2019 SAJC H2 Chem Prelim P1 ANS

1 NAME Class ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examination H2 Chemistry (9729) Paper 1 Multiple Choice 19 Sep 2019 1 hour Additional Materials: Multiple Choice Answer Sheet, Data Booklet READ THESE INSTRUCTIONS: Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. There are thirty questions on 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 in soft pencil on the separate Answer sheet. 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 booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 14 printed pages (including this page).

2 1 Which statement about one mole of sodium metal is always true? A It has the same mass as one mole of 12C. B It has the same number of atoms as 18 g of water. C It has the same number of atoms as 1 12 mole of 12C. D It has the same number of atoms as 12 dm3 of fluorine gas at r.t.p. Ans: D 12/24 = 0.5 mol of F2 = 1.0 mol of F atoms = 6.02 x 1023 H atoms 2 0.84 g of an oxide MO of a metal M was dissolved in excess sulfuric acid. 25.0 cm 3 of 0.12 mol dm3 potassium manganate(VII) solution was required to oxidise M2+ to M3+. What is the relative atomic mass of M? A 36.0 B 40.0 C 52.0 D 56.0 Ans: B Amt of MnO4 = 25/1000 x 0.12 = 0.003 mol Amt of e gained by MnO4 = Amt of e lost by M2+ = 5 x 0.003 = 0.015 mol Mr of MO = 0.84/0.015 = 56 Ar of M = 56-16 = 40 3 X and Y are elements found in the first three periods of the Periodic Table. The outermost shell electronic configurations of two species are given as follows: X2: Y2+: What can best be deduced from the above information? A X has a larger proton number than Y. B X has more unpaired electrons than Y at the ground state. C X exists as a gas while Y is a solid at standard condition. D X2 and Y2+ are isoelectronic. Ans: B Since X and Y are elements in the first three periods and their valence electronic configurations consist of s and p subshells, there are likely to be in period 2 or period 3. X has the valence shell electronic configuration ns2 np4 so it is a Group 16 element and it can be in Period 2 or Period 3. X can be O or S. Y has the valence shell electronic configuration of (n+1)s2 and it should be in Period 3. Y can be Mg.

3 Option A: Since Y is Mg, X can have a smaller or larger proton than Y, depending on whether X is O or S Option B: X would have 2 unpaired electrons while Y has no unpaired electrons. Option C: X can be oxygen which is a gas but it may also be sulfur which is a solid at standard condition. Option D: If X is S, S2 is not isoelectronic with Mg2+. 4 In an experiment, a sample of gaseous 87Sr2