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Volume of pipette = 25.00 cm\(^{3}\) indicator used = methyl orange colour change at end point yellow to orange/purple Burette Reading Rough (1) First (2nd) Second (3rd) Third Final (cm\(^3\)) 25.20 24.90 24.80 24.82 Initial (cm\(^3\)) 0.00 0.00 0.00 0.00 Volume of acid used 25.20 24.00 24.80 24.85 Average volume of A used = \(\frac{24.90 + 24.80 + 24.85}{3}\) = 24.85 cm\(^3\) The equation for the reaction X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) 2 x 2NO\(_{3(aq)}\) + CO\(_{2(g)}\) + 11HNO\(_{(l)}\) (b)(i) Mole ratio of Acid to Base is 2:1 concentration of B in mol. dm\(^{-3}\) \(\frac{C_AV_A}{C_BV_B} = \frac{2}{1}\) C\(_B\) = \(\frac{C_AV_A}{2V_B}\) C\(_B\) = \(\frac{0.0950 \times 24.85}{2 \times 25}\) = \(\frac{2.36075}{50}\) = 0.0472 mol. dm\(^{-3}\) (ii) Molar mass B in g mol\(^{-1}\) Molar mass X\(_2\)CO\(3\).10H\(_2\)O = \(\frac{\text{mass concentration gdm}^{-3}}{\text{molar concentration of B in moldm}^{-3}}\) = \(\frac{13.5 gdm^{-3}}{0.0472 moldm^{-3}}\) = 286 g mol\(^{-1}\) (iii) Percentage by mass of X X\(_2\)CO\(_3\).10H\(_2\)O = 286g i.e. relative molar mass of X\(_2\)CO\(_3\) = 2x + 12 + (16 x 3) + 10(2 x 1) + 16 = 286 = 2x + 12 + 48 + 180 = 286 = 2x + 240 = 286 2x = 286 - 240 = 46 x = \(\frac{46}{2}\) = 23 gmol\(^{-1}\) Percentage by mass of X = \(\frac{2 \times 23}{286} \times \frac{100}{1}\) = 16.08%

(a) (i) to some dil. HCI in a beaker, add Zn granules and shake. (ii) continue to add Zn granules un effervescence stops and the Zn is in excess. (iii) filter off undissolved Zn granules (iv) put the filtra in an evaporating dish and heat to concentrate. (v) cool to obtain crystals (vi) filter and dry the crysta between filter papers. (ii) Zn\(_{(s)}\) + 2HCI\(_{(aq)}\) \(to\) ZnCl\(_{2(aq)}\) + H\(_{2(g)}\) (b)(i) CH\(_3\)COOK\(_{(aq)}\) (ii) Na\(_2\)SO\(_{4(aq)}\) (iii) Pb(NO\(_3\))\(_{2(aq)}\) and MgCl\(_{2(aq)}\) (c)(i) I. MnO\(_2\) Mn + (-2 x 2) = O Mn — 4 = 0 Mn = + 4 (ii) KMnO\(_4\) 1 + Mn + (– 2 x 4) = 0 1 + Mn – 8 = 0 Mn = + 7 (ii) MnO\(_2\) – as a catalyst – as an oxidizing agent KMnO\(_4\) – as an oxidizing agent (d) Oxidation is a process of electron loss, while reduction is a process of electron gain. A species can only be oxidized if there is another species that would readily accept the donated electron(s)/ Hence every oxidation must be accompanied by a reduction. (e) (i) I. Fe\(_{(s)}\) \(\to\) F\(^{2+}_{(aq)}\) + 2e\(^-\) II. Ag\(^+_{(aq)}\) + e\(^-\) \(\to\) Ag\(_{(s)}\) (ii) I. Fe is oxidized II. Ag\(^+\) is reduced (iii) +1 to 0.

(i) Boyle's law states that the volume of a given mass of gas is inversely proportional to its pressure if the temperature remains constant. (ii) P \(\propto\) 1 or PV = K or V \(\propto\) \(\frac{1}{P}\) (iii) \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) \(V_2 = \frac{6.5 \times 10^4 \times 500 \times 273}{(273 + 30) \times 1.01 \times 10^5}\) \(V_2 = 290 cm^3\) (c) An inflated balloon expands in warm water because in air, particles in the balloon acquire more kinetic energy. There is an increase in the pressure inside the balloon. In order to keep the pressure constant the balloon expands. (d) (i) Vapour density is the number of times a given volume of a gas or vapour is as heavy as an equal volume of hydrogen gas at the same temperature and pressure. (ii) molar mass of SO\(_2\) = 32 + 16 x 2 = 64g mol\(^{-1}\). Vapour density x 2 = relative molar mass Vapour density = \(\frac{64}{32}\) = 32 (e) (i) I. upward delivery = NH\(_3\) II. downward delivery = SO\(_2\) (ii) I. acidic II. basic/alkaline (iii) SO\(_2\) – oxidizing agent NH\(_3\) - reducing agent. (f) (i) – bronze – typemetal – soft solder (ii) bronze – for making coins, - metals/other ornaments typemetal – for printing soft solder – for plumbing/welding.

(a) (i) Polymerization is a process whereby many small molecules (monomers) link together to form a giant molecule (polymer) (ii) - addition polymerization -condensation polymerization. (iii) I. starch/cellulose/ proteins/natural rubber/latex. II. polythene(polythene)/PVC/polystyrene/nylon/perplex. (b) (i) Ethyne is prepared in the laboratory by the action of water on cc!cium dicarbide. The gas evolved is passed through a solution of copper II tetraoxosulphate (VI) to remove impurities and the ethyne is collected over water by upward delivery. (ii) I. 3C\(_2\)H\(_{2(g)}\) \(\to\) C\(_6\)H\(_{6(g)}\) II. C\(_2\)H\(_2\) + 2HBr \(\to\) C\(_2\)H\(_4\)Br\(_2\)/ H - C\(\equiv\)C – H + 2HBr \(\to\) CH\(_3\)CHBr\(_2\). (iii) I. Benzene II. 1,1 – dibromoethane (c) I. Pollution is the presence of harmful substances in the environment in concentrations that cause injury to life and property. (II) Poil spillage prevention - enlightenment programme - tree planting -proper sewage disposal. (ii) - Air pollutions - water pollutgion - land/soil pollution (iii) - Use of anti-pollutants in automobiles - promulgation of laws with effective monitoring - oil spoilage prevention - elightenment programme - tree planting - proper sewage disposal

(a)(i) X (ii) R (iii) VV and X (iv) R (b) (i) I. group VII II. period III (ii) eleven (iii) one (iv) ZY. (c) (i) ionic/electrovalent bond (ii) covalent co-ordinate covalent and ionic (iii) covalent. (d) - pour some quantity of each solution into a separate beaker. - insert electrodes into the solutions and close the circuit. take the readings on the ammeter. - rinse the electrodes properly with distilled water and dip them into the second solution. Close the circuit and take the reading on the ammeter again. - the solution with the higher reading is NaOH\(_{(aq)}\) while the one with lower reading is ammonia. OR NaOH solution is a strong electrolyte and completely ionizes to produce a lot of ions. NH\(_3\) solution is a weak electrolyte and produces very few ions. The ions produced by NaOH far out-number those produced by NH\(_3\) solution. Hence NaOH solution conducts more than NH\(_3\) solution. (e) (i) Esterification (ii) Neutralization. (f) 2H\(_2(g)\) + O\(_{2(g)}\) —> 2H\(_2\)O (I) from the equation 2 vols of H\(_2\) reacts with 1 vol. of O\(_2\)/ 20cm\(^3\) of H\(_{2(s)}\) reacts with 10cm\(^3\) of O\(_{2(g)}\) residual gas = (15 — 10)cm\(^3\) = 5cm\(^3\) .

(a) (i) I. Deliquescence is a process by which a compound absorbs moisture/water from the atmosphere and dissolves in it to form a solution. II. Efflorescence is a process whereby a hydrated salt loses all or part of its water of crystallization when exposed to the air. (ii) Deliquescence: NaOH/FeCl\(_3\)/CaCl\(_3\)/CaCl\(_2\)/POl\(_5\) etc. Efflorescence: Na\(_2\)CO\(_3\).10H\(_2\)O/Na\(_2\)SO\(_4\).10H\(_2\)O/MgSO\(_4\).7H\(_2\)O etc. (b) — To a sample of each of the gases, add acidified KMnO\(_4\)/Br, water (vapour). — No visible reaction shows ethane OR decolourization of KMnO\(_4\)/Br\(_2\) water (vapour) shows ethene and ethyne. —To each of the two remaining samples, add ammoniacal copper (I) chloride/ammoniacal AgNO\(_3\)/Ag\(_2\)O — No visible reaction shows ethene/formation of reddish-brown precipitate/white precipitate indicate the presence of ethyne. (c) (i) Water of crystallization is the definite amount of water some substances chemically combine with when they form salt from their aqueous solution. (ii) % of anhydrous CaCl\(_2\) is 100 — 49.32 = 50.68%. Number of moles of CaCl\(_2\) H\(_2\)O \(\frac{50.68}{111}\) \(\frac{49.32}{18}\) 0.457 2.74 0.457 0.457 1 : 6 x = 6 (d) (i) Calcium trioxocarbonate (IV) (ii) Calcium chloride (iii) Calcium tetraoxosulphate (VI) (iv) Calcium oxide/calcium hydroxide. (e) (i) Na\(^+\) and CI\(^-\) H\(^+\) and OH\(^-\) (ii) I. chlorine II. hydrogen (iii) Sodium hydroxide solution/NaOH

a) Gay Lussac's law states that when gases react, they do so in volumes which simple ratios to one another and to the volumes of the products, if gaseous, provided that the temperature and pressure remain constant. (b) (i) 2CO\(_{(g)}\) + O\(_{2(g)}\) \(\to\) 2CO\(_{2(g)}\) (ii) I. from the equation: 2 volumes of CO\(_{(g)}\) and 1 volume of O\(_{2(g)}\) —> 2 volumes of CO\(_2\). ratio of reacting volumes 40cm\(^3\) + 20cm\(^3\) —> 40cm\(^3\) 40cm\(^3\) of CO requires 20cm\(^3\) of oxygen to form 40cm\(^3\) of CO\(_2\). 40cm\(^3\) of CO\(^2\) is produced II. excess reactant is oxygen Volume is 30cm\(^3\)— 20cm\(^3\) = 10cm\(^3\). III. total volume of gaseous mixture 10cm\(^3\) of O\(^2\) + 40cm\(^3\) CO\(_2\) = 50cm\(^3\). (c) (i) SiO\(_2\). (ii) CaO (iii) ZnO (iv) CO (v) CO and NO\(_2\). (d) (i) Conc. HNO\(_3\). slowly decomposes releasing brown NO, which dissolves in the acid to give the yellow colour. (ii) HNO\(_3\). is an oxidizing acid. Any hydrogen produced is oxidized to water. (e) (i) 2Cu(NO\(_3\))\(_2\) \(\to\) 2CuO\(_{(s)}\) + 4NO\(_{2(g)}\) + O\(_{2(g)}\) (ii) NH\(_4\)NO\(_3{(s)}\) \(\to\) N\(_2\)O(_{(g)}\) + 2H\(_2\)O\(_{(g)}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

(a) Vaporating dish - Bunsen Burner - Tripod stand / wire guaze (b)(i) Na\(_2\)CO\(_3\) - K\(_2\)CO\(_3\) - CH\(_3\)COONa (ii) hydrolysis (c)(i) White precipitate, effervescence, precipitate dissolves (ii) SO\(_2\)/H\(_2\)S/unsaturated gaseous hydrocarbon

Tes Observation Inference (a)(i) C + H\(_2\)O, then filter (ii) Filterate + NaOH\(_{(aq)}\) in drops, then in excess (iii) Filterate + NH\(_{3(aq)}\) in drops, then in excess (iv) Filterate + BaCl\(_{2(aq)}\) + dil. HCl (b)(i) First portion of residue + heat (i) Second portion of residue + dil. HCl Green residue, blue filterate Blue gelatinous precipitate insoluble in excess Pale blue (gelatinous) precipitate ppt dissolves/soluble in excess to give a deep blue solution White precipitate insoluble in dil. HCl Colour changed from green to black powder Effervescence, colourless, odourless gas which turned lime water milky C is a mixture of soluble and insoluble salts Cu\(^{2+}\) present Cu\(^{2+}\) confirmed SO\(_4^{2-}\) confirmed Gas is CO\(_2\) from CO\(_3^2\)

Volume of pipette = 25.00 cm\(^{3}\) indicator used = methyl orange colour change at end point yellow to orange/purple Burette Reading Rough (1) First (2nd) Second (3rd) Third Final (cm\(^3\)) 25.20 24.90 24.80 24.82 Initial (cm\(^3\)) 0.00 0.00 0.00 0.00 Volume of acid used 25.20 24.00 24.80 24.85 Average volume of A used = \(\frac{24.90 + 24.80 + 24.85}{3}\) = 24.85 cm\(^3\) The equation for the reaction X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) 2 x 2NO\(_{3(aq)}\) + CO\(_{2(g)}\) + 11HNO\(_{(l)}\) (b)(i) Mole ratio of Acid to Base is 2:1 concentration of B in mol. dm\(^{-3}\) \(\frac{C_AV_A}{C_BV_B} = \frac{2}{1}\) C\(_B\) = \(\frac{C_AV_A}{2V_B}\) C\(_B\) = \(\frac{0.0950 \times 24.85}{2 \times 25}\) = \(\frac{2.36075}{50}\) = 0.0472 mol. dm\(^{-3}\) (ii) Molar mass B in g mol\(^{-1}\) Molar mass X\(_2\)CO\(3\).10H\(_2\)O = \(\frac{\text{mass concentration gdm}^{-3}}{\text{molar concentration of B in moldm}^{-3}}\) = \(\frac{13.5 gdm^{-3}}{0.0472 moldm^{-3}}\) = 286 g mol\(^{-1}\) (iii) Percentage by mass of X X\(_2\)CO\(_3\).10H\(_2\)O = 286g i.e. relative molar mass of X\(_2\)CO\(_3\) = 2x + 12 + (16 x 3) + 10(2 x 1) + 16 = 286 = 2x + 12 + 48 + 180 = 286 = 2x + 240 = 286 2x = 286 - 240 = 46 x = \(\frac{46}{2}\) = 23 gmol\(^{-1}\) Percentage by mass of X = \(\frac{2 \times 23}{286} \times \frac{100}{1}\) = 16.08%

% of H2 = 100 - 92.3 = 7.7% mole = Carbon [92.3 / 12] = 7.7 Hydrogen [7.7 / 1] = 7.7 divide with the lowest: Carbon - 7.1 / 7.1 = 1 Hydrogen - 7.1 / 7.1 = 1 empirical formula = CH (CH)n = 26 (12 + 1)n = 26 (13)n = 26 n = 26/13 = 2 molecular formula = C\(_2\)H\(_2\)

element : C | H Atomic Mass : 12.011 | 1.008 mass in 100 g of compound 88.9 | 11.100 we calculate the number of moles by dividing the mass by the atomic mass # moles in 100 g : 7.40155 | 11.01190 molar ratio 7.40155 : 11.01190 smallest No : 7.4015 divide by the smallest No. : 1.000 : 1.488 empirical formula is CH\(_2\)

An insoluble salt can be prepared by a precipitation method through a double decomposition reaction. Two soluble metal salts react with each other to form one soluble salt and one insoluble salt . Lead nitrate solution and sodium chloride solution react to produce solid lead chloride , leaving soluble sodium nitrate in solution. The lead chloride can be recovered by filtration.

Potash alum [kAl(S\({O}_4\))\(_2\)] n sodium aluminate (ii) (NaHCO\(_2\)) are added to cause coagulation. As fine particles to clump together, forming large particles heavy enough to settle at the bottom of the tanks as impurities.