2192 2HCl (g)
\n(ii) 3 BaCl2<\/sub> + Al2<\/sub>(SO4<\/sub>)3<\/sub> \u2192 3 BaSO4<\/sub> + 2AlCl3<\/sub>
\n(iii) Na (s) + H2<\/sub>O (l) \u2192 NaOH (aq) + H2<\/sub>(g)<\/p>\n

Question 3.
\nWrite a balanced chemical equation with state symbols for the following reactions :
\n(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
\n(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride and water.
\nAnswer:
\n(i) BaCl2<\/sub> (aq) + Na2<\/sub>SO4<\/sub> (aq) \u2192 BaSO4<\/sub> (\u2193) + 2NaCl (aq)
\n(ii) NaOH (aq) + HCl (aq) \u2192 NaCl (aq) + H2<\/sub>O (l)<\/p>\n

In-text Questions (Page 10)<\/span><\/p>\n

Question 1.
\nA solution of a substance \u2018X is used for white-washing:
\n(i) Name the substance ‘X’ and write its formula.
\n(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
\nAnswer:
\n(i) The substance ‘X’ is quick lime or calcium oxide.
\n\"NCERT<\/p>\n

Question 2.
\nWhy 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.
\nAnswer:
\nWhen electricity is passed through the water in presence of an acid., water is electroysed and produces hydrogen and oxygen gas.
\n\\(2 \\mathrm{H}_{2} \\mathrm{O}(l) \\stackrel{\\text { Electrolysis }}{\\longrightarrow} 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)\\)
\nIt 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.
\nThe double amount gas is hydrogen.<\/p>\n

\"NCERT<\/p>\n

In-text Questions (Page 13)<\/span><\/p>\n

Question 1.
\nWhy does the colour of copper sulphate solution change when an iron nail is dipped in it?
\nAnswer:
\nIron is more reactive that copper so it displaces copper metal from copper sulphate solution and produces sulphate of iron.
\n\"NCERT<\/p>\n

Displacement calculator<\/a> that determines the size of an engine based on Bore, Stroke and number of cylinders.<\/p>\n

Question 2.
\nGive an example of a double displacement reaction other than the one given in Activity 1.10.
\nAnswer:
\n\"NCERT<\/p>\n

Question 3.
\nIdentify the substances that are oxidised and the substances that are reduced in the
\nfollowing reactions.
\n(i) Na(s) + O2<\/sub>(g) \u2192 2Na2<\/sub>O(s)
\n(ii) CuO(s) + H2<\/sub> (g) \u2192 CU(s) + H2<\/sub>O(l)
\nAnswer:
\n(i) 2 Na (s) + O2<\/sub> (g) \u2192 2 NO2<\/sub>O (s)
\nNa (s) \u2192 Oxidised
\nO2<\/sub>(g) \u2192 Reduced
\n(ii) CuO (s) + H2<\/sub>(g) \u2192 Cu (S) + H2<\/sub>O (l)
\nCuO (s) \u2192 Reduced
\nH2 <\/sub>(g) \u2192 Oxidised<\/p>\n

Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Questions and Answers<\/h3>\n

Page No. 14<\/span><\/p>\n

Question 1.
\nWhich of the statements about the reaction below are incorrect?
\n2PbO(s) + C(s) \u2192 2Pb(s) + CO2<\/sub>(g)
\n(a) Lead is getting reduced.
\n(b) Carbon dioxide is getting oxidised.
\n(c) Carbon is getting oxidised.
\n(d) Lead oxide is getting reduced.
\n(i) (a) and (b)
\n(ii) (a) and (c)
\n(iii) (d), (b) and (c)
\n(iv) all
\nAnswer:
\n(i) a and b<\/p>\n

Question 2.
\nFe2<\/sub>O3<\/sub> + 2Al \u2192 Al2<\/sub>O3<\/sub> + 2Fe
\nThe above reaction is an example of a
\n(a) combination reaction.
\n(b) double displacement reaction.
\n(c) decomposition reaction.
\n(d) displacement reaction.
\nAnswer:
\n(d) displacement reaction.<\/p>\n

Question 3.
\nWhat happens when dilute hydrochloric add is added to iron fillings? Tick the correct answer.
\n(a) Hydrogen gas and iron chloride are produced.
\n(b) Chlorine gas and iron hydroxide are produced.
\n(c) No reaction takes place.
\n(d) Iron salt and water are produced.
\nAnswer:
\n(a) Hydrogen gas and iron chloride are produced.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nWhat is a balanced chemical equation? Why should chemical equations be balanced?
\nAnswer:
\nA 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.
\ne.g. H2<\/sub> + O2<\/sub> \u2192 H2<\/sub>O
\nThe above equation is not a balanced equation because it does not have the equal no. of atoms of oxygen in both sides.
\nBut 2H2<\/sub> + O2<\/sub> \u2192 2 H2<\/sub>O is a balanced equation because it does not have the equal no. atoms of different elements in both the sides.<\/p>\n

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.<\/p>\n

Question 5.
\nTranslate the following statements into chemical equations and then balance them.
\n(a) Hydrogen gas combines with nitrogen to form ammonia.
\n(b) Hydrogen sulphide gas bums in air to give water and sulpur dioxide.
\n(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
\n(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
\nAnswer:
\n(a) H2<\/sub> (g) + N2<\/sub>(g) \u2192 NH3<\/sub> (g)
\nStep I. Look more closely the number of atoms of different elements present in the unbalanced chemical equation.<\/p>\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
H<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
N<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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.<\/p>\n\n\n\n\n\n
Atoms of Hydrogen<\/td>\nIn reactants<\/td>\nIn Products<\/td>\n<\/tr>\n
Initial<\/td>\n2 (in H2<\/sub>)<\/td>\n3 (in NH3<\/sub>)<\/td>\n<\/tr>\n
To balance<\/td>\n2 \u00d7 3<\/td>\n3 \u00d7 2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Now the partly balanced equation becomes as follows :
\n3H2<\/sub> + N2<\/sub> \u2192 2NH3<\/sub><\/p>\n

Step III. But nitrogen in the above equation is automatically balanced. So the balanced equation will be
\n3 H2<\/sub> + N2<\/sub> \u2192 NH3<\/sub>
\nThe above equation balanced because it contains the equal no. of atoms of hydrogen and nitrogen on both sides.
\n(b) H2<\/sub>S + O2<\/sub> \u2192 H2<\/sub>O + H2<\/sub>O
\nThe above equation can be balanced by algebraic sum method.<\/p>\n

Let ‘a’, ‘b’, \u2018c’ and ‘d’ are the no. of molecules of ‘H2<\/sub>S’, ‘O2<\/sub>‘, ‘H2<\/sub>O’ and ‘SO2<\/sub>‘ respectively in the balanced chemical equation. So the above equation can be written as
\na H2<\/sub>S + b O2<\/sub> \u2192 c H2<\/sub>O + d SO2<\/sub>
\nNow in a balanced chemical equation.
\nL.H.S. R.H.S.
\nHydrogen 2 a= 2c …(i)
\nSulphur a = d …(ii)
\nOxygen 2b = c + 2d ….(iii)
\nLet a = 1 …..(iv)<\/p>\n

So from equations (i) and (iv)
\n2 \u00d7 1 = 2 c
\n\u21d2 c = 1
\nFrom equation (iii)
\n2b = c + 2d
\n2b = 1 + 2 \u00d7 1
\n\u21d2 2b = 3
\n\u21d2 b = \\(\\frac {3}{2}\\)
\nNow by putting the value of a, b, c and d in the given equation, we get
\n1 \u00d7 H2<\/sub>S + \\(\\frac {3}{2}\\)O2<\/sub> \u2192 1 \u00d7 H2<\/sub>O + 1 \u00d7 SO2<\/sub>
\nOR 2 H2<\/sub>S + 3O2<\/sub> \u2192 2 H2<\/sub>O + 2SO2<\/sub><\/p>\n

(c)
\n\"NCERT<\/p>\n

Step I:<\/p>\n\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Ba<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cl<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
Al<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
S<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n
O<\/td>\n12<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II: Now start from Al2<\/sub> (SO4<\/sub>)3<\/sub> 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<\/sub> (SO4<\/sub>)3<\/sub> by ‘1’ ‘BaCl2<\/sub>‘ by ‘3’, we get
\nBaCl2 + Al2 (SO4)3 \u2192 AlCl3 + 3BaSO4<\/p>\n

Step III: Now balance ‘Al’ be multiplying ‘2’ in the product side (i.e. in AlCl3<\/sub>)
\nBaCl2<\/sub> + Al2<\/sub> (SO4<\/sub>)3<\/sub> \u2192 2 AlCl3<\/sub> + 3 BaSO4<\/sub><\/p>\n

Step IV : Balance barium by multiplying ‘3’ in the reactant to BaCl2<\/sub>.
\n3 BaCl2<\/sub> + Al2<\/sub> (SO4<\/sub>)3<\/sub> \u2192 2 AlCl3<\/sub> + 3 BaSO4<\/sub>
\n‘Cl’ is automatically balanced in the above equation.<\/p>\n

(d) K + H2<\/sub>O \u2192 KOH + H2
\n<\/sub>Step I.<\/p>\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
K<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
H<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
O<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II: Now select ‘KOH’ in the reactant side because it contains the maximum no. of elements.
\nK + 2H2<\/sub>O \u2192 2 KOH + H2<\/sub><\/p>\n

Step III: Balance potassium metal in the above partially balanced equation by multiplying ‘2’ in the potassium i.e. reactant side.
\n2K + 2H2<\/sub>O \u2192 2KOH + H2<\/sub><\/p>\n

\"NCERT<\/p>\n

Question 6.
\nBalance the following chemical equations.
\n(a) HNO3<\/sub> + Ca(OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + H2<\/sub>O
\n(b) NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O
\n(c) NaCl + AgNO3<\/sub> \u2192 AgCl + NaNO3<\/sub>
\n(d) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + HCl
\nAnswer:
\n(a)
\nStep I.<\/p>\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
H<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n
N<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
O<\/td>\n5<\/td>\n7<\/td>\n<\/tr>\n
Ca<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II. Select Ca (NO3<\/sub>)2<\/sub> because it contains the maximum no. of elements. To balance oxygen multiply Ca (OH)2 <\/sub>by two, we get
\nHNO3<\/sub>+ 2Ca (OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\n

Step III. Now balance calcium atoms by multiplying two in reactants sides. We get
\nHNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 Ca (NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\n

Step IV : Balance nitrogen in the above partially balanced equation, multiply HNO3<\/sub> by ‘4’. We get
\n4 HNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 2 Ca (NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\n

Step V : Balance hydrogen by multiplying ‘4’ in the product side. We get,
\nHNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 2 Ca (NO3<\/sub>)2<\/sub> + 4H2<\/sub>O
\nThis is a balanced chemical equation.<\/p>\n

(b) NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O
\nStep I.<\/p>\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Na<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
O<\/td>\n5<\/td>\n5<\/td>\n<\/tr>\n
H<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n
S<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II: Balance sodium in the above equation by multiplying ‘2’ in NaOH.
\n2 NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O<\/p>\n

Step III: Balance hydrogen by multiplying ‘2’ in H2<\/sub>O.
\n2 NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + 2 H2<\/sub>O
\nThis is a balanced equation.<\/p>\n

(c) NaCl + AgNO3<\/sub> \u2192 AgCl + NaNO3<\/sub>
\nStep I.<\/p>\n\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Na<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cl<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Ag<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
N<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
O<\/td>\n3<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The above equation contains equal no. of atoms of different elements in reactants and products sides.<\/p>\n

(d) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + HCl
\nStep I.<\/p>\n\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Ba<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cl<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
H<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
S<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
O<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II : Now balance chlorine atoms in the above equation by multiplying ‘2’ in the product side to HCl
\nBaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + 2HCl
\nThis is a balanced chemical equation.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nWrite the balanced chemical equations for the following reactions.
\n(a) Calcium hydroxide + Carbon dioxide \u2192 Calcium carbonate + Water
\n(b) Zinc + Silver nitrate \u2192 Zinc nitrate + Silver
\n(c) Aluminium + Copper chloride \u2192 Aluminium chloride + Copper
\n(d) Barium chloride + Potassium sulphate \u2192 Barium sulphate + Potassium chloride.
\nAnswer:
\n(a) Ca (OH2<\/sub>) + CO2<\/sub> \u2192 CaCO3<\/sub> + H2<\/sub>O
\nStep I.<\/p>\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Ca<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
O<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n
H<\/td>\n2<\/td>\n2<\/td>\n<\/tr>\n
C<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The above equation is balanced.<\/p>\n

(b) Zn + AgNO3<\/sub> \u2192 Zn (NO3<\/sub>)2<\/sub> + Ag
\nStep I.<\/p>\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Zn<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Ag<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
N<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
O<\/td>\n3<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step II: Balance oxygen by multiplying ‘2’ in AgNO3<\/sub>, we get
\nZn + 2 AgNO3<\/sub> \u2192 Zn (NO3<\/sub>)2<\/sub> + Ag<\/p>\n

Step III ; Now Balance ‘Ag’ metal by multiplying ‘2’ in product side.
\nZn + 2 AgNO3<\/sub> \u2192 Zn (NO3<\/sub>)2<\/sub> + 2Ag<\/p>\n

(c) Al + CuCl2<\/sub> AlCl3<\/sub> + Au
\nStep I.<\/p>\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Al<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cu<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cl<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

By multiplying ‘AlCl3<\/sub>‘ with \u20182\u2019 and ‘CuCl2<\/sub>‘ with ‘3’, ‘Al’ with \u20182′ and ‘Cu’ with ‘3’ we get
\n2Al + 3 CuCl2<\/sub> \u2192 2AlCl3<\/sub> + 3 Cu
\nThis is a balanced equation<\/p>\n

(d) BaCl2<\/sub> + K2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + KCl
\nStep I.<\/p>\n\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
Ba<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Cl<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
K<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
S<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
O<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Balance chlorine by multiplying ‘2’ in NaCl, we get
\nBaCl2<\/sub> + K2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub>+ 2KCl<\/p>\n

Question 8.
\nWrite the balanced chemical equation for the following reactions and identify the type of reaction in each case.
\n(a) Potassium bromide(aq) + Barium iodide(aq) \u2192 Potassium iodide(aq) + Barium bromide(s)
\n(b) Zinc carbonate(s) \u2192 Zinc oxide(s) + Carbon dioxide(g)
\n(c) Hydrogen(g) + Chlorine(g) \u2192 Hydrogen chloride(g)
\n(d) Magnesium(s) + Hydrochloric acid(aq) \u2192 Magnesium chloride(aq) + Hydrogen(g)
\nAnswer:
\n(a) KBr (aq) + Bal2<\/sub> (aq) \u2192 KI (aq) BaBr2<\/sub> (aq)
\nStep I.<\/p>\n\n\n\n\n\n\n\n
Element<\/td>\nNo. of atoms in Reactants<\/td>\nNo. of atoms in Products<\/td>\n<\/tr>\n
K<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
Br<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
Ba<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
I<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Balance bromine and iodine in the above equation.
\n2 KBr (aq) + Bal2<\/sub> (aq) \u2192 2 Kl (aq) + BaBr2<\/sub> (aq)
\nIt is a double displacement reaction.
\n(b) ZnCO3<\/sub> (s) \u2192 ZnO (s) + CO2<\/sub> (g)
\nIt is decomposition reaction.
\n(c) H2<\/sub>(g) + Cl2<\/sub>(g) \u2192 2 HCl (g)
\nIt is combination reaction.
\n(d) Mg (s) + 2 HCl (aq) \u2192 MgCl2<\/sub> (aq) + H2<\/sub> (g)
\nIt is a displacement reaction.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nWhat does one mean by exothermic and endothermic reactions? Give examples.
\nAnswer:
\nExothermic Reaction: A reaction in which heat energy is given out along with the product is called exothermic reaction.
\nC (s) O2<\/sub> (g) \u2192 CO2<\/sub> (g) + Heat energy (394 kJ)
\nN2<\/sub> (g) + 3H2<\/sub> \u2192 (g) 2 NH3<\/sub> (g) + Heat energy (92.4 kJ)<\/p>\n

Endothermic Reaction : A reaction in which heat energy is absorbed along with the product is called endothermic reaction.
\nCaCO3<\/sub> (s) \u2192 CaO (s) CO2<\/sub> (g)-Heat energy
\n2HgO (s) \u2192 2 Hg (l) + O2<\/sub> (g)-Heat energy<\/p>\n

Question 10.
\nWhy is respiration considered an exothermic reaction? Explain.
\nAnswer:
\nIt 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.
\nC6<\/sub>H12<\/sub>O6<\/sub> (aq) + 6O2<\/sub> (aq) \u2192 6 CO2<\/sub> (aq) 6H2<\/sub>O (l) + energy
\nBecause energy is released during respiration so it is an exothermic reaction.<\/p>\n

Question 11.
\nWhy are decomposition reactions called opposite of combination reactions? Write equations for these reactions.
\nAnswer:
\nIn decomposition reactions a single reactant breaks down to give two or more simpler products.
\n\"NCERT
\nIn combination reaction two or more than two reactants combine with each other and to give simple product.
\n\"NCERT
\nSo it is dear from above discussion that decomposition reactions are opposite of the combination reactions,<\/p>\n

Question 12.
\nWrite equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
\nAnswer:
\nIn a decomposition reaction a single reactant breaks down to give two or more simpler products. These reactions need energy in different forms to proceed.<\/p>\n

The decomposition of CaCO3<\/sub> (s) takes place by supplying energy in the form of heat.
\n\\(\\mathrm{CaCO}_{3}(s) \\stackrel{\\Delta}{\\rightarrow} \\mathrm{CaO}(\\mathrm{s})+\\mathrm{CO}_{2}(g)\\)
\nIn the presence of sunlight white silver chloride decomposes and turns grey because silver metal is formed.
\n\"NCERT
\nWater decomposes into H, and O, by passing electricity in an electrolytic cell.
\n\\(2 \\mathrm{H}_{2} \\mathrm{O}(l) \\stackrel{\\text { Electricity }}{\\rightarrow} 2 \\mathrm{H}_{2}(g) \\mathrm{O}_{2}(g)\\)<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nWhat is the difference between displacement and double displacement reactions ? Write equations for these reactions.
\nAnswer:
\nDisplacement Reaction : In a displacement reaction, more active element displaces or removes another element from its compound.
\ne.g. Zn (s) + CuSO4<\/sub> (aq) \u2192 ZnSO4<\/sub> (aq) + Cu (s)<\/p>\n

Double displacement reactions: In a double displacement reaction two different atoms or groups of atoms are displaced by other atoms or group of atoms.
\ne.g
\n\"NCERT<\/p>\n

Question 14.
\nIn the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
\nAnswer:
\n\"NCERT<\/p>\n

Question 15.
\nWhat do you mean by a precipitation reaction? Explain by giving examples.
\nAnswer:
\nPrecipitation reaction : A reaction in which precipitates are formed is called precipitation reaction.
\ne.g.
\n\"NCERT<\/p>\n

Question 16.
\nExplain the following in terms of gain or loss of oxygen with two examples each.
\n(a) Oxidation
\n(b) Reduction
\nAnswer:
\n(a) Oxidations A process which involves the gain of oxygen is called oxidation. In oxidation reaction oxygen is added to the reactant.
\ne.g. 2 Mg (s) + O2<\/sub> (g) \u2192 2 MgO (s)
\n2 H2<\/sub>S (g) + O2<\/sub> (g) \u2192 2 S (s) + 2 H2<\/sub>O (l)<\/p>\n

(b) Reduction: A process which involves the loss of oxygen is called reduction.
\n2 Pb (NO3<\/sub>)2<\/sub> (S) \u2192 2 PbO (s) + 2NO2<\/sub> (g) + \\(\\frac {1}{2}\\)O2<\/sub> (g)
\n2 KNO3<\/sub> \u2192 2 KNO2<\/sub> + O2<\/sub><\/p>\n

Question 17.
\nA shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
\nAnswer:
\nThe shiny brown coloured element ‘X’ is copper.
\nWhen copper is heated in air it forms copper oxide which is black in colour.
\n\"NCERT<\/p>\n

Question 18.
\nWhy do we apply paint on iron articles?
\nAnswer:
\nWhen 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.<\/p>\n

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.<\/p>\n

\"NCERT<\/p>\n

Question 19.
\nOil and fat containing food items are flushed with nitrogen. Why?
\nAnswer:
\nWhen 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.<\/p>\n

Question 20.
\nExplain the following terms with one example each.
\n(a) Corrosion (b) Rancidity
\nAnswer:
\n(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.<\/p>\n

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. Fe2<\/sub>O3<\/sub>. xH2<\/sub>O.<\/p>\n

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.<\/p>\n

(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.<\/p>\n

Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Activities<\/h3>\n

Activity 1.1 (Page 1)<\/span><\/p>\n

Caution : This activity needs teacher’s assitsance. It would be better if students wear eye
\n\"NCERT
\nFig : Burning of a magnesium ribbon in air and collection of magnesium oxide in watch glass<\/p>\n