Correct Answer: B. di-butylamine

Correct Answer: B. Pb₂₊ salts

Correct Answer: A. Reducing agent

Correct Answer: D) level of oxygen reduces

Correct Answer: C. has excess electrons

Explanation

(a) Solubility is the maximum amount of solute that would dissolve / be contained in 1 dm\(^3\) of solution at a given temperature.

(b)(i)

(ii) At 35° solubility = 0.50 to 0.55 moldm\(^{-3}\) range

(iii) Solubility @ 55°C = 0.95 to 0.98 moldm\(^{-3}\)

1 moldm\(^{-3}\) of salt Z = 0.97; 100 = 97g

100cm\(^{-3}\) will contain = \(\frac{97 \times 100}{100}\) = 9.7g

Solubility @ 35°C = 0.54moldm\(^{-3}\)

1 drn\(^{-3}\) contain = 0.54 x 100 = 54g

100cm\(^{-3}\) will contain 54 x100 = 5.4g

10 Mass of salt that would crystallize out = 9.7 - 5.4 = 4.3g

(c)(i) (I) Al\(_2\)O\(_{3(s)}\) + 6HCl\(_{(aq)}\) \(\to\) 2AlCl\(_{3(s)}\) + 3H\(_2\)O\(_{(l)}\)

(II) Al\(_2\)O\(_{3(g)}\) 2NaOH\(_{(aq)}\) + 3H\(_2\)O\(_{(l)}\) \(\to\) 2NaAl(OH)\(_4\)

(ii) Amphoteric

(iii) ZnO / PbO / SnO /Al\(_2\)O\(_3\)

d)(i) I [Al(H\(_2\)O)\(_6\)]\(^{3+}\) = x + (0) = +3

x = +3

Al = +3

(II) NaH = 1+ x = 0

x = —1

H = — 1

(ii)(I) Copper (II) tetraoxosulphate VI pentahydrate

(II) Calcium tripxocarbonate IV

(III) Potassium tetraoxomanganate VII.

Correct Answer: D) 150 cm³

Explanation

\(\begin{array}{llllcr}

& \mathrm{2NO} & + & \mathrm{O_2} & \longrightarrow & \mathrm{2NO_2} \\

\text{Initially} & \mathrm{150cm^3} & & \mathrm{100cm^3} & & \\

\text{Gay Lussac's Law} & \text{2 mols} & & \text{1 mol} & & \text{2 mols} \\

\text{Vol. used up for reac.} & \mathrm{150cm^3} & & \mathrm{75cm^3} & & \mathrm{150cm^3} \\

\end{array}\)

Vol. of oxygen left after reac. \(= \mathrm{100cm^3} - \mathrm{75cm^3} = \mathrm{25cm^3}\)

Vol. of nitrogen (IV) oxide prod. = \(\mathrm{150cm^3}\)

Note: According to Gay Lussac's Law, 2 volumes of nitrogen II oxide combine with 1 volume of oxygen to form 2 volumes of nitrogen IV oxide.

Correct Answer: B. oxgyen

Explanation

(a) Industrial preparation of ammonia (NH\(_{3(g)}\))

(i) Dry nitrogen (N\(_{2(g)}\)) mixed with dry hydrogen (H\(_{2(g)}\)) in the volume proportion of 1 : 3 over a pressure of 200 — 500 atm at temperature of 450°C — 500°C. Finely divided iron used as catalyst.

(ii) N\(_{7(g)}\) 3H\(_{2(g)}\) 2NH\(_{3(g)}\) Balanced equation of the reaction written above.

(iii) Way of increasing the yield of ammonia (NH\(_{3(g)}\)) is continuous removal of ammonia as it is formed or liquefying of the ammonia gas at a low temperature or high pressure.

(iv) Two uses of ammonia: (1) used in manufacture of fertilizers. (2) used as a precipitating agent. Other uses: Manufacture of trioxonitrate (v) acid; manufacture of sodium trioxocarbonate.

(iv) textiles, pesticides, used in detergent/cleasing agent, manufacture of plastics, laundry works, pharmaceuticals as a refrigerant, softening temperarily hard water, removal of insect stings.

(b)(i) Chemical test for hydrogen H\(_{2(g)}\), a lighted wooden splint is brought to the tip of a test tube containing the gas. The gas gives a pop sound shows the presence of Hydrogen H\(_{2(g)}\).

(ii) CO\(_{2(g)}\), The gas bubble through limewater Ca(OH)\(_{2}\) solution. The lime water turns milky shows the presence of CO\(_{2(g)}\)

(iii) Oxygen O\(_{2(g)}\), a glowing (wooden) splint dipped into a test tube of the gas. The (wooden) splint rekindles or relights shows that O\(_{2(g)}\) is present.

(c)(i) Composition of water gas: Carbon (II) oxide (CO\(_{g}\)) and Hydrogen gas H\(_{2(g)}\)

(ii) 2 Uses of water gas:

(1) industrial manufacture of hydrogen.

(2) industrial manufacture of methanol butanol.

(3) used as fuel.

(d) A simple experiment to determine type of hardness in a sample of water. Sample of water boiled, soap added and shaken, if it forms lather easily it is temporary hardness, if it forms lather with difficultly or does not form lather easily it is permanent hardness.

Correct Answer: D. All of the above