H1 Chem P1 SAJC 2024 Ans

1 ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 1 CANDIDATE NAME CLASS 2 3 S CHEMISTRY Paper 1 Multiple Choice Candidate answer on the Optical Answer Sheet Additional Materials: Data Booklet 8873/01 11 September 2024 1 hour READ THESE INSTRUCTIONS FIRST 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 Optical 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 13 printed pages (including this cover page).

2 1 Oxygen exist as two isotopes; 16O and 18O respectively. Which of the following particles contain more neutrons than protons and more protons than electrons respectively? 1 12C18O32– 2 1H316O+ 3 14N18O+ A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 3 only Ans: D Option 1 is wrong because the particle have a negative charge, implying that the number of electrons is more than the number of protons. Option 2 is wrong because 16O has similar number of protons and neutrons and 1H has more proton than neutron even though there is more proton than electron due to the positive charge. Option 3 is correct because 18O in N 18O+ has 2 more neutrons than protons, and 1 more proton than electron. 2 The first seven successive ionisation energies for element X are as shown. X is found in Period 3. 1st 2nd 3rd 4th 5th 6th 7th Ionisation energy / kJ mol–1 1010 1900 2900 5000 6300 21300 25400 Which compound can be formed using X? A XO B XO2 C XO3 D X4O10 Ans: D There is a large jump in ionisation energy from 5th to 6th, which implies that the 6th electron comes from an inner shell that is closer to nucleus and will experience stronger nuclear attraction. Hence, X has 5 valence electrons and is from Group 15. As a period 3 element, it is able to form X4O10 (ie P4O10).

3 3 Use of the Data Booklet is relevant to this question. Tl81 204 can undergo natural radioactive decay, where one of its electrons enters the nucleus to change a proton into a neutron, to form a new element X. When X is put in an ionisation chamber, it emits a high energy -particle (which is a 4He nucleus). What is the identity of the element X and the path of the emitted -particle in an electric field? X Deflection Path A X80 204 I B X82 204 II C X80 205 I D X82 205 II Ans: A Based on the radioactive decay, the proton number will drop by 1. But however, the nucleon number will remain the same as there is now 1 more neutron. Hence, X is X80 204 . Angle of deflection α |q/m| q/m for He nucleus = 2/4 = 0.5 q/m for electron = 1/(1/1840) = 1840 Hence, the angle of deflection for the α-particle will be smaller than that of electron. Hence I is the deflection path for the α-particle.

4 4 Which statements about cyanogen molecule, (CN)2, are correct? 1 (CN)2 is polar. 2 (CN)2 is bent at the central carbon atoms. 3 A (CN)2 molecule has 3 σ and 4 π bonds. 4 A (CN)2 molecule has a total of 26 electrons. A 1, 2, 3 and 4 B 1 and 2 only C 2 and 3 only D 3 and 4 only Ans: D Option 1 is wrong as the molecule is linear and the individual dipole moment from C to N are opposite and equal to each other, thus resulting in its cancellation. Option 2 is wrong as the molecule is linear (ie. 2 bond pairs and 0 lone pair around each C). Option 3 is correct. Option 4 is correct as 7 + 7 + 6 + 6 = 26.

5 5 The structure of ice is as shown. Which statement is incorrect? A The open structure causes ice to be less dense than liquid water. B The open structure gives ice a larger mass than liquid water. C Four electrons from each oxygen are involved in forming hydrogen bonds. D Each oxygen atom in a water molecule is tetrahedrally bonded to 4 hydrogen atoms. Ans: B Option A is correct as the open structure (owing to the hydrogen bonds between water molecules) results in a larger volume occupied than liquid water and hence, density of ice is smaller than water. Option B is incorrect as the open structure does not change the mass. Option C is correct as each O atom will use its 2 lone pairs of electrons to form 2 hydrogen bonds in ice. Option D is correct as around each O atom, there are 2 covalent bonds to H atoms and 2 hydrogen bonds to H atoms, resulting in a tetrahedral arrangement.

6 6 The structure of histamine is as shown. Which is the correct order of bond angle from smallest to largest? Smallest bond angle Largest bond angle A x y z B z y x C y z x D x z y Ans: A x = 107ο, y = 109.5ο, z = 120ο 7 Use of the Data Booklet is relevant to this question. Copper metal, copper(II) ions and water are formed when dilute sulfuric acid is added to copper(I) oxide. Which option is correct? number of moles of Cu+ reacted number of moles of Cu formed number of moles of Cu2+ formed A 1 1 1 B 1 2 1 C 2 1 1 D 2 2 1 Ans: C Cu2O + H2SO4 → H2O + SO42– + Cu + Cu2+ To get 1 mol of Cu and 1 mol of Cu2+, refer to Data Booklet for the following half equations.

7 Cu+ + e– → Cu --- (1) Cu+ → Cu2+ + e– -- (2) (1) + (2): 2Cu+ → Cu + Cu2+ (2Cu+ came from 1 mol of Cu2O) 8 Use of the Data Booklet is relevant to this question. Which statement is correct? A 2.00g of hydrogen gas contains 3.00 x 1023 atoms. B 4.00g of helium gas contains 6.00 x 1023 molecules. C 28.0g of carbon monoxide gas contains 6.00 x 1023 molecules. D 88.0g of carbon dioxide gas contains 2.40 x 1024 atoms. Ans: C Option A is wrong because 1 mol of H2 contains 6.02 x 1023 x 2 = 1.204 x 1024 atoms. Option B is wrong because 1 mol of He contains 6.02 x 1023 atoms. He is a noble gas and does not exist as molecules. Option C is correct because 1 mol of CO contains 6.02 x 1023 molecules. Option D is wrong because 2 mol of CO2 contains 2 x 3 x 6.02 x 1023 = 3.612 x 1024 atoms. 9 Which compound has the same empirical formula as its molecular formula? A dinitrogen tetraoxide B ethanoic acid C propanone D tetrafluoroethene Ans: C Molecular formula Empirical formula Option A is wrong. N2O4 NO2 Option B is wrong. C2H4O2 CH2O Option C is correct. C3H6O C3H6O Option D is wrong. C2F4 CF2

8 10 Use of the Data Booklet is relevant to this question. Which compound contains 54.1% by mass of calcium? A Calcium oxide B Calcium nitrate C Calcium sulfate D Calcium hydroxide Ans: D For 1 mol of compound; % by mass of Ca CaO 40.1/56.1 x 100% = 71.5% Ca(NO3)2 40.1/164.1 x 100% = 24.4% CaSO4 40.1/136.2 x 100% = 29.4% Ca(OH)2 40.1/74.1 x 100% = 54.1% 11 10.0 cm 3 of 0.30 mol dm –3 thallium nitrate, T lNO3, required 20.00 cm 3 of 0.10 mol dm –3 acidified NH4VO3 for oxidation to Tl3+. Vanadium is the only element which is reduced. What is the final oxidation state of vanadium? A 0 B +2 C +3 D +4 Ans: B Tl+ → Tl3+ + 2e– Amount of Tl+ = 0.01 x 0.3 = 0.003 mol Amount of e– = 0.003 x 2 = 0.006 mol Amount of VO3– = 0.02 x 0.1 = 0.002 mol e– : VO3– = 3 : 1 Final oxidation state of vanadium = + 5 – 3 = + 2

9 12 Use of the Data Booklet is relevant to this question. In an energetics experiment, 2.00 g of a fuel is completely burnt. 55% of the energy released is absorbed by 200 g of water and the temperature rose from 18 οC to 66 οC. What is the energy released per gram of fuel burnt? A 20 064 J B 36 480 J C 36 845 J D 72 960 J Ans: B q = 200 x 4.18 x (66 – 18) = 40 128 J q (100%) = (100 x 40 128) / 55 = 72 960 J Total energy released per gram of fuel burnt = 72