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

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

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

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

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

(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.

(a)(i)- alpha particle, (doubly charged) helium nucleus/high ionizing/low penetrating power/deflected towards negative plate. - beta particle, (fast moving) electron/moderate ionizing/moderate penetrating power/deflected towards positive plate. - gamma radiation, (high energy) electromagnetic radiation/low ionizing/high penetrating power effect in magnetic field. (ii) Chemical Reaction Nuclear Reaction - atoms are rearranged by the breaking and forming of chemicarbondlno new atom is formed/atomic number or mass number of each element does not change. - only (valence) electrons are involved. - reactions are accompanied by absorption /release of (relatively) small amount of energy. - rate of reaction is influenced by temperature, pressure, concentration and catalyst. - elements are converted to other elements/new elements may be formed/atomic number or mass number of each element may change - protons, neutrons, electrons and other elementary particles may be involved. - reactions are accompanied by release of (relatively) large amount of energy. - rate of reaction normally is not affected by pressure, temperature and catalyst. (iii) I. \(^{212}_{84}Po\) \(to\) \(^{208}_{82}Pb\) + \(^{4}_2X\) (X is \(\alpha\)-particle/helium nucleus/\(^{4}_2He\)) II. \(^{137}_{55}Cs\) \(\to\) \(^{137}_{56}Ba\) + \(^{0}_{-1}Y\) (Y is \(\beta\)-particle/electron/ \(^{0}_{-1}He\)) (b).(i) atomic number = 17 (ii) group 7 (iii) - formation of diatomic molecules. - formation of monovalent ions/same valence electrons, - they are coloured substances. - they are oxidizing agents., - most electronegative in its period/smallest atomic size in its period. - they are non-metals. (iv) chlorine . (v) 3Cl\(_2\) +6NaOH ----> 5NaCl + NaCIO\(_3\) + 3H\(_2\)O. (c)(i) I. Graphite consists of planar layers (in which all carbon-carbon bonds extend in two dimensions) which can slide over each other. Hence it is used as lubricant. II. Diamond consist of (large) three dimensinoal network of carbon-carbon bonds which Makes it hard hence, used as a cutting tool. (ii) HCI\(_{aq}\) + NH\(_{3(aq)}\) --> NH\(_4\)CI\(_{(aq)}\) OR HCI\(_{aq}\) + NH\(_4\)HO\(_{aq}\) --> NH\(_4\)Cl\(_{aq}\) + H\(_2\)\(O_l\) (iii) Combination OR Neutralisation.

(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^\circC = 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\(^{3}\)

(a)(i) A base is a substance which produces hydroxyl (OH\(^-\)) ions in its aqueous solution. (ii) NaOH/Ca(OH)\(_2\)/ KOH. (iii) I. non-electrolyte II. weak electrolyte HI. strong electrolyte. (iv) CH\(_3\)COOH + KOH --> CH\(_3\)COOK + H\(_2\)O (b) \(\frac{C_AV_A}{C_BV_B}\) = 1 V\(_A\) = \(\frac{C_BV_B}{C_A}\) \(\frac{0.30 \times 20.50}{0.50}\) cm\(^3\) (c)(i) sodium oxochiorate 1 (ii) magnesium (iii) hydrogen trioxocarbonate(IV). (d)(i) A standard solution is a solution whose concentration is (accurately) known. (ii) Na\(_2\)CO\(_3\); Na\(_2\)CO\(_3\) is easy to weigh accurately because it is stable in air/non-deliquescent. (iii) I. Fe\(_{(s)}\) + 2HCl\(_{(aq)}\) \(\to\) FeCl\(_{2(aq)}\) + H\(_{2(aq)}\). II. number of moles of HCI used = 2.20 x \(\frac{50.0}{1000}\) = 0.11 mole ratio of HCI : Fe = 2 : 1 number of moles of Fe used = \(\frac{0.11}{2}\) = 0.055 mole 0.055 mole of Fe = 3.08g 0.1 mole of Fe = \(\frac{3.08 \times 1.0}{0.055}\) = 56g Relative atomic mass of Fe = 56.

(a)(i) - sodalime does not attack glass apparatus, unlike caustic soda - sodalime is not, deliquescent unlike caustic soda. (ii) - it is a gas at room temperature. - colourless gas. - odourless gas. - slightly soluble in water. -less dense than air. (iii) I. % Carbon Hydrogen 82.76 17.24 82.76 17.24 12 1.00 6.90 17.24 6.90 6.90 1 2.5 2 5 II. Emperical formula = C\(_2\)H\(_5\) Molecular formula = C\(_2\)H\(_5\)n = VD x 2 [(12 x 2) + (1 x 5)]n = 58 29n = 58 n = 2 molecular formula = (C\(_2\)H\(_5\))\(_2\) = C\(_4\)H\(_{10}\) (b) (i) Tha existence of two or more compounds with the same molecular formula but different molecular structures/different arrangement of atoms. (iii) ethanol methoxymethane (iv) The two isomers belong to different homologous series/they have different functional groups. (c) - functional groups - number of atoms of different elements - types of elements - types of bonds - percentage composition of the elements in a compound - spatial arrangements of atoms in a molecule (d) Pass each of the gases into (acidified) KMO\(_4\)/ bromine water/bromine inCCl\(_4\)/ bromine propene decolourizes bromine water/KMnO\(_4\)/bromine in CCI\(_4\)/bromine whereas propane does not.

(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.

(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.

(a) The solution of salt Q was acidic/acid solution - CO\(_2\)/carbon (IV) oxide was given off - the reaction was exothermic (i.e. Heat is given off) - an insoluble compound was formed (b)(i) PbSO\(_4\) and ZnCO\(_3\) (ii) NaHCO\(_3\) and ZnCO\(_3\) (iii) NH\(_4\)Cl, NaHCO\(_3\), Cu(NO\(_3\))\(_2\) and ZnCO\(_3\) (v) NH\(_4\)Cl

Test Observation Inference (a)(i) C\(_{(s)}\) C dissolved to give a colourless solution (ii) C\(_{(aq)}\) + X\(_{(aq)}\) X decolourized/turned colourless C is a reducing agent (iii) C\(_{(aq)}\) + BaCl\(_{2(aq)}\) White precipitate SO\(_4^{2-}\), SO\(_3^{2-}\) or S\(^{2-}\) or SO\(_3^{2-}\) + excess HCl\(_{(aq)}\) precipitate dissolved/precipitate soluble SO\(^{2-}_3\) confirmed (b) D\(_{(s)}\) + distilled water D dissolved to give a colourless solution (i) D\(_{(aq)}\) NaOH\(_{(aq)}\) in drops + excess NaOH White precipitate precipitate dissolved/precipitate Pb\(^{2+}\) or Ca\(^{2+}\) Pb\(^{2+}\) (ii) D\(_{(aq)}\) + K\(_2\)CrO\(_{(aq)}\) Yellow precipitate Pb\(^{2+}\) confirmed

(a) Volume of base used (pipette) = 25.00 cm\(^3\); indicator - methyl orange Burette reading cm\(^3\) Rough 1st Titration 2nd Titration Final burette readings (cm\(^3\)) 21.70 43.30 31.90 Initial burette readings (cm\(^3\)) 0.00 21.60 10.30 Volume of acid used (cm\(^3\)) 21.70 21.60 21.60 Average volume of acid = \(\frac{21.60 + 21.60}{2}\) V\(_A\) = 21.60cm\(^3\) (b)(i) Concentration of A in mol dm\(^{-3}\) Molar mass of NaHSO\(_4\) = 23 + 1 + 32 + (4 x 16) = 120g mol\(^{-1}\) Concentration of A I mol/dm3 = \(\frac{ mass concentration of A}{Molar mass of A}\) = \(\frac{12.0}{120}\) = 0.100mol\(dm^3\) (ii) concentration of B in mol dm\(^{-3}\) \(\frac{C_AV_A}{C_BV_B}\) = 1 C\(_B\) = \(\frac{C_AV_A}{V_B}\) = \(\frac{0.10 \times 21.6}{25}\) = 0.0864 mol dm\(^{-3}\) (3 s.fig) iii) Mass of \(Na^+\) formed in solution . Amount of NaOH in the solution = \(\frac{25 \times0.0864}{1000}\) = 0.00216 moles Mole ratio NaOH : \(Na_{2}SO_{4}\) = 1 : 1 Amount of \(Na_{2}SO_{4}\) = 0.00216 moles. Atomic mass of \(Na^+\) = 23 Amount of \(Na^+\) = 2 x 0.00216. Mass of \(Na^+\) formed = 23 x 2 x 0.00216 = 0.9936g. ⇒ 0.10g