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a)(i) X = Vapour density of a gas (ii) Y = Free energy of change (\(\Delta\) G) (b) Primary cell cannot be recharged after it has used up the chemicals of which it is composed. The secondary cell can be charged. In the primary cell, the cathode is a zinc rod and the electrolyte in which they are s ammonium chloride solution. In the secondary cell, the cathode and the anode are both grids of lead-antimony alloy immersed in diluted H\(_2\)SO\(_4\) solution.

(i) The spreading of a smell is called diffusion which is simply the movement of parbcies of a substance from region of higher concentration to that of lower concentration. (ii) Isotopy. (b) The relative atomic mass = (18 x 0.75) + (20 x 0.25) = 13.5 + 5 = 18.5.

(a) (i) Yellow zinc (II) oxide is the residue when Zinc (II) trioxocarbonate (IV) is strongly heated in an open crucible. (ii) White powdery mass of anhydrous copper(II) tetraoxosulphate (VI) is left as the residue in the desiccator (b)(a)(i) above when hot, it is yellow and white when cold. (ii) above, when hot, it is white and blue when cold

(a) The hardness of diamond. (b) The characteristic lustre of diamond (c) The electrical conductivity of graphite (d) Soft nature of graphite (e) The porous nature of wood charcoal makes it good absorbent for gases.

(a) Physical change (b) Chemical change (c) Physical change (d) Physical change

(a)(i) Wasteful use of soap is uneconomical (ii) Soft water lacks calcium and such water can ease tooth decay. (b) Clear lime water contains high content of soluble calcium which can precipitate soap in a higher degree than tl:a sample of lime water already turned milky by carbon(IV) oxide. The milkiness of course is caused by the presence of insoluble calcium trioxocarbonate (IV)

(a) and (d) can exhibit geometric isomerism because of the restriction to movement of the molecule about the double bond.

(a) (i) Element Y (ii) Element Y (iii) Element Y (b) The element Y is in group VII and so it has a smaller atomic volume than element X which is in group I. This size of the atom is a measure of the atomic radius. The smaller the atom is, the more is the attractive force between the outermost electron and the nucleus and this is the reason why element Y will be a better acceptor of electron than X as oxidizing agent.

(a) It will be hardened (b) 3 moles of hydrogen molecules (c) Sodium salt of a fatty acid which is soap.

(a) (i) Strong acid can ionize completely E.g. HCI\(_{(aq)}\) —> H\(^{+}_{(aq)}\) + CI\(^-_{(aq)}\) Concentrated acid means great amount of the acidic compound dissolved in little amount of water. (ii) Amphoteric oxide is a metallic oxide which is both acidic and basic. E.g Al\(_2\)O\(_3\) + 6HCl —> 2AICl\(_3\) + 3H\(_2\)O\(_{(l)}\) Al\(_2\)O\(_{3(5)}\) + 2NaO\(_{(aq)}\) + 3H\(_2\)O\(_1\) ---> 2NaAl(OH)\(_{4(aq)}\) (b)(i) The manufacture of H\(_2\)SO\(_4\) by contact process sulphur(VI) oxide, the anhydride of H\(_2\)SO\(_4\) can be obtained by contact process which entails SO\(_{2(g)}\) + O\(_{2(g)}\) SO\(_{3(g)}\) The sulphur(VI) To dissolved in concentrated H\(_2\)SO\(_4\) to form, a fuming liquid called oleurn (H\(_2\)S\(_2\)O\(_{7(aq)}\)) H\(_2\)SO\(_{4(aq)}\) + SO\(_{3(g)}\) \(\to\) (H\(_2\)S\(_2\)O\(_{7(aq)}\) The oleum is carefully diluted with correct amount of water to .give ordinary concentrated H\(_2\)S\(O_4\).H\(_2\)S\(_2\)O\(_{7(aq)}\) + H\(_2\)O\(_{(l)}\) --> 2H\(_2\)SO\(_{4(aq)}\) (ii) H\(_2\)SO\(_4\) as an oxidizing agent C\(_{(S)}\) + 2H\(_2\)SO\(_{4(aq)}\) -->2H\(_2\)O\(_{(l)}\) + 2SO\(_{2(g)}\) + CO\(_{2(s)}\) H\(_2\)SO\(_4\) as a dehydration agent H\(_2\)C\(_2\)O\(_{4(g)}\) + H\(_2\)SO\(_{4(l)}\) --> CO\(_{(g)}\) + CO\(_{2(g)}\) + (H\(_2\)O + H\(_2\)SO\(_{4(aq)}\) (iii) It is commonly used in most industries.

(a) (i) Polymerization is the combination of many molecules of the same kind to form a complex molecule Conc. H\(_2\)SO\(_4\) without gain or loss of material. (b)(i) \(CH_3CH_2CH_2OH_{(aq)}\) \(CH_2CHCH_3\) + (\(H_2SO_4 + H_2O\)). (ii) \(CH_3CH_2CH_2OH_{(aq)}\) + CHOOH --> \(CH_3CH_2CH_2OH_{(l)}\) + H\(_2\)O\(_{(l)}\) (i) & (ii) are respectively dehydration and condensation type of reactions. (c)(i) Carbon 40.O% = \(\frac{40}{12}\) = 3.33 moles ; Hydrogen 6.7% moles \(\frac{6.7}{1} = 6.7\) moles Oxygen 53.3% = \(\frac{53.3}{16}\) = 3.33 moles Divide through by smaller number i.e. C \(\frac{33}{3.33}\) : H \(\frac{6.7}{3.33}\) : O \(\frac{3.33}{3.33}\) C = 1: H = 2.0: O = 1 Empirical formula = CH\(_2\)O(CH\(2\)O)x = 180 i.e. (30)x = 180 x = \(\frac{180}{30}\) = 6 Molecular formula = \(C_6H_{12}O_6\) (ii) The intermolecular bonding in ethanoic acid is hydrogen bonding while the butane molecules are only aggregated by weak van der Waals.

(a)(i) How fast a reaction can take place. (ii) Collision theory states that the reactant molecules must collide with one another before forming the product. The number of collisions is directly proportional to the rate of reaction. This value of rate of reaction is however limited by the number of molecules attaining the activation energy and having appropriate orientation of the molecules. Increase in temperature only favours endothermic process and low temperature will favour the exothermic process. Low temperature however slows the rate of reaction and therefore catalyst is applied. (b)(i) 2H\(_2\)O\(_{2(l)}\) \(\to\) 2H\(_2\)O\(_{(l)}\) + O\(_{2(g)}\) (ii) Thermodynamically, the decomposition of hydrogen peroxide is spontaneous. When hydrogen peroxide is put in a conical flask, nothing seem to happen. However. the addition of hydrogen peroxide to a little amount of manganese(IV) oxide (MnO\(_2\)) generates effervescence and a gas is vigorously given off. The reaction of MnO\(_2\) on the decomposition of hydrogen peroxide is catalystic. (iii) Energy profile. (c)(i) Certain endothermic processes can be exothermic at a certain condition of temperature.

(a)(i) Calcium and Magnesium. (b)(i) 2Cr - 14 = -2; 2Cr = -2 + 14 = 12 Cr = \(\frac{12}{2}\) = 6 the oxidation number of Cr = +6 (ii) On addition of few drops of dilutes H\(_2\)SO\(_4\) the equilibrium system, the solution colour changes from yellow to green because Cr\(^{+6}\) is reduced to Cr\(^{3+}\). (i) The solution changes from yellow colour to green and sulphur is precipitated because H\(_2\)SO\(_4\) is oxidized to elementary sulphur. \(Cr_2O^{2-}_{7(aq)} + 3H_2S_{(g)} + 8H^+_{(aq)} \to 2Cr^{3+}_{(aq)} + 7H_2O(11 + 3S_{(s)})\) (ii) The gas should be cautiously generated in the fume cupboard.

(a)(i) Faster rate of obtaining product (ii) Product is less pure. There is risk of decomposition of product. (b)(i) NH\(_4\)Cl, FeCl\(_3\), AlCl\(_3\), CuSO\(_4\), FeCl\(_2\), NH\(_4\)NO\(_3\), etc (any two) (ii) Hydrolysis (c)(i) Add distilled water, stir and filter, the filtrate obtained is glucose while the residue is startch (ii) Add fehling's solution to a solution suspension of each sample Bricks - red ppt. formed No visible reaction Sample is starch Sample is glucose or sucrose Add few drops of iodine soln. to a solution/suspension of each sample Blue - black colouration No visible reaction Sample is starch Sample is glucose or sucrose

Test Observation Influence (a)(i) C + D Blue precipitate formed Cu\(^{2+}\) is probably present (ii) Mixed filtered Colourless filtrate, Blue residue Cu\(^{2+}\) probably present (b)(i) Filtrate + BaCl\(_{2(aq)}\) + dil. HCl in excess White precipitate formed. Precipitate insoluble SO\(_4^{2-}\) or SO\(_3^{2-}\) or CO\(_3^{2-}\) present (ii) Residue + dil. HCl Effervescence gas evolved. The gas was odourless and turned lime water milky. Residue dissolved to give a pale blue solution SO\(_4^{2-}\) present CO\(_4^{2-}\) from CO\(_3^{2-}\) Cu\(^{2+}\) probably present (c)(i) Soln. from (b)(i) + excess NaOH\(_{(aq)}\) light blue precipitate insoluble in excess Cu\(^{2+}\) present (ii) Soln. from (b)(ii) + NH\(_{3(aq)}\) in drops until in excess Pale blue precipitate soluble in excess NH\(_{3(aq)}\) to give a deep blue solution Cu\(^{2+}\) confirmed