# The Language of Chemistry

## 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.

Chloride of a metal X is XCl2

By interchanging subscript and writing as superscript:

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

Therefore, valency of metal X = 2.

Formula of the sulphate:

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

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 $\bold{XSO}_{\bold{4}}$

Formula of the hydroxide:

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

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

Therefore, we get

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

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

#### Question 1(1987)

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

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.

Nitride of a metal X is XN.

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

Formula of the sulphate:

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

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

Formula of the hydroxide:

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

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

Therefore, we get

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

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

#### Question 1(1992)

What is the valency of nitrogen in :

(i) NO

(ii) N2O

(iii) NO2

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

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

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

Therefore, valency of Nitrogen is 2.

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

Therefore, valency of Nitrogen is 1.

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

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

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

Therefore, valency of Nitrogen is 4.

#### Question 1

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

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 $\dfrac{1}{12}$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.

(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.

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.

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

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:

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

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

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

Therefore, valency of Sulphur is 4.

(b) Valency of sulphur in SO3 is 6

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

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

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

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

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

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+
(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.

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.

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

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

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

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

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

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

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

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

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] chloride Copper [II] chloride Copper [I] oxide Copper [II] oxide Copper [I] sulphide Copper [II] sulphide Copper [II] nitrate Copper [II] sulphate Tetra amine copper [II] sulphate

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] chloride Iron [III] chloride Iron [II] oxide Iron [III] oxide Iron [II] sulphate Iron [III] sulphate Iron [II] sulphide Iron [III] sulphide Iron [II] hydroxide Iron [III] hydroxide

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 :

CompoundFormula

#### Question 12(10)

Write the formula of the following compounds :

Silver

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

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

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.

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

(a) NH3 + H2O ⟶ NH4OH

(b) C2H5Br + NaOH ⟶ C2H5OH + NaBr

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

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

#### Question 16

$2\text{KClO}_3 \xrightarrow{\text{MnO}_2} 2\text{KCl} + 3\text{O}_2[\text{g}]$ 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.

(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.)

(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.

(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

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
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

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

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.

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

$\begin{array}{l} 2\text{NH}_3 + 3\text{Cl}_2 \longrightarrow \cancel{6\text{HCl}} + \text{N}_2 \\ 6\text{NH}_3 + \cancel{6\text{HCl}} \longrightarrow 6\text{NH}_4\text{Cl} \\ \hline 8\text{NH}_3 + 3\text{Cl}_2 \longrightarrow 6\text{NH}_4\text{Cl} + \text{N}_2 \end{array}$

2. Oxidation of Lead [II] Sulphide by Ozone

$\begin{array}{l} \text{O}_3 \longrightarrow \text{O}_2 + \cancel{\text{[O]}} \times 4 \\ \text{PbS} + \cancel{4\text{[O]}} \longrightarrow \text{PbSO}_4 \\ \hline \text{PbS} + 4\text{O}_3 \longrightarrow \text{PbSO}_4 + 4\text{O}_2 \end{array}$

#### 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.

(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.

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

(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.

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

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]

(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]

(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 $\dfrac{40}{110}$ x 100 % = 36.36% of Calcium [percentage].

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

∴ 100 g of CaCl2 will contain $\dfrac{70}{110}$ 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 $\dfrac{40}{164}$ 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 $\dfrac{28}{164}$ 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 $\dfrac{96}{164}$ 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 Hypochlorite 2 Permanganate 3 Plumbous 4 Zincate 5 Nitride 6 Mercuric 7 Stannic 8 Nitrite 9 Sulphite 10 Dichromate

NameIons/
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)3SO4 B: NaZnO2 C: KCr2O7 D: NaCO3 E: Ca2(PO4)3 F: Mg(SO4)2 G: KNO3 H: NaClO I: NaO J: BaCl2

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]

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

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.