Chemistry 2 Nov/Dec 2010

Chemistry Paper 2,Nov/Dec 2010

Questions:		1	2	3	4	5	6	7	8		Main

General Comments
Weakness/Remedies
Strength

Question 2
(a) (i) Mention the various states of matter (ii) Arrange the states of matter in decreasing order of: I. kinetic energy of particles; II. disorderliness between particles; III. forces of cohesion between particles. (iii) A pure sample of a substance melts at 35°C and boils at 100°C. I. Sketch a graph to show the heating curve II. What is the state of the substance at normal room temperature? [9 marks] (b) Consider the following equation: 2KHC03(s) What volume of CO2(g) measured at s.t.p. would be produced when 25.0 g of KHC03(s) was completely decomposed? [K = 39.0 = 16.0; C = 12.0; H = 1.00; Molar volume of a gas ats.t.p. = 22.4 drn" mol"] [ 5 marks] (c) When ammonium chloride was heated in a dry test tube, it was observed that the damp red litmus paper placed at the tip of the test tube immediately turned blue and after some seconds turned red. (i) State the gas that was responsible for the: I. red litmus paper turning blue; II. blue colour of the litmus paper changing back to red. (ii) Give reason for your answer in 2 (c)(i) (iii) Write an equation for the decomposition [6 marks] (d) (i) Name two salts that could react with dilute acids to produce sulphur (IV) oxide (ii) Mention I. two uses of sulphur (IV) oxide; II. one chemical property of sulphur (IV) oxide. [5 marks]
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OBSERVATION
The question was attempted by most candidates and the performance was fair. In(a)(i) and (ii), Candidates correctly mentioned solid, liquid and gas as the states of matter and gave the arrangement of the states of matter as follows: (ii) I Gas > liquid > solid II Gas > liquid > solid III Solid > liquid > Gas In (a)(iii) I and II, most candidates could not sketch the required heating curve graph and hence, did not know the substance. The expected answer from candidates were as follow: II. Solid. In part (b), candidates correctly determined the volume of CO₂(g) at S.T.P thus: Molar mass of KHCO₃ = 39 + 1 + 12 + ( 3 x 16) = 100gmorl^-1 No. of mole of KHCO₃ = 25.0 = 0.25 mol 100 From the equation 2 moles of KHC0₃ = Imole CO₂ No. of mole of CO₂ = ½ x 0.25 mol = 0.125 mol. Volume of CO₂ at s.t.p = 0.125 mol x 22.4 dm³ mol^-1 = 2.8 dm3 OR Alternate Method Molar mass of KHCO₃ = 39 + 1 + 12 + (3 x 16) = 100gmol^-1 2 x 100g KHCO₃ = 22.4 dm³ ., 25g = 22.4 x 25 200 = 2.8dm³ In part (c), most candidates correctly inferred ammonia and HCL in (i) I and II. However, most of them could not explain their answer in (ii). In (c)(iii), most candidates correctly wrote an eqation for the decomposition The expected responses from candidates in each of (ii) and (iii) were as follows: (ii) NH₃ being lighter than HCI/NH₃ travels faster than HCL (iii) NH₄CI(s) ~ NH₃(g) + HCI(g) In part (d) (i), candidates named potassium trioxosulphate(IV) and sodium trioxosulphate(IV) as the two salts that would react with dilute acids to produce sulphur(IV)oxide. In(d)(ii), candidates mentioned two uses and one chemical property of sulphur(IV)oxide from among the following: - as germicide - manufacture of tetraoxosulphate(VI) acid - as preservative - as refrigerant - as bleaching agent. II. Sulphur (IV) oxide combines with water to form solution of trioxosulphate (IV) acid reacts with oxygen in the presence of catalyst to form sulphur(VI)oxide reacts with alkali to form salt and water reacts with H2S to form water bleaching action. .