SRJC H2 CHEM P1 ANS

1 SRJC 9647/01/PRELIM/2014 Turn Over] SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 Candidate Name Class CHEMISTRY 9647/01 Preliminary Examination 29 August 2014 Paper 1 Multiple Choice 1 hour Additional Materials: Data Booklet Optical Mark Sheet (OMS) READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, FIN/NRIC number and class on the OMS in the spaces provided. Shade correctly FIN/NRIC number and your class. Eg. If your NRIC is S9306660Z, shade S9306660Z for the item “index number”. There are forty 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 correct 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. Any rough working should be done in this question paper. This document consists of 22 printed pages and 0 blank page.

2 SRJC 9647/01/PRELIM/2014 Turn Over] Section A For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct. 1 The hydrolysis of S2Cl2 proceeds by two reactions. Reaction 1 S2Cl2 + 2H2O H2S + SO2 + 2HCl Reaction 2 2H2S + SO2 8 3 S8 + 2H2O Which of the following correctly describe the reactions? A A weakly acidic solution is formed in Reaction 1. B Reverse disproportionation occurs in Reaction 2. C The oxidation state of chlorine and oxygen changes in both reactions. D The products of Reaction 1 require 2 mol of NaOH for complete neutralisation. Ans: B  Strong acid HCl is formed.  H2S is oxidised to S 8 while SO 2 is reduced to S 8 (the reverse of disproportionation reaction).  Oxidation states of chlorine and oxygen remain as –1 and –2 in both reactions.  2HCl + 2NaOH  2NaCl + 2H2O SO2 + H2O  H2SO3 H2SO3 + 2NaOH  Na2SO3 + 2H2O Total amount of NaOH required = 4 mol 2 Sodium carbonate peroxyhydrate with the formula (Na 2CO3)x y H2O2 is used in eco-friendly cleaning products and as a laboratory source of anhydrous hydrogen peroxide. When 20.0 cm 3 of 0.100 mol dm –3 sodium carbonate peroxyhydrate is titrated with 0.200 mol dm –3 acidified KMnO 4, it requires 12.0 cm 3 of acidified KMnO 4 before the first pink colour appears. 2MnO4 – + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2 When an identical sample is acidified, it releases 96.0 cm 3 of carbon dioxide at room conditions.

3 SRJC 9647/01/PRELIM/2014 Turn Over] 3 The successive ionisation energies, in kJ mol -1, of an element E are given below. 578 1820 2750 11600 14800 18400 Which of the following could be the electronic configuration of the outermost shell in E? A ns2 B ns2 np1 C ns2 np2 D ns2 np3 Ans: B Largest increase is between the 3 rd and 4 th IE, hence this is a Group III element. Therefore, the outermost shell electronic configuration is ns2 np1 What is the ratio of x:y ? A 1 : 3 B 2 : 3 C 2 : 1 D 3 : 1 Ans: B Amt of MnO4 - used = 0.002400 mol Amt of H2O2 present = 0.002400 / 2 x 5 = 0.006000 mol Amt of sodium carbonate peroxyhydrate = 0.002000 mol y = 0.006 / 0.002 = 3 Na2CO3 + 2H+  CO2 + H2O + 2Na+ Amt of CO2 = 0.096 / 24 = 0.004000 mol x = 0.004 / 0.002 = 2

4 SRJC 9647/01/PRELIM/2014 Turn Over] 4 The diagram represents the melting points of four consecutive elem ents in the third period of the Periodic Table. The sketches below represent another two properties of the elements. Which of the following represents the properties F and G? property F property G A third ionisation energy electronegativity B number of valence electrons boiling point C ionic radius effective nuclear charge D electrical conductivity atomic radius Ans: A Based on the melting point data, the four elements are Si (Group IV) (high melting point due to giant molecular structure), P 4 (Group V), S8 (Group VI) and Cl2 (Group VII). Property F is 3rd IE as there is an anomaly at Group V: Si2+ → Si3+ + e 3s2 3s1 P2+ → P3+ + e 3s2 3p1 3s2     proton number 0 melting point / K     proton number 0 property F     proton number 0 property G

5 SRJC 9647/01/PRELIM/2014 Turn Over] Electron from P 2+ is removed from an outer subshell and hence less energy is required. Property G is electronegativity as electronegativity increases across the period. 5 Two identical bulbs at the same temperature contain ideal gases J and K separately. The density of gas J is twice that of gas K and the molecular mass of gas J is half that of gas K. What is the ratio of the pressure of gas J to that of gas K? A 1 : 2 B 1 : 1 C 2 : 1 D 4 : 1 Ans: D PV = nRT PM = m V RT P = RT M  2 1 42 1 J K J JK K MP P M   6 Which of the following statements about an ideal gas are correct? A One mole of any ideal gas occupies the same volume under the same conditions of temperature and pressure. B The density of an ideal gas at constant pressure is directly proportional to the temperature. C The volume of a given mass of an ideal gas is doubled when its temperature is raised from 25 °C to 50 °C. D The temperature of a given mass of an ideal gas is doubled when its volume is raised from 0.05 cm3 to 0.1 dm3.

6 SRJC 9647/01/PRELIM/2014 Turn Over] Ans: A A  PV =nRT B   = PM RT density is inversely proportional to temperature. C  Temperature is in kelvin so it is not doubled, thus volume is not doubled also. D  0.1 dm3 is 2000 times of 0.05 cm 3 so temperature should increase by 2000 times. 7 The type of bonding between two elements can be rationalised and even predicted using a van Arkel triangle. The triangle is based on electronegativity values. Difference in electronegativity is plotted along the y –axis and average electronegativity is plotted along the x–axis. What is the type of bonding present at each of these bonding extremes, labelled L, M and N on the triangle? L M N A covalent metallic ionic B metallic covalent ionic C covalent ionic metallic D ionic covalent metallic Ans: B Difference in electronegativity is zero for L & M  metallic bonding or covalent bonding. Since metals have low electronegativity  L is metallic bonding. L M N average electronegativity difference in electronegativity 

7 SRJC 9647/01/PRELIM/2014 Turn Over] 8 Lead is the final product formed by a series of changes in which the rate–determining stage is the radioactive decay of uranium –238. This radioactive decay is a first–order reaction with a half–life of 4.5 x 109 years. What woul d be the age of a rock sample, originally lead –free, in which the molar proportion of uranium to lead is now 1:31? A 2.25 x 1010 years B 2.70 x 1010 years C 3.15 x 1010 years D 3.60 x 1010 years Ans: A Uranium–238 is reactant; lead is product. Use formula: Since Uranium:lead = 1:31  1+31 = 32 parts 5n 2 1 32 1 2 1              where n = no. o