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Chapter 1

The Language of Chemistry

Class 9 - Dalal Simplified ICSE Chemistry Solutions


Questions

Question 1(1985)

XCl2 is the chloride of a metal X. State the formula of the sulphate and the hydroxide of the metal X.

Answer

Chloride of a metal X is XCl2

By interchanging subscript and writing as superscript:

X11  Cl2X22  Cl1\underset{\phantom{1}{1}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{Cl}} \Rightarrow \overset{\phantom{2}{2}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{Cl}} \\[0.5em]

Therefore, valency of metal X = 2.

Formula of the sulphate:

X2+SO42X22  S2O4X22  S2O4\text{X}^{2+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{S}}\text{O}_{4} \\[0.5em]

As valency of both X and SO4 is 2 so dividing by 2 we get 1 but 1 is never written so we get the formula as XSO4\bold{XSO}_{\bold{4}}

Formula of the hydroxide:

X2+OH1X22  O1HX11  O2H\text{X}^{2+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}}\text{H} \\[0.5em]

Dropping 1 and enclosing OH in brackets, we get the formula as X(OH)2\bold{X(OH)}_{\bold{2}}

Therefore, we get

Formula of Sulphate : XSO4\bold{XSO}_{\bold{4}}

Formula of Hydroxide : X(OH)2\bold{X(OH)}_{\bold{2}}

Question 1(1987)

An element X is trivalent. Write the balanced equation for the combustion of X in oxygen.

Answer

Valency of X is 3+ and that of Oxygen is 2-

By interchanging the valency number and shifting it to the lower right side of the atom or radical, we get the formula : X2O3

Equation for the combustion of X in oxygen: 4X + 3O2 ⟶ 2X2O3

Question 1(1991)

The formula of the nitride of a metal X is XN, state the formula of :

(i) it's sulphate (ii) it's hydroxide.

Answer

Nitride of a metal X is XN.

Since valency of nitrogen is 3-
so valency of X is 3+

Formula of the sulphate:

X3+SO42X33  S2O4X22  S3O4\text{X}^{3+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{S}}\text{O}_{4} \\[0.5em]

So, we get the formula as X2(SO4)3\bold{X}_2\bold{(SO}_{\bold{4}}\bold{)}_\bold{3}

Formula of the hydroxide:

X3+OH1X33  O1HX11  O3H\text{X}^{3+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}}\text{H} \\[0.5em]

Dropping 1, we get the formula as X(OH)3\bold{X(OH)}_{\bold{3}}

Therefore, we get

Formula of Sulphate : X2(SO4)3\bold{X}_2\bold{(SO}_{\bold{4}}\bold{)}_\bold{3}

Formula of Hydroxide : X(OH)3\bold{X(OH)}_{\bold{3}}

Question 1(1992)

What is the valency of nitrogen in :

(i) NO

(ii) N2O

(iii) NO2

Answer

(i) NO=N11  O1N11  O1\text{(i)} \space \text{NO} = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]

But valency of O is 2. Multiplying by 2, we get:

N2×1  O2×1N22  O2\overset{{2 \times 1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Nitrogen is 2.

(ii) N2O=N22  O1N11  O2\text{(ii)} \space \text{N}_2\text{O} = \underset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Nitrogen is 1.

(ii) NO2=N11  O2N22  O1\text{(ii)} \space \text{NO}_2 = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]

But valency of O is 2. Multiplying by 2, we get:

N2×2  O2×1N44  O2\overset{{2 \times 2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Nitrogen is 4.

Additional Questions

Question 1

What is meant by the term 'symbol'. Give the qualitative and quantitative meaning of the term 'symbol'.

Answer

Symbol represents the short form of an element.

Qualitative meaning — A symbol represents specific element or one atom of an element. E.g., 'S' represents one atom of the element sulphur.

Quantitative meaning — A symbol also represents the weight of the element equal to it's atomic weight i.e. it represents how many times an atom is heavier than one atomic mass unit [a.m.u.] which is defined as 112\dfrac{1}{12}th the mass of a carbon atom C12

Question 2

Name three metals whose symbols are derived from :

(a) the first letter of the name of the element

(b) from their Latin names.

Answer

(a) Elements whose symbols are derived from first letter of the name of the element :

  • Carbon - C
  • Sulphur - S
  • Oxygen - O

(b) Elements whose symbols are derived from their Latin names :

  • Kalium [Potassium] - K
  • Ferrum [Iron] - Fe
  • Natrium [Sodium] - Na

Question 3

Explain the meaning of the term 'valency'. State why the valency of the metal potassium is +1 and of the non-metal chlorine is -1.

Answer

Valency is the number of hydrogen atoms which can combine with or displace one atom of the element or radical so as to form a compound.

Valency of a metal is the number of electrons lost per atom of the metal. Valency of all metals and hydrogen is considered positive. Therefore, valency of potassium is 1+ as it has 1 electron in outer most shell which it loses and becomes K+.

Valency of non-metal is the number of electrons gained per atom of the non-metal. Valency of all non-metals/groups of non-metals is taken as negative. Therefore, valency of non-metal chlorine is -1, because chlorine [Cl] has 7 electrons in valence shell and gains 1 electron and becomes [Cl].

Question 4

What is meant by the term 'variable valency'. Give a reason why silver exhibits a valency of +1 and +2.

Answer

Certain elements exhibit more than one valency hence it is said that these elements have variable valency.

Reasons for exhibiting variable valency — An atom of an element can sometimes lose more electrons than are present in it's valence shell. This happens when it loses electrons from the penultimate [i.e., last but one] shell and hence exhibit more than one or variable valency.

Variable valency of Silver — Atomic number of Silver (Ag) is 47. Its electronic configuration is [2, 8, 18, 18, 1]. The outermost shell has one electron and the penultimate shell has 18 electrons. However, the penultimate shell has not attained stability and one more electron sometimes jumps to the outermost shell there by increasing valence electrons and the new configuration [2, 8, 18, 17, 2] loses two electrons and has valency [+2]. Therefore, Silver exhibits varibale valency forming Ag1+ and Ag2+.

Question 5

Give examples of eight metals which show variable valency. State the valency of sulphur in :

(a) SO2
(b) SO3

Answer

Eight metals which show variable valency are :

1+2+3+4+
CuCu
HgHg
AgAg
Au  Au
  FeFe
  Pb  Pb
  Sn  Sn
  Mn  Mn

(a) Valency of sulphur in SO2 is 4

By interchanging subscript and writing as superscript:

S11  O2S22  O1\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]

But valency of O is 2. Multiplying by 2, we get:

S2×2  O2×1S44  O2\overset{{2 \times 2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Sulphur is 4.

(b) Valency of sulphur in SO3 is 6

S11  O3S33  O1\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{O}} \Rightarrow \overset{\phantom{3}{3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]

But valency of O is 2. Multiplying by 2, we get:

S2×3  O2×1S66  O2\overset{{2 \times 3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{6}{6}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Sulphur is 6.

Question 6

State the valency in each case and name the following elements or radicals given below :

  1. K
  2. Cr2O7
  3. Cl
  4. Ni
  5. ClO3
  6. CO3
  7. Ba
  8. HCO3
  9. NO2
  10. Na
  11. Br
  12. Zn
  13. Mg
  14. O
  15. Co
  16. CrO4
  17. ClO
  18. MnO4
  19. Li
  20. I
  21. OH
  22. O2
  23. ZnO2
  24. SiO3
  25. NO3
  26. SO3
  27. SO4
  28. PO4
  29. N
  30. C
  31. PO3
  32. Al
  33. Ca
  34. H
  35. PbO2
  36. HSO3
  37. AlO2
  38. Cr
  39. HSO4
  40. NH4

Answer

Sl.
No.
ElementValencyName
1.KK1+Potassium
2.Cr2O7Cr2O72-Dichromate
3.ClCl1-Chloride
4.NiNi2+Nickle
5.ClO3ClO31-Chlorate
6.CO3CO32-Carbonate
7.BaBa 2+Barium
8.HCO3HCO3 1-Hydrogen [Bi] Carbonate
9.NO2NO21-Nitrite
10.NaNa1+Sodium
11.BrBr1-Bromide
12.ZnZn2+Zinc
13.MgMg2+Magnesium
14.OO2-Oxide
15.CoCo2+Cobalt
16.CrO4CrO42-Chromate
17.ClOClO1-Hypochlorite
18.MnO4MnO41-Permanganate
19.LiLi1+Lithium
20.II1-Iodide
21.OHOH1-Hydroxide
22.O2O2 2-Peroxide
23.ZnO2ZnO22-Zincate
24.SiO3SiO32-Silicate
25.NO3NO31-Nitrate
26.SO3SO32-Sulphite
27.SO4SO42-Sulphate
28.PO4PO43-Phosphate
29.NN3-Nitride
30.CC4-Carbide
31.PO3PO33-Phosphite
32.AlAl3+Aluminium
33.CaCa2+Calcium
34.HH1+Hydrogen
35.PbO2PbO22-Plumbite
36.HSO3HSO31-Hydrogen [Bi] Sulphite
37.AlO2AlO21-Aluminate
38.CrCr3+Chromium
39.HSO4HSO41-Hydrogen [Bi] Sulphate
40.NH4NH41+Ammonium

Question 7

State the variable valencies of the following elements and give their names. (a) Cu, (b) Ag, (c) Hg, (d) Fe, (e) Pb, (f) Sn, (g) Mn, (h) Pt, (i) Au

Answer

Sl.
No.
ElementValencyName
(a)CuCu1+
Cu2+
Copper [I] i.e., cuprous
Copper [II] i.e., cupric
(b)AgAg1+
Ag2+
Silver [I] Argentous
Silver [II] Argentic
(c)HgHg1+
Hg2+
Mercury [I] Mercurous
Mercury [II] Mercuric
(d)FeFe2+
Fe3+
Iron [II] Ferrous
Iron [III] Ferric
(e)PbPb2+
Pb4+
Lead [II] Plumbous
Lead [IV] Plumbic
(f)SnSn2+
Sn4+
Tin [II] Stannous
Tin [IV] Stannic
(g)MnMn2+
Mn4+
Manganese [II] Manganous
Manganese [IV] Manganic
(h)PtPt2+
Pt4+
Platinum [II] Platinous
Platinum [IV] Platinic
(i)AuAu1+
Au3+
Gold [I] Aurous
Gold [III] Auric

Question 8

State which of the following elements or radicals are divalent –

(a) Lithium, (b) Nickel, (c) Ammonium, (d) Bromide, (e) Sulphite, (f) Nitride, (g) Carbide, (h) Chromium, (i) Bisulphite, (j) Dichromate, (k) Permanganate.

Answer

The divalent elements or radicals are:

  • Nickel
  • Sulphite
  • Dichromate

Question 9

Explain the meaning of the term 'compound' with a suitable example. State the main characteristics of a compound with special reference to the compound iron [II] sulphide.

Answer

Compound is a pure substance composed of two or more elements combined chemically in a fixed proportion. For example, water (H2O) is composed of two elements — Hydrogen and Oxygen in a fixed proportion by weight.

Main characteristics - with special reference to the compound iron [II] sulphide:

  1. Components are present in - a definite proportion.
  • Elements iron and sulphur combine in a definite proportion.
  1. Compound is always homogenous.
  2. Particles in a compound are of one kind.
  • Composition of iron [II] sulphide, - cannot be varied, hence is uniform composition.
  • Component - may not be seen separately.
  1. Compound [iron [II] sulphide] has a definite set of properties.
  2. Components in [Iron [II] Sulphide] do not retain their original properties and can be separated only by chemical means.
  • Particles in iron [II] sulphide are chemically combined and hence -
  • iron in iron [II] sulphide cannot be attracted by a magnet and does not give H2 with dil. acid.
  • sulphur present in iron [II] sulphide - is insoluble in solvent CS2.

Question 10

Name the elements in the compound and give the formula – of the following compounds :

(a) Nitric acid, (b) Carbonic acid, (c) Phosphoric acid, (d) Acetic acid, (e) Blue vitriol, (f) Green vitriol, (g) Glauber's salt, (h) Ethane, (i) Ethanol

Answer

Sl.
No.
CompoundName of elementsFormula
(a)Nitric acidHydrogen, Nitrogen, OxygenHNO3
(b)Carbonic acidHydrogen, Carbon, OxygenH2CO3
(c)Phosphoric acidHydrogen, Phosphorous, OxygenH3PO4
(d)Acetic acidCarbon, Hydrogen, OxygenCH3COOH
(e)Blue vitriolCopper, sulphur, Oxygen, HydrogenCuSO4.5H2O
(f)Green vitriolIron, Sulphur, Oxygen, HydrogenFeSO4.7H2O
(g)Glauber's saltSodium, Sulphur, Oxygen, HydrogenNa2SO4.10H2O
(h)EthaneCarbon, HydrogenC2H6
(i)EthanolCarbon, Hydrogen, OxygenC2H5OH

Question 11

Explain the term 'chemical formula'. State why the molecular formula of zinc carbonate is ZnCO3

Answer

A molecule of a substance i.e., element or compound could be represented by symbols. This representation is known as chemical formula.

As molecular formula also indicates the number of units of the radicals present in a compound with the proper subscript outside the unit of the radical and in the case of ZnCO3, there are 1 atom of Zn and C whereas 3 atoms of O. Therefore, we get ZnCO3.

Question 12(1)

Write the formula of the following compounds :

Potassium

(a)chloride(b)nitrate
(c)carbonate(d)bisulphate
(e)sulphite(f)dichromate
(g)permangnate(h)zincate
(i)plumbite(j)sulphate
(k)bicarbonate(l)aluminate
(m)hydroxide(n)iodide
(o)nitrite(p)bisulphite

Answer

Potassium

Sl.
No.
CompoundFormula
(a)chlorideKCl
(b)nitrateKNO3
(c)carbonateK2CO3
(d)bisulphateKHSO4
(e)sulphiteK2SO3
(f)dichromateK2Cr2O7
(g)permangnateKMnO4
(h)zincateK2ZnO2
(i)plumbiteK2PbO2
(j)sulphateK2SO4
(k)bicarbonateKHCO3
(l)aluminateKAIO2
(m)hydroxideKOH
(n)iodideKI
(o)nitriteKNO2
(p)bisulphiteKHSO3

Question 12(2)

Write the formula of the following compounds :

Sodium

(a)chloride(b)nitrate
(c)carbonate(d)bisulphate
(e)sulphite(f)dichromate
(g)permangnate(h)zincate
(i)plumbite(j)sulphate
(k)bicarbonate(l)aluminate
(m)hydroxide(n)iodide
(o)nitrite(p)bisulphite

Answer

Sodium

Sl.
No.
CompoundFormula
(a)chlorideNaCl
(b)nitrateNaNO3
(c)carbonateNa2CO3
(d)bisulphateNaHSO4
(e)sulphiteNa2SO3
(f)dichromateNa2Cr2O7
(g)permangnateNaMnO4
(h)zincateNa2ZnO2
(i)plumbiteNa2PbO2
(j)sulphateNa2SO4
(k)bicarbonateNaHCO3
(l)aluminateNaAlO2
(m)hydroxideNaOH
(n)iodideNaI
(o)nitriteNaNO2
(p)bisulphiteNaHSO3

Question 12(3)

Write the formula of the following compounds :

Calcium

(a)chloride(b)nitrate
(c)carbonate(d)bisulphite
(e)sulphite(f)sulphate
(g)bicarbonate(h)hydroxide

Answer

Sl.
No.
CompoundFormula
(a)chlorideCaCl2
(b)nitrateCa(NO3)2
(c)carbonateCaCO3
(d)bisulphiteCa(HSO3)2
(e)sulphiteCaSO3
(f)sulphateCaSO4
(g)bicarbonateCa(HCO3)2
(h)hydroxideCa(OH)2

Question 12(4)

Write the formula of the following compounds :

Magnesium

(a)chloride(b)nitrate
(c)carbonate(d)sulphate
(e)bicarbonate(f)hydroxide
(g)oxide  

Answer

Sl.
No.
CompoundFormula
(a)chlorideMgCl2
(b)nitrateMg(NO3)2
(c)carbonateMgCO3
(d)sulphateMgSO4
(e)bicarbonateMg(HCO3)2
(f)hydroxideMg(OH)2
(g)oxideMgO

Question 12(5)

Write the formula of the following compounds :

Zinc

(a)chloride(b)nitrate
(c)carbonate(d)sulphate
(e)hydroxide(f)oxide

Answer

Sl.
No.
CompoundFormula
(a)chlorideZnCl2
(b)nitrateZn(NO3)2
(c)carbonateZnCO3
(d)sulphateZnSO4
(e)hydroxideZn(OH)2
(f)oxideZnO

Question 12(6)

Write the formula of the following compounds :

Aluminium

(a)chloride(b)nitrate
(c)carbonate(d)sulphate
(e)hydroxide(f)oxide

Answer

Sl.
No.
CompoundFormula
(a)chlorideAlCl3
(b)nitrateAl(NO3)3
(c)carbonateAl2(CO3)3
(d)sulphateAl2(SO4)3
(e)hydroxideAl(OH)3
(f)oxideAl2O3

Question 12(7)

Write the formula of the following compounds :

Copper

Copper [I] chlorideCopper [II] chloride
Copper [I] oxideCopper [II] oxide
Copper [I] sulphideCopper [II] sulphide
Copper [II] nitrateCopper [II] sulphate
Tetra amine copper [II] sulphate 

Answer

CompoundFormula
Copper [I] chlorideCuCl
Copper [II] chlorideCuCl2
Copper [I] oxideCu2O
Copper [II] oxideCuO
Copper [I] sulphideCu2S
Copper [II] sulphideCuS
Copper [II] nitrateCu(NO3)2
Copper [II] sulphideCuS
Tetra amine copper [II] sulphate[Cu(NH3)4]SO4

Question 12(8)

Write the formula of the following compounds :

Iron

Iron [II] chlorideIron [III] chloride
Iron [II] oxideIron [III] oxide
Iron [II] sulphateIron [III] sulphate
Iron [II] sulphideIron [III] sulphide
Iron [II] hydroxideIron [III] hydroxide

Answer

CompoundFormula
Iron [II] chlorideFeCl2
Iron [III] chlorideFeCl3
Iron [II] oxideFeO
Iron [III] oxideFe2O3
Iron [II] sulphateFeSO4
Iron [III] sulphateFe2(SO4)3
Iron [II] sulphideFeS
Iron [III] sulphideFe2S3
Iron [II] hydroxideFe(OH)2
Iron [III] hydroxideFe(OH)3

Question 12(9)

Write the formula of the following compounds :

Lead

Lead [II] chlorideLead [II] oxide
Lead [II] hydroxideLead [II] nitrate
Lead [II] sulphate 

Answer

CompoundFormula
Lead [II] chloridePbCl2
Lead [II] oxidePbO
Lead [II] hydroxidePb(OH)2
Lead [II] nitratePb(NO3)2
Lead [II] sulphatePbSO4

Question 12(10)

Write the formula of the following compounds :

Silver

Silver [I] chlorideSilver [II] chloride
Diamine silver chloride 

Answer

CompoundFormula
Silver [I] chlorideAgCl
Silver [II] chlorideAgCl2
Diamine silver chloride[Ag(NH3)2]Cl

Question 13

Write the names of the following compounds :

(a)KClO(b)HClO
(c)NaClO3(d)AlN
(e)K2Cr2O7(f)KMnO4
(g)Ca3N2(h)Ca3(PO4)2
(i)H2SO3(j)HCl
(k)HNO3(l)H2SO4
(m)NH4OH(n)NaOH
(o)H2CO3(p)HNO2
(q)Mg(HCO3)2(r)NaAlO2
(s)K2PbO2(t)Cr2(SO4)3
(u)Na2O  

Answer

Sl.
No.
FormulaName
(a)KClOPotassium Hypochlorite
(b)HClOHydrogen Hypochlorite
(c)NaClO3Sodium Chlorate
(d)AlNAluminium Nitride
(e)K2Cr2O7Potassium Dichromate
(f)KMnO4Potassium Permanganate
(g)Ca3N2Calcium Nitride
(h)Ca3(PO4)2Calcium phosphate
(i)H2SO3Sulphurous acid
(j)HClHydrogen chloride
(k)HNO3Nitric acid
(l)H2SO4Sulphuric acid
(m)NH4OHAmmonium hydroxide
(n)NaOHSodium hydroxide
(o)H2CO3Carbonic acid
(p)HNO2Nitrous acid
(q)Mg(HCO3)2Magnesium Bicarbonate
(r)NaAlO2Sodium aluminate
(s)K2PbO2Potassium plumbite
(t)Cr2(SO4)3Chromium sulphate
(u)Na2OSodium oxide

Question 14

Explain the term 'chemical equation'. What is meant by 'reactants' and 'products' in a chemical equation.

Answer

Chemical equation is a shorthand form for a chemical change. It shows the result of a chemical change in which the reactants and the products are represented by:

  • symbols [in the case of elements]
  • formula [in the case of compounds]

The substances which take part in a chemical change and are written on the left side of the equation are called the reactants.

The substances formed as a result of a chemical change and are written on the right side of the equation are called the products.

E.g., NH3 + H2O ⟶ NH4OH

Here, NH3 and H2O are reactants and NH4OH is the product.

Question 15

Give an example of a chemical equation in which two reactants form –

(a) one product
(b) two products
(c) three products
(d) four products

Answer

(a) NH3 + H2O ⟶ NH4OH

(b) C2H5Br + NaOH ⟶ C2H5OH + NaBr

(c) 6CO2 + 12H2O ⟶ C6H12O6 + 6H2O + 6O2

(d) 2KMnO4 + 16HCl ⟶ 2KCl + 2MNCl2 + 8H2O + 5Cl2

Question 16

2KClO3MnO22KCl+3O2[g]2\text{KClO}_3 \xrightarrow{\text{MnO}_2} 2\text{KCl} + 3\text{O}_2[\text{g}] is a balanced equation.

(a) State what is a 'balanced equation'.

(b) Give a reason why the above equation is balanced.

(c) State why the compound MnO2 is written above the arrow.

Answer

(a) A balanced equation is one in which the number of atoms of each element is the same on the side of the reactants and on the side of the products.

(b) As atoms of each reactant [K, Cl, O] on the L.H.S. is equal to the number of the atoms of products on the R.H.S., hence the given equation is balanced.

(c) MnO2 is a catalyst [simply increases the rate of reaction] and it does not take part in the reaction and so it is written on the top of the arrow.

Question 17

What do the following symbols present in a chemical equation mean.

(i) ⟶

(ii) ⇌

(iii) (s)

(iv) (I)

(v) (g)

(vi) (aq.)

Answer

(i) ⟶ : The direction of the reaction is irreversible

(ii) ⇌ : The direction of the reaction is reversible

(iii) (s) : The state of the matter is solid

(iv) (I) : The state of the matter is liquid

(v) (g) : The state of the matter is gas

(vi) (aq.) : The substance is in aqueous form.

Question 18

CaCO3 + 2HCl [dil.] ⟶ CaCl2 + H2O + CO2 [g]

(a) State the information provided by the above chemical equation.

(b) State the information not conveyed by the above chemical equation.

Answer

(a) Information provided by the equation:

  • One molecule of calcium carbonate reacts with two molecules of [dil.] hydrochloric acid to produce one molecule of calcium chloride, one molecule of water and one molecule of carbon dioxide.
  • The reaction is irreversible.
  • Hydrochloric acid is in dil. form.
  • Carbon dioxide produced is in gaseous form.

(a) Information not provided by the equation:

  • The concentration of the reactants and products.
  • The speed of the reaction
  • Change in colour occurring
  • Completion of the reaction i.e., whether the reaction is complete or not
  • Changes in evolution of light or sound energy during the occurrence of the reaction.

Question 19

Balance the following simple equations :

  1. C + O2 ⟶ CO

  2. N2 + O2 ⇌ NO

  3. ZnS + O2 ⟶ ZnO + SO2

  4. Al + O2 ⟶ Al2O3

  5. Mg + N2 ⟶ Mg3N2

  6. Al + N2 ⟶ AlN

  7. NO + O2 ⟶ NO2

  8. SO2 + O2 ⇌ SO3

  9. H2 + Cl2 ⟶ HCl

  10. Fe + Cl2 ⟶ FeCl3

  11. H2S + Cl2 ⟶ S + HCl

  12. FeCl2 + Cl2 ⟶ FeCl3

  13. CO2 + C ⟶ CO

  14. KHCO3 ⟶ K2CO3 + H2O + CO2

  15. K + CO2 ⟶ K2O + C

  16. Ca(OH)2 + HNO3 ⟶ Ca(NO3)2 + H2O

  17. K + H2O ⟶ KOH + H2

  18. Ca + H2O ⟶ Ca(OH)2 + H2

  19. Al + H2O ⟶ Al2O3 + H2

  20. Fe + H2O ⇌ Fe3O4 + H2

  21. Zn + NaOH ⟶ Na2ZnO2 + H2

  22. Zn + HCl ⟶ ZnCl2 + H2

  23. Al + H2SO4 ⟶ Al2(SO4)3 + H2

  24. H2 + O2 ⟶ H2O

  25. N2 + H2 ⟶ NH3

  26. Fe2O3 + H2O ⟶ Fe + H2O

  27. KBr + Cl2 ⟶ KCl + Br2

  28. NaOH + Cl2 ⟶ NaCl + NaClO + H2O

  29. NaHCO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2

  30. Mg + CO2 ⟶ MgO + C

  31. Fe2O3 + CO ⟶ Fe + CO2

  32. CaO + HCl ⟶ CaCl2 + H2O

Answer

  1. 2C + O2 ⟶ 2CO

  2. N2 + O2 ⇌ 2NO

  3. 2ZnS + 3O2 ⟶ 2ZnO + 2SO2

  4. 4Al + 3O2 ⟶ 2Al2O3

  5. 3Mg + N2 ⟶ Mg3N2

  6. 2Al + N2 ⟶ 2AlN

  7. 2NO + O2 ⟶ 2NO2

  8. 2SO2 + O2 ⇌ 2SO3

  9. H2 + Cl2 ⟶ 2HCl

  10. 2Fe + 3Cl2 ⟶ 2FeCl3

  11. H2S + Cl2 ⟶ S + 2HCl

  12. 2FeCl2 + Cl2 ⟶ 2FeCl3

  13. CO2 + C ⟶ 2CO

  14. 2KHCO3 ⟶ K2CO3 + H2O + CO2

  15. 4K + CO2 ⟶ 2K2O + C

  16. Ca(OH)2 + 2HNO3 ⟶ Ca(NO3)2 + 2H2O

  17. 2K + 2H2O ⟶ 2KOH + H2

  18. Ca + 2H2O ⟶ Ca(OH)2 + H2

  19. 2Al + 3H2O ⟶ Al2O3 + 3H2

  20. 3Fe + 4H2O ⟶ Fe3O4 + 4H2

  21. Zn + 2NaOH ⟶ Na2ZnO2 + H2

  22. Zn + 2HCl ⟶ ZnCl2 + H2

  23. 2Al + 3H2SO4 ⟶ Al2(SO4)3 + 3H2

  24. 2H2 + O2 ⟶ 2H2O

  25. N2 + 3H2 ⟶ 2NH3

  26. Fe2O3 + 3H2 ⟶ 2Fe + 3H2O

  27. 2KBr + Cl2 ⟶ 2KCl + Br2

  28. 2NaOH + Cl2 ⟶ NaCl + NaClO + H2O

  29. 2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO2

  30. 2Mg + CO2 ⟶ 2MgO + C

  31. Fe2O3 + 3CO ⟶ 2Fe + 3CO2

  32. CaO + 2HCl ⟶ CaCl2 + H2O

Question 20

Write balanced equations for the following word equations :

  1. Potassium nitrate ⟶ Potassium nitrite + Oxygen
  2. Calcium + Water ⟶ Calcium hydroxide + Hydrogen
  3. Iron + Hydrochloric acid ⟶ Iron [II] chloride + Hydrogen
  4. Nitrogen dioxide + Water + Oxygen ⟶ Nitric acid
  5. Lead dioxide [lead (IV) oxide] ⟶ Lead monoxide + Oxygen
  6. Aluminium + Oxygen ⟶ Aluminium oxide
  7. Iron + Chlorine ⟶ Iron [III] chloride
  8. Potassium bromide + Chlorine ⟶ Potassium chloride + Bromine
  9. Potassium bicarbonate ⟶ Potassium carbonate + Water + Carbon dioxide
  10. Calcium hydroxide + Ammonium chloride ⟶ Calcium chloride + Water + Ammonia

Answer

  1. 2KNO3 ⟶ 2KNO2 + O2
  2. Ca + 2H2O ⟶ Ca(OH)2 + H2
  3. Fe + 2HCl ⟶ FeCl2 + H2
  4. 4NO2 + 2H2O + O2 ⟶ 4HNO3
  5. 2PbO2 ⟶ 2PbO + O2
  6. 4Al + 3O2 ⟶ 2Al2O3
  7. 2Fe + 3Cl2 ⟶ 2FeCl3
  8. 2KBr + Cl2 ⟶ 2KCl + Br2
  9. 2KHCO3 ⟶ K2CO3 + H2O + CO2
  10. Ca(OH)2 + 2NH4Cl ⟶ CaCl2 + 2H2O + 2NH3

Question 21

Balance the following important equations :

  1. NaHCO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2

  2. NaOH + H2SO4 ⟶ Na2SO4 + H2O

  3. Pb(NO3)2 + NaCl ⟶ NaNO3 + PbCl2

  4. FeSO4 + NaOH ⟶ Na2SO4 + Fe(OH)2

  5. FeCl3 + NaOH ⟶ NaCl + Fe(OH)3

  6. CuSO4 + NaOH ⟶ Na2SO4 + Cu(OH)2

  7. FeCl3 + NH4OH ⟶ NH4Cl + Fe(OH)3

  8. ZnO + NaOH ⟶ Na2ZnO2 + H2O

  9. Pb(OH)2 + NaOH ⟶ Na2PbO2 + H2O

  10. Al2O3.2H2O + NaOH ⟶ NaAlO2 + H2O

  11. NaAlO2 + H2O ⟶ NaOH + Al(OH)3

  12. Al(OH)3 ⟶ Al2O3 + H2O

  13. ZnS + O2 ⟶ ZnO + SO2

  14. Fe2O3 + Al ⟶ Al2O3 + Fe

  15. Al + Cl2 ⟶ AlCl3

  16. NaCl + H2SO4 ⟶ Na2SO4 + HCl

  17. Fe + HCl ⟶ FeCl2 + H2

  18. Na2CO3 + HCl ⟶ NaCl + H2O + CO2

  19. Pb(NO3)2 + HCl ⟶ PbCl2 + HNO3

  20. AgCl + NH4OH ⟶ Ag(NH3)2Cl + H2O

  21. MnO2 + HCl ⟶ MnCl2 + H2O + Cl2

  22. Pb3O4 + HCl ⟶ PbCl2 + H2O + Cl2

  23. KMnO4 + HCl ⟶ KCl + MnCl2 + H2O + Cl2

  24. K2Cr2O7 + HCl ⟶ KCl + CrCl3 + H2O + Cl2

  25. NH4Cl + Ca(OH)2 ⟶ CaCl2 + H2O + NH3

  26. (NH4)2SO4 + NaOH ⟶ Na2SO4 + H2O + NH3

  27. Mg3N2 + H2O ⟶ Mg(OH)2 + NH3

  28. AlN + H2O ⟶ Al(OH)3 + NH3

  29. NH3 + O2 ⟶ N2 + H2O [burning of NH3]

  30. NH3 + O2 ⟶ NO + H2O [Catalytic oxidation of NH3]

  31. NH4OH + H2SO4 ⟶ (NH4)2SO4 + H2O

  32. NH3 + CuO ⟶ Cu + H2O + N2

  33. NH3 + Cl2 ⟶ HCl + NCl3 [nitrogen trichloride]

  34. HNO3 ⟶ H2O + NO2 + O2

  35. Ca(HCO3)2 + HNO3 ⟶ Ca(NO3)2 + H2O + CO2

  36. C + HNO3 [conc.] ⟶ CO2 + H2O + NO2

  37. S + HNO3 [conc.] ⟶ H2SO4 + H2O + NO2

  38. Cu + HNO3 [conc.] ⟶ Cu(NO3)2 + H2O + NO2

  39. C + H2SO4 [conc.] ⟶ CO2 + H2O + SO2

  40. S + H2SO4 [conc.] ⟶ SO2 + H2O

  41. Cu + H2SO4 [conc.] ⟶ CuSO4 + H2O + SO2

Answer

  1. 2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO2

  2. 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O

  3. Pb(NO3)2 + 2NaCl ⟶ 2NaNO3 + PbCl2

  4. FeSO4 + 2NaOH ⟶ Na2SO4 + Fe(OH)2

  5. FeCl3 + 3NaOH ⟶ 3NaCl + Fe(OH)3

  6. CuSO4 + 2NaOH ⟶ Na2SO4 + Cu(OH)2

  7. FeCl3 + 3NH4OH ⟶ 3NH4Cl + Fe(OH)3

  8. ZnO + 2NaOH ⟶ Na2ZnO2 + H2O

  9. Pb(OH)2 + 2NaOH ⟶ Na2PbO2 + 2H2O

  10. Al2O3.2H2O + 2NaOH ⟶ 2NaAlO2 + 3H2O

  11. NaAlO2 + 2H2O ⟶ NaOH + Al(OH)3

  12. 2Al(OH)3 ⟶ Al2O3 + 3H2O

  13. 2ZnS + 3O2 ⟶ 2ZnO + 2SO2

  14. Fe2O3 + 2Al ⟶ Al2O3 + 2Fe

  15. 2Al + 3Cl2 ⟶ 2AlCl3

  16. 2NaCl + H2SO4 ⟶ Na2SO4 + 2HCl

  17. Fe + 2HCl ⟶ FeCl2 + H2

  18. Na2CO3 + 2HCl ⟶ 2NaCl + H2O + CO2

  19. Pb(NO3)2 + 2HCl ⟶ PbCl2 + 2HNO3

  20. AgCl + 2NH4OH ⟶ Ag(NH3)2Cl + 2H2O

  21. MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2

  22. Pb3O4 + 8HCl ⟶ 3PbCl2 + 4H2O + Cl2

  23. 2KMnO4 + 8HCl ⟶ 2KCl + 2MnCl2 + 4H2O + Cl2

  24. K2Cr2O7 + 14HCl ⟶ 2KCl + 2CrCl3 + 7H2O + 3Cl2

  25. 2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3

  26. (NH4)2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O + 2NH3

  27. Mg3N2 + 6H2O ⟶ 3Mg(OH)2 + 2NH3

  28. AlN + 3H2O ⟶ Al(OH)3 + NH3

  29. 4NH3 + 3O2 ⟶ 2N2 + 6H2O [burning of NH3]

  30. 4NH3 + 5O2 ⟶ 4NO + 6H2O [Catalytic oxidation of NH3]

  31. 2NH4OH + H2SO4 ⟶ (NH4)2SO4 + 2H2O

  32. 2NH3 + 3CuO ⟶ 3Cu + 3H2O + N2

  33. NH3 + 3Cl2 ⟶ 3HCl + NCl3 [nitrogen trichloride]

  34. 4HNO3 ⟶ 2H2O + 4NO2 + O2

  35. Ca(HCO3)2 + 2HNO3 ⟶ Ca(NO3)2 + 2H2O + 2CO2

  36. C + 4HNO3 [conc.] ⟶ CO2 + 2H2O + 4NO2

  37. S + 6HNO3 [conc.] ⟶ H2SO4 + 2H2O + 6NO2

  38. Cu + 4HNO3 [conc.] ⟶ Cu(NO3)2 + 2H2O + 2NO2

  39. C + 2H2SO4 [conc.] ⟶ CO2 + 2H2O + 2SO2

  40. S + 2H2SO4 [conc.] ⟶ 3SO2 + 2H2O

  41. Cu + 2H2SO4 [conc.] ⟶ CuSO4 + 2H2O + SO2

Question 22

Give balanced equations for (1) & (2) by partial equation method, [steps are given below]

  1. Reaction of excess ammonia with chlorine – Ammonia as a reducing agent
    (a) Ammonia first reacts with chlorine to give hydrogen chloride and nitrogen.
    (b) Hydrogen chloride then further reacts with excess ammonia to give ammonium chloride.
  2. Oxidation of Lead [II] Sulphide by Ozone
    (a) Ozone first decomposes to give molecular oxygen & nascent oxygen.
    (b) Nascent oxygen then oxidizes lead [II] sulphide to lead [II] sulphate.

Answer

1. Reaction of excess ammonia with chlorine – Ammonia as a reducing agent

2NH3+3Cl26HCl+N26NH3+6HCl6NH4Cl8NH3+3Cl26NH4Cl+N2\begin{array}{l} 2\text{NH}_3 + 3\text{Cl}_2 \longrightarrow \cancel{6\text{HCl}} + \text{N}_2 \\ 6\text{NH}_3 + \cancel{6\text{HCl}} \longrightarrow 6\text{NH}_4\text{Cl} \\ \hline 8\text{NH}_3 + 3\text{Cl}_2 \longrightarrow 6\text{NH}_4\text{Cl} + \text{N}_2 \end{array}

2. Oxidation of Lead [II] Sulphide by Ozone

O3O2+[O]×4PbS+4[O]PbSO4PbS+4O3PbSO4+4O2\begin{array}{l} \text{O}_3 \longrightarrow \text{O}_2 + \cancel{\text{[O]}} \times 4 \\ \text{PbS} + \cancel{4\text{[O]}} \longrightarrow \text{PbSO}_4 \\ \hline \text{PbS} + 4\text{O}_3 \longrightarrow \text{PbSO}_4 + 4\text{O}_2 \end{array}

Question 23

Define the terms – (a) Relative atomic mass (b) Relative molecular mass. State why indirect methods are utilized to determine the absolute mass of an atom. Explain in brief the indirect method used.

Answer

(a) The number of times one atom of an element is heavier than 1⁄12th the mass of an atom of carbon [C12] is known as the Relative Atomic Mass [RAM] of the element.

Mass of one atomRAM=of the element(112) Mass of one atomof carbon [C12]\begin{matrix} & \text{Mass of one atom} \\ \text{RAM} = & \text{of the element} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}

(b) The number of times one molecule of the substance is heavier than 1⁄12th the mass of an atom of carbon [C12] is known as Relative Molecular Mass [RMM] of the element.

Mass of one moleculeRMM=of the substance(112) Mass of one atomof carbon [C12]\begin{matrix} & \text{Mass of one molecule} \\ \text{RMM} = & \text{of the substance} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}

As atoms are extremely small and very light, hence cannot be weighed directly. So indirect methods are utilized to determine the absolute mass of an atom.

The relative mass of an atom or molecule is hence considered by considering a mass of a light atom and relating the mass of other atoms or molecules to it.

Question 24(1)

Calculate relative molecular mass of

(a) ZnCO3
(b) CaSO4

[Zn = 65, S = 32, O = 16, Ca = 40, C = 12]

Answer

(a) Relative Molecular Mass of ZnCO3:

Molecular wt. of ZnCO3 = At. wt. of Zn + At. wt. of C + 3(At. wt. of O)
= 65 + 12 + 3(16)
= 65 + 12 + (48)
= 65 + 60
= 125

Hence, Relative Molecular Mass of ZnCO3 = 125

(b) Relative Molecular Mass of CaSO4:

Molecular wt. of CaSO4 = At. wt. of Ca + At. wt. of S + 4(At. wt. of O)
= 40 + 32 + 4(16)
= 40 + 32 + 64
= 136

Hence, Relative Molecular Mass of CaSO4 = 136

Question 24(2)

Calculate the percentage composition of

(a) calcium chloride
(b) calcium nitrate

[Ca = 40 , Cl = 35.5 , N = 14 , O = 16]

Answer

(a) Percentage composition of Calcium Chloride:

Chemical Formula of Calcium Chloride = CaCl2

Mol. wt. of calcium chloride = At. wt. of Ca + 2(At. wt. of Cl)
= 40 + 2(35.5)
= 40 + 70
= 110

110 g of CaCl2 contains 40 g of Calcium,

∴ 100 g of CaCl2 will contain 40110\dfrac{40}{110} x 100 % = 36.36% of Calcium [percentage].

Similarly, 110 g of CaCl2 will contain 70 g of Chlorine,

∴ 100 g of CaCl2 will contain 70110\dfrac{70}{110} x 100 % = 63.64% of Chlorine.

Hence, Calcium Chloride contains 36.36% of Calcium and 63.64% of Chlorine.

(b) Percentage composition of Calcium Nitrate:

Chemical Formula of Calcium Nitrate = Ca(NO3)2

Mol. wt. of Calcium Nitrate = At. wt. of Ca + 2[At. wt. of N] + 6[At. wt. of O]
= 40 + 2(14) + 6(16)
= 40 + 28 + 96
= 164

164 g of Ca(NO3)2 contains 40 g of Calcium,

∴ 100 g of Ca(NO3)2 will contain 40164\dfrac{40}{164} x 100 % = 24.39% of Calcium.

Similarly, 164 g of Ca(NO3)2 will contain 28 g of Nitrogen,

∴ 100 g of Ca(NO3)2 will contain 28164\dfrac{28}{164} x 100 % = 17.07% of Nitrogen.

Similarly, 164 g of Ca(NO3)2 will contain 96 g of Oxygen,

∴ 100 g of Ca(NO3)2 will contain 96164\dfrac{96}{164} x 100 % = 58.54% of Oxygen.

Hence, Calcium Nitrate contains 24.39% of Calcium, 17.07% of Nitrogen and 58.54% of Oxygen.

Unit Test Paper 1 — The Lang of Chem

Question 1

Match the names of ions and radicals from 1 to 10 with their correct answer from A to Q.

A:Hg2+B:MnO41-C:Sn4+
D:Pb2+E:Sn2+F:Pb4+
G:SO32-H:N3-I:NO21-
J:MnO42-K:Hg1+L:SO42-
M:ClO1-N:ZnO22-O:Cr2O72-
P:CrO42-Q:ClO31-  
1.Hypochlorite2.Permanganate
3.Plumbous4.Zincate
5.Nitride6.Mercuric
7.Stannic8.Nitrite
9.Sulphite10.Dichromate

Answer

NameIons/
Radicals
1. HypochloriteM: ClO1-
2. PermanganateB: MnO41-
3. PlumbousD: Pb2+
4. ZincateN: ZnO22-
5. NitrideH: N3-
6. MercuricA: Hg2+
7. StannicC: Sn4+
8. NitriteI: NO21-
9. SulphiteG: SO32-
10. DichromateO: Cr2O72-

Question 2

State which of the following formulas of compounds A to J are incorrect. If incorrect write the correct formula.

A:(NH4)3SO4B:NaZnO2
C:KCr2O7D:NaCO3
E:Ca2(PO4)3F:Mg(SO4)2
G:KNO3H:NaClO
I:NaOJ:BaCl2

Answer

CompoundCorrect/
Incorrect
Correct
Compound
A: (NH4)3SO4Incorrect(NH4)2SO4
B: NaZnO2IncorrectNa2ZnO2
C: KCr2O7IncorrectK2Cr2O7
D: NaCO3IncorrectNa2CO3
E: Ca2(PO4)3IncorrectCa3(PO4)2
F: Mg(SO4)2IncorrectMgSO4
G: KNO3Correct
H: NaClOCorrect
I: NaOIncorrectNa2O
J: BaCl2Correct

Question 3

Fill in the blanks with the correct word from the words in brackets :

  1. A symbol represents a short form of a/an ............... [atom/element/molecule]
  2. Compounds are always ............... [heterogeneous/homogeneous] in nature.
  3. Variable valency is exhibited, since electrons are lost from an element from the ............... [valence/penultimate] shell.
  4. A chemical equation is a shorthand form for a ............... [physical/chemical] change.
  5. Relative molecular mass of an element/compound is the number of times one ............... of the substance is heavier than -1⁄12th the mass of an atom of carbon [C12]. [atom/ion/molecule]

Answer

  1. A symbol represents a short form of an element.
  2. Compounds are always homogeneous in nature.
  3. Variable valency is exhibited, since electrons are lost from an element from the penultimate shell.
  4. A chemical equation is a shorthand form for a chemical change.
  5. Relative molecular mass of an element/compound is the number of times one molecule of the substance is heavier than -1⁄12th the mass of an atom of carbon [C12].

Question 4

Underline the compound in each equation given below, which is incorrectly balanced and write the correct balancing for the same.

  1. Na2SO3 + HCl ⟶ 2NaCl + H2O + SO2
  2. CaC2 + N2 ⟶ 2CaCN2 + C
  3. Fe2O3 + 2H2 ⟶ 2Fe + 3H2O
  4. Cl2 + 2H2O + SO2 ⟶ 4HCl + H2SO4
  5. 6NaOH + 3Cl2 ⟶ 6NaCl + NaClO3 + 3H2O
  6. C2H5OH + 3O2 ⟶ 2CO2 + 2H2O
  7. NaOH + CO2 ⟶ Na2CO3 + H2O
  8. 2H2O + 2Cl2 ⟶ 2HCl + O2
  9. 3CuO + NH3 ⟶ 3Cu + 3H2O + N2
  10. PbO2 + 4HCl ⟶ PbCl2 + H2O + Cl2

Answer

  1. Na2SO3 + HCl ⟶ 2NaCl + H2O + SO2
    Corrected Balancing — 2HCl
  2. CaC2 + N22CaCN2 + C
    Corrected Balancing — CaCN2
  3. Fe2O3 + 2H2 ⟶ 2Fe + 3H2O
    Corrected Balancing — 3H2
  4. Cl2 + 2H2O + SO24HCl + H2SO4
    Corrected Balancing — 2HCl
  5. 6NaOH + 3Cl26NaCl + NaClO3 + 3H2O
    Corrected Balancing — 5NaCl
  6. C2H5OH + 3O2 ⟶ 2CO2 + 2H2O
    Corrected Balancing — 3H2O
  7. NaOH + CO2 ⟶ Na2CO3 + H2O
    Corrected Balancing — 2NaOH
  8. 2H2O + 2Cl22HCl + O2
    Corrected Balancing — 4HCl
  9. 3CuO + NH3 ⟶ 3Cu + 3H2O + N2
    Corrected Balancing — 2NH3
  10. PbO2 + 4HCl ⟶ PbCl2 + H2O + Cl2
    Corrected Balancing — 2H2O

Question 5

With reference to a chemical equation, state which of the statements 1 to 5 pertain to A or B.

A : Information provided by a chemical equation.
B : Limitations of a chemical equation

  1. The nature of the individual elements.
  2. The speed of the reaction.
  3. The state of matter in which the substance is present.
  4. The completion of the reaction.
  5. The direction of the reaction.

Answer

  1. A
  2. B
  3. A
  4. B
  5. A
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