Question Library

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.

Aliya is a smart and brilliant student who has A grades in all her subjects except chemistry where she has C.

Her Biology Teacher [See Page 46]

First book [Read about the novelist; Bolaji Abdullahi]

The Hunter's Story [See Page 51 - 53]

This is a narrative techniques/point of view where the story is narrated by one character at a time. The character may be speaking about himself or sharing events that he or she is expecting.

Aliya's Father [See Page 34 - 36]