2024 ACJC Prelim H1 Chem Paper 1 (Solutions)

This document consists of 13 printed pages and 1 blank page. Anglo-Chinese Junior College JC2 Preliminary Examination Higher 1 CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 8873/01 9 September 2024 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, Centre number and index number on the Answer Sheet in the spaces provided unless this has been done for you. 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. Read the instructions on the Answer Sheet very carefully. 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.

2 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over ACJC H1 Chem Prelim 2024 Paper 1 Answers 1 C 11 D 21 C 2 B 12 B 22 A 3 C 13 C 23 B 4 B 14 A 24 C 5 B 15 A 25 A 6 D 16 B 26 D 7 C 17 C 27 C 8 A 18 A 28 B 9 D 19 D 29 B 10 B 20 A 30 D 1 Carbon sulfide, CS2, is a volatile flammable liquid used in the manufacture of cellophane. On combustion, CS2 is oxidised as follows. CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) A 20 cm3 sample of carbon disulfide vapour is ignited with 100 cm 3 of oxygen. The final volume of gas after burning is treated with an excess of aqueous alkali. What percentage of this final volume dissolves in the alkali? [all volumes measured at the same temperature and pressure.] A 20% B 40% C 60% D 80% Answer: C Thinking process only acid will react & dissolve in alkali. From the above equation, CO2 & SO2 are both acidic gases [learn in periodicity] since the reactants & products are all gases, we can assume volume ratio as mole ratio. CS2(g) ≡ CO2(g) ≡ 2SO2(g) 20 : 20 : 2(20) So the reaction produces 20 + 40 = 60 cm3 of acidic gases.

3 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over Volume of O2 required to burn 20cm3 of CS2 = 20 x 3 = 60cm3 Volume of O2 unreacted = 100 – 60 = 40cm3 Total volume of gas at the end of reaction = (acidic gases CO2 + SO2 + unreacted O2) = 60 + 40 = 100 cm3 % of final vol that dissolve in alkali = 60/ 100 x 100 % = 60 % 2 2 moles of nitric acid, HNO3, a powerful oxidising agent, reacts with 3 moles of hydrogen sulfide, H2S, to form three products, one of which is water. In this reaction, the oxidation number of nitrogen decreases by 3. What are the other two products of this reaction? A N2O2 and H2SO4 C NO and SO2 B NO and S D N2O2 and H2SO3 Answer: B O.S of N in HNO3 =+5 →+2 Looking for N with OS=2, → NO 1 mol of HNO3 gains 3 moles of electrons 2 mol of HNO3 gains 6 moles of electrons 6 moles of electrons lost by 3 moles of H2S 2 moles of electrons lost by 1 mole of H2S O.S of S in H2S =-2 →0 Looking for S with OS=0, → S Products are NO and S 3 Consider the following half-equations. MnO4– + 8H+ + 5e– → Mn2+ + 4H2O Fe2+ → Fe3+ + e– C2O42– → 2CO2 + 2e– What volume of 0.01 mol dm–3 KMnO4 is required to oxidise 15 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4?

4 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over A 3 cm3 B 6 cm3 C 9 cm3 D 15 cm3 Answer: C The oxidation of FeC2O4 is represented by this overall equation [O]: Fe2+ + C2O42– → Fe3+ + 2CO2 + 3e– Therefore 3 mol of MnO 4– will react completely with 5 mol of FeC 2O4 (balance out the electrons). Overall Eqn: 3MnO4– + 5Fe2+ + 5C2O42– + 24H+ → 3Mn2+ + 5Fe3+ + 10CO2 + 12H2O No. of moles of FeC2O4 = 15 1000 x 0.01 = 1.5 x 10–4 mol No. of moles of MnO4– = 3 5 x 1.5 x 10-4 = 9.0 x 10–5 mol Volume of MnO4– = 9.0 x 10-5 0.01 x 1000 = 9 cm3 4 Use of the Data Booklet is relevant to this question. The table shows statements made by three students about the s, p and d electrons in the atoms of the element with atomic number 30. student statement X There are s electrons in 4 different quantum shells. Y There are p electrons in 2 different quantum shells. Z The d electrons have the same principal quantum number as the outermost s electrons. Which students are correct? A X, Y and Z B X and Y C Y and Z D X and Z Answer: B Keyword: atoms of the element with atomic number 30 => up to period 4 e.g. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 (30 electrons) Student X obviously correct as each period/ quantum shells will have s electrons. (there are 1s, 2s, 3s and 4s electrons) Student Y stated that there are p electrons in 2 different shells: there are 2p and 3p electrons so he is correct.

5 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over Student Z is wrong as in Period 4, d electrons are the 3d and outermost s electrons are the 4s 5 Alpha particles, He 2+, are commonly emitted by larger radioactive nuclei. The path of a mixture of an unknown ion, A, O2+ and alpha particles after passing through an electric field is as shown below. What could ion A be? A H+ B Be2+ C C+ D Al3+ Answer: B In general, we expect that the angle of deflection ( ) of a charged particle passing through an electric field to be directly proportional to the size of the charge (q), and inversely proportional to the mass (m). For A, the q/m value is in between 0.5 and 0.125. From the working shown in above table, angle of deflection for Be2+ is in between that of O2+ and He2+. q/m Alpha particle: He2+ 2/4 = 0.5 O2+ 2/16=0.125 A H+ 1/1=1 B Be2+ 2/9=0.22 C C+ 1/12=0.083 D Al3+ 3/27=0.11 source alpha particle O2+ A q m

6 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over 6 Which molecular structure has the smallest overall dipole? A C B D Answer: D The structure in D is the most symmetrical, so the individual dipole moment of the bonds, in this case C=O will cancel out. 7 Some car paints contain small flakes of silica, SiO2. In the structure of solid SiO2 • each silicon atom is bonded to x oxygen atoms, • each oxygen atom is bonded to y silicon atoms, • each bond is a z type bond. What is the correct combination of x, y and z in these statements? x y z A 2 1 covalent B 2 1 ionic C 4 2 covalent D 4 2 ionic Answer: C SiO2 exists in the form of a giant molecular diamond-like structure, with each silicon atom being tetrahedrally bonded to four oxygen atoms, and each oxygen forming two bonds to silicon. The bonds are covalent in nature.

7 © ACJC 2024 8873/01/Preliminary Examination/2024 [Turn over 8 What is the order of increasing volality at room temperature? 1 2,3-dimethylbut-2-ene 2 cis-hex-3-ene 3 trans-hex-3-ene A 2, 3, 1 C 3, 2, 1 B 1, 2, 3 D 1, 3, 2 Answer: A Volatility is inversely proportio