Correct Answer: B. ethanol and carbon (IV) oxide

Explanation

(a)(i) Water is a universal solvent because it can dissolve many substances. water is polar and as a result, it can dissolve ionic solutes as well as partially ionic/polar covalent substances

(ii) Add water to anhydrous copper

(II) tetraoxosulphate (VI) salt, colour changes from white to blue. OR Add water to anhydrous cobalt

(II) ch!oric!e, colour changes from blue to pink.

(b)(i) I. OH\(^-{_(aq)}\) \(\to\) 2H\(_2\)O\(^-{_(l)}\) + O\(_{2(g)}\) + 4e\(^-\)

II. 2H\(^+_{(aq)}\) + 2e\(^-\) --> H\(_{2(g)}\)

OR 4H\(^+_{(aq)}\) + 4e\(^-\) --> 2H\(_{2(g)}\)

Ill. 2H\(_2\)O\(_{(l)}\) -> OH\(_{2(g)}\) + O\(_{2(g)}\)

OR 4OH\(^-_{(aq)}\) + 4H\(^+_{(g)}\) --> 2H\(_2\)O\(_{(l)}\) + 2H\(_{2(g)}\) + O\(_{2(g)}\)

(ii) Quantity of electricity = 1 x t = 1.25 x 40 x 60 = 3000C

For 1 mole of O\(_2\) formed at the anode, 4e\(^-\) are generated Volume of O\(_2\) formed by 3000C = \(\frac{3000}{96500}\) x \(\frac{1}{4}\) mole O\(_2\)

= 7.77 x 10\(^{-3}\) moles

1 mole O\(_2\) occupies 22.4 dm\(^{3}\) at s.t.p. mole of O\(_2\) = 7.77 x 10\(^{-3}\) x 22.4 dm\(^{3}\) = 0.174dm\(^{3}\)/174 cm\(^{3}\)

OR From the equation at anode; 4 x 96500C = 22.4 dm\(^{3}\)

3000C = \(\frac{22.4 \times 3000}{4 \times 96500}\)

= 0.174dm\(^{3}\) / 174 cm\(^{3}\)

(c)(i) 2Au\(_{(s)}\) + 3Cl\(_{2(g)}\) \(\to\) 2AuCI\(_{3(s)}\)

(ii) Amount of Au= \(\frac{1.2}{197}\) = 0.00634 mole

Amount of Cl\(_2\) = \(\frac{1.744}{71}\) = 0.0246 mole.

From equation: 2 moles of Au = 3 moles of Cl\(_{(2)}\)

.-. 0.00634 mole of Au = \(\frac{0.00634 \times 3}{2}\) = 0.00951 mole

Amount of CI\(_2\) available = 0.0246 mole but amount required by all the Au for the reaction = 0.00951

CI\(_2\) is in excess.

(iii) Excess amount of CI\(_2\) = amount available - amount used = 0.0246 - 0.00951 = 0.015 mole.

Correct Answer: C. remove impurities

Explanation

(a)(i) A saturated solution is that which contains the maximum amount of solute it can dissolve at a given temperature (in the presence of undissolved solute).

(ii) solubility of KNO\(_3\) in gdm\(^{-3}\) = 3.00 x 101. = 303

1000cm\(^{3}\) of saturated solution ---- 303g

250cm\(^{3}\) of the solution = \(\frac{303 \times 250}{1000}\) = 75.8 g.

Since the quantity of KNO\(_3\) added (67.0) to 250 cm\(^{3}\) of water is less than maximum amount required to form a saturated solution, then the solution is unsaturated.

(b)(i) In dative bond, only one of the participating atoms/species donates the electrons to be shared by both atoms while in covalent bond both participating atoms/species contribute equally the electrons being shared.

(ii) Sugar is covalent while common salt (NaCI) is electrovalent/ionic. Electrical conductivity (in compounds) depends on presence of mobile ions. Sugar does not conduct electricity because it does not contain ions (they are molecules) while solid common salt does not conduct electricity because its ions are not mobile.

(iii) I. Hydrogen bond.

(iv) H. Van der Waals forces.

(iv) O - H --- N Because oxygen is more electronegative and smaller in size than sulphur.

(c)(i) The total pressure exerted by a mixture of gases which do not-react chemically is equal to the sum of the individual partial pressures of the gases in the mixture.

(ii) pressure of the dry gas (P\(_1\) = 700 - 15 = 685 mmHg, V\(_1\) = 200 cm\(^3\), T\(_1\) = 18°C = 273 + 18 = 291K, P\(_2\) = 760 mmHg, T\(_2\) = 273

\(\frac{P_1V_1}{T_1}\) =\(\frac{P_2V_2}{T_2}\)

V\(_2\) = \(\frac{P_1V_1T_2}{P_2T_1}\)

\(\frac{685 \times 200 \times 273}{760 \times 291}\)

= 169.1 cm\(^{3}\)

Correct Answer: C.₁₃M³⁺

Correct Answer: A. AgNO3

Correct Answer: B. electron affinity

Explanation

(a)(i) (1) Oxidation agent is a substance which accepts electrons or is an electron acceptor.

(2) Reducing agent is a substance which donates electrons or is an electron donor.

(ii) 2CuO\(_{(s)}\) +C\(_{(s)}\) \(\to\) 2Cu\(_{(s)}\) + CO\(_{2(g)}\)

H\(_{2}\)O\(_{(g)}\) + C\(_{(s)}\) ----> CO\(_{(g)}\) + H\(_{2(g)}\)

CO\(_{2(g)}\) --> 2CO\(_{(g)}\)

(ii) Cu in CuO from +2 to 0

H in H\(_2\)O from +1 to 0

C in CO\(_2\) from +4 to + 2.

(b)(i) X is sulphur (IV) oxide or sulphur dioxide.

(ii)(a) It is heavier than air. (b) It is acidic

(c) Reducing or oxidizing agent.

(d) Colourless (poisonous) gas with irritating smell.

(iii) (1) Used for bleaching

(2) For the manufacture of H\(_2\)SO\(_4\)

(3) as germicide and fumigant.

(4) for preservation.

(5) as refrigerant.

(c) (i) sodium magnesium (ii) Hydrogen (iii) Lead(II) iodide (iv) Magnesium.

(d) (i) A salt is a compound formed when all or part of hydrogen of an acid is replaced by metal or ammonium ion.

(ii) (a) Normal salt (b) Acid salt (c) Basic salt (d) Double salt (e) Complex salt.

(iii) NaCl/ZnSO\(_4\)/NaH\(_2\)PO\(_4\)/Zn(OH)Cl/Mg(OH)NO\(_3\)/NH\(_4\).Fe(SO\(_4\))\(_2\).6H\(_2\)O).

(e) (I) H\(_2\)SO\(_{4(aq)}\) + 2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)SO\(_4\)\(_{(aq)}\) + 2H\(_2\)O\(_{(l)}\)

(ii) From the reaction 1 mole of H\(_2\)SO\(_4\) = 2 moles of NaOH\(_{(l)}\).

Therefore 0.50 mole of H\(_2\)SO\(_4\) = \(\frac{2 \times 0.5}{1}\) of NaOH

= 1 mole of NaOH.

Correct Answer: C. it is used for the purification of copper

Explanation

(a)(i) coke (ii) ammoniacal liquor/coal tar (iii) coal gas

(b) (i) methane (ii) because sulphur is removed in the gasification process/it does not contain sulphur or sulphur compounds.

(c)(i) C\(_{(S)}\) + O\(_{2(g)}\) --> CO\(_{2(g)}\)

(ii) I. is a gas at room temperature - it is a colourless gas - it is an odourless gas - it is denser than air - it readily liquefies and solidifies/easily compressible.

II. - reacts with alkali to form trioxocarbonate (IV) - it is soluble in water - reacts with (burning) Mg to form MgO - reacts with red hot carbon to form CO - It is weakly acidic in water - it does not support combustion.

(d)(i) I. diamond graphite II. -coal/ lampblack/soot/carbon black - coke - charcoal.

(ii) I. Diamond - as a cutting tool - as jewellery;

Graphite - as a lubricant - as an electrode - as pigment in paints - as neutron moderators in nuclear reactors.

II. Coke - as a fuel -as a reducing agent/in the extraction of metals - preparation of carbides - for making water gas/producer gas.

Coal - as a fuel -as a starting material for chemicals.

Charcoal - as a fuel - as absorbent/in gas mask - in the purification of gases - in the recovery of industrial solvent - for removing colouring matter from substances

Lampblack/Soot/carbon black: - to colour rubber tyres - to make boot/shoe polish - to make printers ink - to make black paint - typewriting ribbons - carbon paper.

(e)(i) I. S\(_{(s)}\) + O\(_{2(g)}\) --> SO\(_{2(g)}\)

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

III. SO\(_{3(g)}\) + H\(_2\)SO\(_{4(g)}\) -> H\(_2\)S\(_2\)O7\(_{7(l)}\)

IV. H\(_2\)SO7\(_{7(l)}\) + H\(_2\)O\(_{(l)}\) -> 2H\(_2\)SO\(_{4(l)}\)

(ii) I. - dehydrating agent

II. precipitating agent/displacement agent.