SRJC H1 CHEM P1 ANS

1 Turn Over] SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 Candidate Name Class CHEMISTRY 8872/01 JC2 Preliminary Examination 28 Aug 2015 Paper 1 Multiple Choice 50 min Additional Materials: Data Booklet Optical Mark Sheet (OMS) READ THESE INSTRUCTIONS FIRST On the separate multiple choice OMS given, write your name, subject title and class in the spaces provided. Shade correctly your FIN/NRIC number. There are 30 questions in this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider c orrect and record your choice using a soft pencil on the separate OMS. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. You are advised to fill in the OMS as you go along; no additional time will be given for the transfer of answers once the examination has ended. Any rough working should be done in this question paper. This document consists of ___ printed pages and __ blank page.

2 Turn Over] Answer all questions 1 Oxides of nitrogen, NxOy, are air pollutants. In a reaction, 0.30 mol of NxOy is reacted with 30 dm 3 of hydrogen gas at room temperature and pressure and passed over a heated catalyst to produce ammonia and water as the only products . At the end of the reaction, 1.20 dm 3 of hydrogen gas remains . The ammonia produced required 0.300 mol of sulfuric acid for complete neutralisation. What is the molecular formula of the oxide? A NO B NO2 C N2O D N2O4 Answer: C NxOy + 2 )23( yx  H2  xNH3 + yH2O 0.30 mol 1.20 mol 0.600 mol No. of moles of H2 = (30-1.2)/24 = 1.20 2NH3 ≡ H2SO4 No. of moles of NH3 = 0.300 x 2 = 0.600 Comparing mole ratio, NxOy : NH3 1 : 2  x = 2 Comparing mole ratio, NxOy : H2 1 : 4 2 )23( yx  = 4  y = 1 Molecular formula is N2O. 2 In an experiment, H 2S was reacted with 28.00 cm 3 of 0.250 mol dm 3 of an unknown arsenate species, H 3AsOx, in a strongly acidic medium to form a yellow solid of mass 0.225 g and ortho–arsenite, AsO3 3. Determine the oxidation state of As in H3AsOx. A +2 B +3 C +4 D +5 Answer: D Unknown yellow solid is S (Deduction). H2S  S + 2H+ + 2e H3AsOx + ne AsO3 3 (unbalanced equation) +3 Mole ratio: 0.250 x 1000 28 H3AsOx ≡ 32.1 0.225 S 0.007 H3AsOx ≡ 0.007 S H3AsOx ≡ S  n = 2 Since final oxidation state of AsO3 3 = +3, initial oxidation state must be +5.

3 Turn Over] 3 The ion E2+ has 86 electrons and 138 neutrons. Which of the following statements is true? A Element E is isoelectronic with radon. B The first ionisation energy of element E is higher than that of Sr. C The oxide of E formed is expected to have a higher melting point than SrO. D In an electric field, the ion E2+ will be deflected at a smaller angle than that of Sr2+. Answer: D A × Element E has 86 + 2 = 88 electrons. ( Ra) B × First ionisation energy decreases down the group. C ×E2+ has the same charge as Sr2+ but a larger cationic radius smaller lattice energy D √ E2+ has a larger mass than Sr2+ and hence deflected at a smaller angle since extent of deflection  m e 4 The second ionisation energies ( I.E.) of seven consecutive unknown elements F to L in Periods 2 and 3 of the Periodic Table are shown in a graphical plot below. Which unknown letter (F to L) is likely to represent the element nitrogen? A Element F B Element G C Element H D Element I Answer: B Look for the highest 2nd I.E. in the graph and that will most likely correspond to a Group I element. In this case, G is most likely deduced to be Na. Hence, G is nitrogen. F G H I J K L 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5 6 7 8 9 10 11 12 13 Second I.E./ kJ mol1 Proton Number

4 Turn Over] 5 Lithium peroxide, Li 2O2, is synthesi sed by reacting lithium hydroxide with hydrogen peroxide. Which diagram correctly shows the bonding of the peroxide anion in Li2O2? Key: electron from the first oxygen atom electron from the second oxygen atom  electron from lithium atom A B C D Answer: D 6 Which pair of compounds satisfies the following conditions? (i) The first compound has a larger bond angle than the second compound. (ii) The first compound is more polar than the second compound. First compound Second compound A ClO2 HCN B NF3 SeF6 C IF3 PH3 D BeCl2 N2H4 Answer: B A ClO2 (105o, polar) HCN (180o, polar) B NF3 (107o, polar) SeF6 (90o, non-polar) C IF3 (90o, polar) PH3 (107o, polar) D BeCl2 (180o, non-polar) N2H4 (107o, polar) x x x x x

5 Turn Over] 7 Which of the following has a positive ∆H value? A CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) B NaOH(aq) + CH3COOH(aq)  CH3COO–Na+(aq) + H2O(l) C 2O(g)  O2(g) D Na(s)  Na(g) Answer: D A enthalpy of combustion has negative ∆H B enthalpy of neutralisation has negative ∆H C bond formation has negative ∆H D enthalpy of atomisation has positive ∆H 8 The table shows the enthalpy change of neutralisation per mole of water formed, ∆H, for the following acids and bases. acid base ∆H /kJ mol-1 hydrochloric acid sodium hydroxide -57.0 P Sodium hydroxide -54.0 hydrochloric acid Q -52.0 nitric acid R -57.0 What are P, Q and R? P Q R A propanoic acid ammonia sodium hydroxide B propanoic acid potassium hydroxide ammonia C sulfuric acid ammonia potassium hydroxide D sulfuric acid sodium hydroxide ammonia Answer: A ∆H = -57 kJ mol-1 indicates neutralisation between strong acid and strong base. Any numerical value lesser than 57 indicates reaction between either strong acid and weak base or weak acid and strong base. As such, P is weak acid: propanoic acid Q is weak base: ammonia R is strong base: sodium hydroxide/ potassium hydroxide

6 Turn Over] 9 The table shows some data on two acid-base indicators. Indicator pH range of colour change Colour change acid alkali thymolphthalein 9-10 colourless blue chlorphenol red 6-7 yellow red Which conclusion can be drawn about a solution in which thymolphthalein is colourless and chlorphenol red is red? A It is weakly acidic. B It is neutral. C It is weakly alkaline. D It is strongly alkaline. Answer: C When thymolphthalein is colourless, pH < 9 When chlorphenol red is red, pH > 7 pH range of solution 7-9 -> Solution is weakly alkaline. 10 Equal volumes of 0.050 moldm-3 of hydrochloric acid was mixed with 0.050 mol dm–3 of calcium hydroxide. Calculate the pH of the resulting solution. A 13.1 B 12.4 C 7.0 D 1.6 Answer: B Ca(OH)2 is diacidic and hence is in excess. [ OH–] remaining = [OH–]int – [OH–]reacted = (0.10 - 0.05) / 2 = 0.0250 mol dm–3 pOH = -lg(0.25) = 1.6 pH = 14 – pOH = 12.4

7 Turn Over] 11 Serine protease (SP) is a biological catalyst which increases the rate of hydrolysis of amide bonds in proteins. Which graph represents the [serine proteases] to rate of reaction A B C