Correct Answer: B. distillation

Explanation

(a)(i) Hard water is water which will not readily form lather with soap.

(ii) 1. Calcium tetraoxosulphate (IV) 2. Calcium hydrogen trioxocarbonate (IV) 3. Magnesium tetraoxosulphate (VI) 4. Magnesium hydrogen trioxocarbonate (IV).

(iii)(a) Heating the water to boiling/distillation (b) Addition of calculated amount of calcium hydroxide [Ca(OH)\(_2\)] (c) Addition of washing soda crystals to the water [Na\(_2\)CO\(_3\)] (d) Ion exchange resin/permutit method.

(b)(i) Sulphur air or oxygen

(ii) 2SO\(_{2(g)}\) +O\(_{2(g)}\) \(\rightleftharpoons\) 2SO\(_{3(g)}\)

(iii) V\(_2\)O\(_5\) / Vanadium (V) oxide.

(c)(i) Use of Na\(_2\)SO\(_4\)

(a) Water softener (b) In glass or paper manufacturing (c) As analytical reagent (d) As purgative (e) In the manufacture of detergents .

(ii) Mass of salt per 30 cm\(^3\) of saturated solution = 1.80g.

1000 cm\(^3\) of saturated solution contains = \(\frac{1000 \times 1.80}{30}\)

= 60g.

Molar mass of Na\(_2\)CO\(_3\) = 23 x 2 + 12 x 1 + 16 x 3 = 106g.

Solubility of Na\(_2\)CO\(_3\) = \(\frac{60}{106}\) = 0.566 mole dm\(^3\).

(d)(i) Activated complex is the intermediate (complex) that reactants must attain in order to form a product or it is a temporary species formed by reactant molecules as a result of collision before the products are formed.

(ii) A collision may fail to produce a chemical reaction if the energy of the colliding particles is less than activation energy / if the reactant are not properly aligned.

(a)(i) Hard water is water which will not readily form lather with soap.

(ii) 1. Calcium tetraoxosulphate (IV) 2. Calcium hydrogen trioxocarbonate (IV) 3. Magnesium tetraoxosulphate (VI) 4. Magnesium hydrogen trioxocarbonate (IV).

(iii)(a) Heating the water to boiling/distillation (b) Addition of calculated amount of calcium hydroxide [Ca(OH)\(_2\)] (c) Addition of washing soda crystals to the water [Na\(_2\)CO\(_3\)] (d) Ion exchange resin/permutit method.

(b)(i) Sulphur air or oxygen

(ii) 2SO\(_{2(g)}\) +O\(_{2(g)}\) \(\rightleftharpoons\) 2SO\(_{3(g)}\)

(iii) V\(_2\)O\(_5\) / Vanadium (V) oxide.

(c)(i) Use of Na\(_2\)SO\(_4\)

(a) Water softener (b) In glass or paper manufacturing (c) As analytical reagent (d) As purgative (e) In the manufacture of detergents .

(ii) Mass of salt per 30 cm\(^3\) of saturated solution = 1.80g.

1000 cm\(^3\) of saturated solution contains = \(\frac{1000 \times 1.80}{30}\)

= 60g.

Molar mass of Na\(_2\)CO\(_3\) = 23 x 2 + 12 x 1 + 16 x 3 = 106g.

Solubility of Na\(_2\)CO\(_3\) = \(\frac{60}{106}\) = 0.566 mole dm\(^3\).

(d)(i) Activated complex is the intermediate (complex) that reactants must attain in order to form a product or it is a temporary species formed by reactant molecules as a result of collision before the products are formed.

(ii) A collision may fail to produce a chemical reaction if the energy of the colliding particles is less than activation energy / if the reactant are not properly aligned.

Correct Answer: D. ethene

Explanation

(a)(i) State Pauli's exclusion principle: Pauli's exclusion principle states that two electrons in the same orbita of an atom cannot have same values for all four quantum numbers/no two electrons can have the same four quantum numbers/no two electrons in the same orbital if an atom can have the same spin.

(ii) Hund's rule of maximum multiplicity. Hund's rule states that electrons occupy each orbital supply first before pairinc takes place in degenerate orbital/most stable arrangement of electrons in sunshells is the one with the greatest number of parallel spins.

(b)(i) (a) Cu\(^+\) — 1s\(^2\)2s\(^2\)2P\(^6\)3P\(^6\)3d\(^{10}\)

(b) Cu\(^{2+}\) --- 1s\(^2\)2s\(^2\)2P\(^6\)3s\(^2\)3P\(^6\)3d\(^9\)

(ii) i. Cu\(^+\) — no unpaired electron

ii. Cu\(^{2+}\) — one unpaired electron

iii. Redox/disproportional

iv. Cu\(_2\)O / CuCI / Cu\(_2\)Cl\(_2\)

(c)(i) (a) gamma rays (b) alpha particles

(ii) (a) gamma (y) - No charge/electrically neutral. (b) alpha \(\alpha\)- positively charge.

(iii) (i) Nuclear fusion (2) Nuclear fusion (3) half-life.

(d) (i) Li\(^+\), Na\(^+\), K\(^+\), size increase down a group as more shells are being added.

(ii) F, O\(^{2-}\), N\(^{2-}\), size increase as nuclear charge decrease in the same period/isoelectronic.

(e) Phosphorus (V) oxide = P\(_{4}\)O\(_{10}\)/P\(_{2}\)Q\(_{5}\)

= (31 x 4) + (16 x 10) or (31 x 2) + (16 x 5) = 124 + 160 or 62 + 80 = 284 or 142.

Therefore % by mass of phosphorus = \(\frac{4P}{P_4Q_{10}}\) x 100% or

\(\frac{2P}{P_2Q_{5}}\) x 100%

\(\frac{124}{284}\) x 100% or \(\frac{62}{142}\) x 100%

= 43.7%

or 100 - 43.7 = 56.3% oxygen.

Correct Answer: C. their ionic radii decrease down the group

Correct Answer: B. saponification

Correct Answer: B. 24

Correct Answer: B. Chlorofluorocarbon

Correct Answer: D. hydrogen bonding

Correct Answer: A. 0.004 mol/dm³