2024 YIJC JC2 H1 CHEM PE P1 (worked solutions)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME SUGGESTED ANSWERS CG H1 GROUP 1 / 2 / 3 CHEMISTRY Paper 1 Additional Materials: Multiple Choice Answer Sheet Data Booklet 8873 /01 13 September 2024 1 hour READ THESE INSTRUCTIONS FIRST This document consists of 19 printed pages. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and class in the spaces at the top of this page and 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.

©YIJC [Turn over 2 1 Which of the species in their gaseous state shown below will be deflected by the smallest angle in an electric field? A 1H2 B 4He2+ C 9Be2+ D 11B3+ Answer: C angle of deflection ∝ charge mass of the particle A 1H2 Not deflected since it is not charged B 4He2+ 2 4 = 0.5 C 9Be2+ 2 9 = 0.222 D 11B3+ 3 11 = 0.273 2 What is the electronic configuration of vanadium, 23V? A 1s2 2s2 2p6 3s2 3p6 3d4 4s1 B 1s2 2s2 2p6 3s2 3p6 3d5 C 1s2 2s2 2p6 3s2 3p6 4s2 4p3 D 1s2 2s2 2p6 3s2 3p6 3d3 4s2 Answer: D V atom has 23 protons and 23 electrons. 1s2 2s2 2p6 3s2 3p6 3d3 4s2 3 Use of the Data Booklet is relevant to this question. How many atoms of gold, 79Au, are there in a pure gold coin with a mass of 3.94g? A 1.20 × 1021 B 2.00 × 1021 C 1.20 × 1022 D 2.00 × 1022

©YIJC [Turn over 3 Answer: C Amount of Au = 3.94  197 = 2 × 10−2 mol No. of atoms = (2 × 10−2) × (6.02 × 1023) = 1.20 × 1022 4 10cm3 of gaseous hydrocarbon is burnt completely in 70cm 3 of oxygen. When cooled to room temperature, the gaseous volume was 50cm 3. A further decrease of 40cm 3 in the gaseous volume was observed when the gaseous mixture is passed through aqueous potassium hydroxide. What is the molecular formula of the hydrocarbon? A C2H6 B C3H8 C C4H8 D C5H12 Answer: C Reduction of gaseous volume due to reaction with KOH(aq) → CO2(g) is an acidic gas and reacts with KOH(aq) via acid-base reaction → volume of CO2(g) = 40 cm3, Volume of oxygen unreacted = 50 − 40 = 10 cm3 Volume of