The Language of Chemistry Class 9 ICSE Important Questions with Answers

Answer: (b) $Al_2(SO_4)_3$Al₂(SO₄)₃.

Aluminium is $Al^{3+}$Al^3+ and sulphate is $SO_4^{2-}$SO₄^2-. Cross-multiplying the valencies gives two aluminium ions and three sulphate ions: $Al_2(SO_4)_3$Al₂(SO₄)₃.

Answer: (c) $Na$Na.

The symbol of sodium comes from its Latin name natrium, so it is $Na$Na. $S$S is sulphur and $Sn$Sn (stannum) is tin.

Answer: (c) $3$3.

Each hydrogen has a valency of $1$1, and three hydrogen atoms combine with one nitrogen atom. Hence nitrogen exercises a valency of $3$3 in $NH_3$NH₃.

Answer: (a) $3,4,4$3,4,4.

Balancing iron needs $3Fe$3Fe; the four oxygen atoms in $Fe_3O_4$Fe₃O₄ need $4H_2O$4H₂O; that supplies $8$8 hydrogen atoms, giving $4H_2$4H₂. Balanced equation: $3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$3Fe + 4H₂O → Fe₃O₄ + 4H₂.

Answer: (a) A balanced equation obeys the law of conservation of mass, which states that matter is neither created nor destroyed; therefore R correctly explains A.

(i) Valency: The combining capacity of an element, i.e. the number of hydrogen atoms that combine with or replace one atom of that element. Example: the valency of oxygen is $2$2 (as in $H_2O$H₂O).

(ii) Radical: An atom or a group of atoms of elements that carries a charge and behaves as a single unit in a chemical reaction. Example: the sulphate radical $SO_4^{2-}$SO₄^2-.

(i) Ammonium carbonate: ammonium is $NH_4^{+}$NH₄^+ and carbonate is $CO_3^{2-}$CO₃^2-. Criss-crossing gives $(NH_4)_2CO_3$(NH₄)₂CO₃.

(ii) Calcium hydrogen carbonate: calcium is $Ca^{2+}$Ca^2+ and hydrogen carbonate (bicarbonate) is $HCO_3^{-}$HCO₃^-. Criss-crossing gives $Ca(HCO_3)_2$Ca(HCO₃)₂.

A radical like $SO_4^{2-}$SO₄^2- is a charged atom or group of atoms that stays together as one unit in compounds and reactions; it is stable as part of a compound, combined with an oppositely charged radical (as in $CuSO_4$CuSO₄).

A free radical like $Cl^\bullet$Cl^• is an atom or group with an unpaired electron. It is formed when a covalent bond breaks, is usually uncharged, and is very reactive and short-lived, because it quickly reacts to pair its electron.

(i) $4Al + 3O_2 \rightarrow 2Al_2O_3$4Al + 3O₂ → 2Al₂O₃

(ii) $2KClO_3 \rightarrow 2KCl + 3O_2$2KClO₃ → 2KCl + 3O₂

(iii) $2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2$2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

In each case the number of atoms of every element is now equal on both sides, satisfying the law of conservation of mass.

Balanced equation:

$Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)\uparrow$Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)↑

Information the symbolic equation adds:

1. The exact chemical formulae of reactants and products.

2. The relative number of atoms/molecules (ratio) taking part and the physical states of each substance.

(It also shows, by the $\uparrow$↑, that hydrogen escapes as a gas.)

Each formula is obtained by balancing the positive and negative valencies of the constituent radicals.

(i) Chlorine radicals ($Cl^\bullet$Cl^•): released when ultraviolet light breaks down CFCs in the upper atmosphere. Nitric oxide ($NO$NO, which has an unpaired electron): formed from nitrogen and oxygen at the high temperatures inside vehicle engines and furnaces.

(ii) Free radicals react with and damage the cells of living things. Chlorine radicals also destroy ozone, so more harmful ultraviolet rays reach the Earth and cause skin cancer and damage to plants.

(iii) Stop using CFCs (replace them with ozone-safe chemicals); fit catalytic converters to vehicles to cut emissions of nitric oxide. (In the body, a diet rich in antioxidants such as vitamins C and E helps neutralise free radicals.)

(a) The relative atomic mass of an element is the number of times one atom of the element is heavier than $\frac{1}{12}$1/12 of the mass of an atom of carbon-12. The relative molecular mass of a substance is the number of times one molecule of it is heavier than $\frac{1}{12}$1/12 of the mass of an atom of carbon-12; it is the sum of the relative atomic masses of all the atoms in the molecule.

(b) $CuSO_4\cdot5H_2O=64+32+(4\times16)+5\times(2\times1+16)=64+32+64+90=250$CuSO₄·5H₂O=64+32+(4×16)+5×(2×1+16)=64+32+64+90=250.
$\%\text{ water}=\frac{90}{250}\times100=36\%.$% water=90/250×100=36%.
(c) $NH_4NO_3=14+4+14+(3\times16)=80$NH₄NO₃=14+4+14+(3×16)=80; it contains $2\times14=28$2×14=28 parts of nitrogen.
$\%\text{ N}=\frac{28}{80}\times100=35\%.$% N=28/80×100=35%.

(a) Variable valency is the property of certain elements of showing more than one valency in their compounds. The lower valency takes the suffix -ous and the higher takes -ic.

Iron: $Fe^{2+}$Fe^2+ (ferrous) and $Fe^{3+}$Fe^3+ (ferric). Copper: $Cu^{+}$Cu^+ (cuprous) and $Cu^{2+}$Cu^2+ (cupric).

(b)

• Ferrous chloride: $FeCl_2$FeCl₂

• Ferric chloride: $FeCl_3$FeCl₃

• Cuprous oxide: $Cu_2O$Cu₂O

• Cupric oxide: $CuO$CuO

(c) $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2\uparrow + 2H_2O$Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O

(i) Skeletal equation: $Fe_2O_3 + Al \rightarrow Fe + Al_2O_3$Fe₂O₃ + Al → Fe + Al₂O₃

(ii) Balanced equation:

$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$Fe₂O₃ + 2Al → 2Fe + Al₂O₃

(iii) Aluminium gains oxygen, so aluminium is oxidised; ferric oxide loses oxygen, so $Fe_2O_3$Fe₂O₃ is reduced.

(iv) Balancing is based on the law of conservation of mass — atoms are neither created nor destroyed, so the number of atoms of each element must be equal on both sides.