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(a) (i) - action of an acid upon an active metal / metals higher than hydrogen in the reactivity series - preparation by double decomposition (reaction between two soluble compounds to produce an insoluble compound) - by neutralization (reaction between acids & bases) - action of an acid on trioxocarbonate (IV) of a metal - direct combination of elements - action of dilute acid on an insoluble base Examples: (ii) - Addition of granulated zinc to a dilute solution of tetraoxosulphate (VI) acid. Zn(s) + H\(_2\)SO\(_{4(aq)}\) → ZnSO\(_{4(aq)}\) + H\(_{2(g)}\) - Pb(NO\(_3\))\(_{2(aq)}\) + H\(_2\)SO\(_{4(aq)}\) → PbSO\(_{4(s)}\) + 2HNO\(_{3(aq)}\) OR Pb(NO\(_3\))\(_{2(aq)}\) + N\(a_2\)SO\(_{4(aq)}\) → PbSO\(_{4(s)}\) + 2NaNO\(_{3(aq)}\) 2NaOH\(_{(aq)}\) + \(H_2\)SO\(_{4(aq)}\) → Na\(_2\)SO\(_{4(aq)}\) + 2H\(_2\)O\(_{(l)}\) ZnCO\(_{3(s)}\) + H\(_2\)SO\(_{4(aq)}\) → ZnSO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\) - CaCO\(_{3(s)}\) + 2HCl\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H2O\(_{(l)}\) + CO\(_{2(g)}\) - 2Fe + 3Cl\(_2\) → 2FeCl\(_3\) Fe + S → FeS - 2HNO\(_3\) + CuO → Cu(NO\(_3\))\(_2\) + H\(_2\)O (iii) Uses of Na\(_2\)CO\(_3\) - used as a water softner - manufacture of glass/paper/rayon/soaps/detergents/NaOH/Borax - food processing aid - pH modifier - swimming pool chemical - electrolyte - analytical reagent / standardization of acids (b) Use pure copper as the cathode and impure copper as anode and then use a named soluble copper salt e.g. CuSO\(_4\) solution as the electrolyte impure copper dissolves /CuCu\(^{2+}\) + 2e\(^-\) and will be deposited at cathode / Cu\(^{2+}\) + 2e\(^-\) Cu Wash the cathode to obtain pure copper. (c) (i) Mass of NaCl deposited = 39.0 – 36.3 g = 2.7 g % NaCl deposited = 2.7 x \(\frac{100}{39.0}\) = 6.92 % or 7.0% Mass of AgNO\(_3\) deposited = 952.0 – 297.0 = 655 g % AgNO\(_3\) = \(\frac{655}{952}\) x 100 = 68.8 % Or 69% (ii) AgNO\(_3\) (Silver trioxonitrate (V)) will be more efficiently crystallized because it has a higher percentage deposited.

(a) (i) - All elements consist of atoms which are indivisible - Atoms can neither be created nor destroyed - The atoms of an element are all alike but differ from those / atoms of other elements - When atoms combine with other atoms they do so in simple ratios - All chemical changes result from the combination or separation of atoms (ii) If A and B form more than one compound, since the atoms are indivisible, combination will be in (simple) integral ratios, as mass of atom of each element is fixed. (b) (i) - soap - margarine - propan – 1,2,3 - triol / glycerine Alkaline hydrolysis gives soap / propan-1,2,3-triol (c) Atomic radii increases down the group and decreases across the period hence the size of Beryllium (Be) is about the same as the size of Aluminium (Al). The charge densities of Be\(^{2+}\) and Al\(^{3+}\) cause almost the same effect. OR Periodic properties increase / decrease from left to right and decrease / increase down the group in the periodic table. A diagonal relationship / similar properties exist between a pair of diagonally adjacent elements in the second and third periods of the periodic table as in the case of Be and Al. (d) I\(_2\) + 2Na\(_2\)S\(_2\)O\(_3\) → 2NaI + Na\(_2\)S\(_4\)O\(_6\)

(a) (i) When water does not easily form lather with soap / when water contains dissolved calcium and magnesium salts. Water that does not easily lather with soap / Water that contains dissolved calcium and magnesium salts (ii) Prepare soap solution Measure volume of sample water from a burette or pipette titrate with shaking to permanent lather. take titre value say x cm\(^3\) Measure same volume of boiled sample water and then titrate with shaking take titre value say y cm\(^3\) Find ratio; \(\frac{\text{permanent}}{\text{Total}}\) = \(\frac{y}{x}\) (iii) Waste of soap - produces scum during washing with soap - blockage of pipes - scales in boilers / furring of kettles - leads to wastage of fuel - not suitable for tanning / laundry / textile/ paper manufacturing industries (b) (i) NaNO\(_3\) or KNO\(_3\) conc H\(_2\)SO\(_4\) (ii) concentrated H\(_2\)SO\(_4\) heat (over a sand bath) all glass apparatus (iii) 2HNO\(_3\) + S → H\(_2\)SO\(_4\) + 2NO OR 6HNO\(_3\) + S → H\(_2\)SO\(_4\) + 6NO\(_2\) + 2H\(_2\)O II. no reaction / reaction prevented / metal passive oxide formed which coats metal (c) Zinc oxide Lead (II) oxide Aluminium oxide Tin(IV)oxide Aluminium hydroxide Tin (II) oxide Lead (IV) oxide Beryllium oxide

(a) (i) 2SO\(_2\) + O\(_2\) + 2H\(_2\)O \(\to\) 2H\(_2\)SO\(_4\) (ii) Use BaCl\(_2\) solution it gives white insoluble solid / precipitate to form bariumtetraoxosulphate (VI) OR Add a solution of bariumtrioxonitrate (V), a white precipitate of bariumtetraoxosulphate (VI) is formed (iii) I. C\(_2\)H\(_2\)O\(_4\) \(\frac{-H_2O}{H_2SO_4}\) CO + CO\(_2\) (iv) I. H\(_2\)SO\(_4\) is acting as a dehydrating agent II. Cu + 2H\(_2\)SO\(_4\) → SO\(_2\) + CuSO\(_4\) + 2H\(_2\)O II. H\(_2\)SO\(_4\) is acting as an oxidizing agent (b) (i) Water is referred to as a universal solvent because it is able to dissolve more substances than any other liquid. This has to do with the polarity of each water molecule. The hydrogen side of each water molecule carries a slight positive electric charge, while the oxygen side carries a slight negative electric charge and this helps water dissociate ionic compounds into their positive and negative ions. (ii) Water will turn cobalt (II) chloride paper from blue to pink / anhydrous copper (II) tetraoxosulphate (VI) crystals from white to blue (c) (i) CO / Carbon (II) Oxide (ii) Because it is combustible / can be used as fuel (d) (i) coke (ii) coal tar / ammoniacal liquor / pitch (iii) coal gas (e) (i) C(s) + O\(_{2(g)}\) →CO\(_{2(g)}\) (ii) I. - It is odourless - It is colourless - It is denser than air - It is soluble in water - It is a gas at room temperature - It is easily compressible (ii) It does not support combustion It reacts with red hot carbon (to form CO) it is weakly acidic in water it reacts with alkali (to form trioxocarbonate (IV) ) / turns lime water milky Reacts with burning magnesium

(a) - high temperatures / stated temperature of 500\(^o\)C or more - catalyst (b) - To prevent corrosion/rusting/tarnishing - To make them shiny/for decoration purpose/beautify/attractive - To make surface harder/thicker - To make more resistant to chemicals - To make it durable - To reduce friction/abrasion (c) (i) Because calcium oxide is basic and hydrogen chloride is acidic ,they will react to form salt and water. (ii) - Fused CaCl\(_2\) - (Conc) H\(_2\)SO\(_4\) (do not accept dil H\(_2\)SO\(_4\)) - Silica gel - Phosphorus(V) oxide/P\(_2\)O\(_5\) (d) Conc. HNO\(_3\) is a strong oxidizing agent. It oxidizes pale green Fe\(^{2+}\) ions to yellow Fe\(^{3+}\) ions (1) and itself reduced to nitrogen (IV) oxide. (e) (i) I. ZnO(s) + H\(_2\)SO\(_{4(aq)}\) \(\to\) ZnSO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) II. ZnO\(_{(s)}\) + 2NaOH\(_{(aq)}\) \(\to\) N\(a_2\) Zn(OH)\(_{4 (aq)}\) OR ZnO + 2NaOH \(\to\) Na\(_2\)ZnO\(_2\) + H\(_2\)O (ii) Amphoterism / Amphoteric properties (f) (i) Same number (17) of protons / same number (17) of electrons/same atomic number (ii) Difference in mass numbers / different number of neutrons / 35Cl has 18 neutrons and 37Cl has 20 neutrons (g) - HCl / hydrochloric acid - O\(_2\) / Oxygen (h) (i) H\(_2\)S (ii) SO\(_2\) (i) It is the mass reference scale in which the atom of the 12C isotope is taken to have twelve units of atomic mass OR It is a scale that uses the 12C isotope as the standard for comparing the atomic masses of other elements. (j) - The rate of forward reaction equals the rate of backward reaction / The system is dynamic - There is no net change in the concentration of each substance - The equilibrium compositions of the substances can be approached from either reactants or products - Gibb’s free energy/ is zero

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

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

(a) (i) X - oxygen Y - sulphur (ii) I. - S(s) + O\(_{2(g)}\) \(\to\) SO\(_{2(g)}\) (1) II. - 2SO\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2SO\(_{3(g)}\) (2) III. - SO\(_{3(g)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) (1) IV. - H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\) 2H\(_2\)SO\(_{4(aq)}\) (2) (iii) Vanadium (V) oxide (1) (iv) It is because the reaction is exothermic (v) - In bleaching wood pulp, wool silk, hay - as germicide and fumigant/ disinfectant - as refrigerant / refrigerating gas/coolant - as preservative for grains and fruits - manufacture of H\(_2\)SO\(_4\) - as a reducing agent in metallurgy - for dechlorination (b) From the equation, 2 mol of H\(_2\) reacts with 1 mol of O\(_2\) hence 40 cm\(^3\) of H\(_2\) would react with 20 cm\(^3\) of O\(_2\) Volume of unused oxygen = (30 – 20) cm\(^3\) = 10 cm\(^3\) (c) (i) CaCO\(_3\) \(\to\) CaO + CO\(_2\) (ii) CaCO\(_3\) = 40 + 12 + (16 x 3) = 100 g CaO = 40 + 16 = 56 g 100 g of CaCO\(_3\) \(\to\) 56 g CaO 1.0 g will give \(\frac{1}{100}\) x 56 = 0.56 g (iii) 1 mole of a gas at s.t.p. 22.4 dm\(^{-3}\) 1 mole of CaCO\(_3\) produces 1 mole of CO\(_2\) 100 g of CaCO\(_3\) produces = 22.4 dm\(_3\) of CO\(_2\) 1.0 g of CaCO\(_3\) will produce = \(\frac{1}{100}\) x 22.4 = 0.224 dm\(^3\) of CO\(_2\) (iv) Volume at 15\(^o\)C and 760 mm Hg Temp = 273 + 15\(^o\)C = 288 \(\frac{V_1}{T_2} = \frac{V_2}{T_2}\) Volume V\(_2\) = 0.224 x \(\frac{288}{273}\) = 0. 236 dm\(^3\)

(a)(i) (ii) - used as germicide/disinfectant/sterilization/antiseptic/insecticides bleaching of materials/wood pulp/textiles commercial production of HCl production of polymers/PVC/plastics production of chlorinated chemicals e.g. CHCl\(_3\), CCl\(_4\) in medicine production of dyes (b) Water gas is CO + H\(_2\) steam is mixed with water gas and passed over heated catalyst of iron (III) oxide or chromium (III) oxide at 450\(^o\) C The CO in water gas is converted to CO\(_2\) with further yield of hydrogen The CO\(_2\) is removed by dissolving the mixture in water / NaOH/KOH/Ca(OH)\(_2\) solutions under pressure (c) (i) Bauxite (ii) - To produce aluminium oxide/alumina. - To prevent poisoning of the electrode. (iii) Carbon/ graphite (iv) - It is added to reduce some of the energy (costs) involved in extracting aluminium - to increase the electrical conductivity of alumina - to lower the melting points to of the alumina (d) Coke - Coal tar - Coal gas - ammoniacal liquor

(a) Rough Tree 1st Titre 2nd Titre 3rd Titre Final reading cm\(^3\) 12.60 25.10 12.40 25.00 Initial reading cm\(^3\) 0.60 12.60 0.00 12.40 Volume of acid cm\(^3\) 12.60 12.50 12.40 12.60 Average Titre = \(\frac{12.50 + 12.40 + 12.60}{3}\) = 12.50cm\(^3\) Equation of the reaction I\(_2\) + 2S\(_2\)O\(_3^{2-}\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\) (b)(i) Concentration of A in moldm\(^{-3}\) = \(\frac{\text{Concentration of A in gdm}^{-3}}{\text{Molar mass of A in g mol}^{-1}}\) = \(\frac{15.8gdm^{-3}}{\text{Molar mass of A}}\) Molar mass Na\(_2\)S\(_2\)O\(_3\) = (23)2 + (32) 2 + (16)3 = 46 + 64 + 48 = 158 g mol\(^{-1}\) concentration of A in moldm\(^{-3}\) = \(\frac{15.8gdm^{-3}}{158gmol^{-1}}\) = 0.100 moldm\(^{-3}\) (ii) Concentration of I\(_2\) in B moldm\(^{-3}\) \(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\) Where; C\(_A\) = 0.100 moldm\(^{-3}\) V\(_A\) = 12.50 cm\(^2\) C\(_B\) = ? V\(_B\) = 25.00 cm\(^2\) C\(_B\) = \(\frac{0.100 \times 12.50 \times 1}{2 \times 25.00}\) C\(_B\) = \(\frac{1.25}{50}\) C\(_B\)= 0.0250 moldm\(^{-3}\) (iii) percentage by ass of I\(_2\) in the sample = \(\frac{\text{Concentration in gdm}^{-3}}{9.0 gdm^{-3} \text{of impure}} \times \frac{100%}{1}\) Concentration in gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{ Molar mass of } I_2}\) 0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{Molar mass of } I_2}\) I\(_2\) = 127 x 2 = 254/mol 0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} (Pure)}{ \text{Molar mass of }I_2}\) Concentration in gdm\(^{-3}\) (Pure) = 0.0250 x 254 = 6.35 gdm\(^{-3}\) Percentage by mass of I\(_2\) in the sample = \(\frac{6.35}{9.0} \times \frac{100%}{1}\) = 70.555 = 70.56% (c) Starch indicator was not added to the titration mixture at the beginning of the titration in order to obtain an accurate end-point or to prevent the formation of the complex which reduces the accuracy of the titre value.

TEST OBSERVATION INFERENCE (a) C + distilled water and filter C dissolved partially forming a green residue with colorless filtrate Fe\(^2+\) or Cu\(^{2+}\) salt is present (b)(i) 2cm\(^3\) of filtrate + AgNO\(_{3(aq)}\) + HNO\(_{3(aq)}\) White precipitate formed Precipitate insoluble or No visible change Cl\(^{-}\), CO\(_3^{2-}\) or SO\(_3^{2-}\), S\(^{2-}\) Cl\(^-\) is present (ii) Resulting mixture in (b)(i) + excess NH\(_{3(aq)}\) Precipitate dissolves or soluble Cu\(^{-}\) is present (c)(i) Residue + dil HCl The residue dissolved to form a blue solution effervescence (bubbling) occurs. Colourlessness, odourless gas evolved that turned blue litmus paper red and turned lime water milky Cu\(^{2+}\) is present CO\(_{2(g)}\) from CO\(_3^{2-}\) (ii) Portion of mixture of (c)(i) + NH\(_{3(aq)}\) in drops then in excess Light (Pale) blue precipitate formed in drop of NH\(_{3(aq)}\) to form a deep blue solution CO\(^{2+}\) is confirmed

(a) BaCl\(_{2(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) followed saturated by dilute HCl in excess - A white precipitate is formed. - Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl (ii) Gas Q that decolourized acidified KMnO\(_2\) solution could be any of these. SO\(_{2(g)}\), H\(_2\)S\(_{(g)}\), unsaturated hydrocarbon (C\(_2\)\(H_2\), C\(_2\)H\(_4\)....) (b) One substance used in the laboratory for drying each of the following: (I) ammonia gas - Calcium oxide (ii) carbon (IV) Oxide - Calcium chloride - Concentrated tetraoxosulphate( VI) acid. (c) Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.

The following factors increase a reaction rate - Increase in concentration of reactants - Increase in temperature - Addition of catalyst - Increase in the surface area of reactant(s)

(a) BaCl\(_{2(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) followed saturated by dilute HCl in excess - A white precipitate is formed. - Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl (ii) Gas Q that decolourized acidified KMnO\(_2\) solution could be any of these. SO\(_{2(g)}\), H\(_2\)S\(_{(g)}\), unsaturated hydrocarbon (C\(_2\)\(H_2\), C\(_2\)H\(_4\)....) (b) One substance used in the laboratory for drying each of the following: (I) ammonia gas - Calcium oxide (ii) carbon (IV) Oxide - Calcium chloride - Concentrated tetraoxosulphate( VI) acid. (c) Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.

Given: P = 4 atm , V = 10 dm\(^3\) , n = ? R = 0.082 (constant) T = 273 + 273 = 546K Using the ideal gas law, PV = nRT n = \(\frac{PV}{RT}\) n = \(\frac{{4}\times{10}}{{0.082}\times{546}}\) n = \(\frac{40}{44.772}\) n = 0.893 mole