NYGH 2021-S3EOY-IP Chem P1 and P2 ans

1 Secondary 3 (IP) EOY Answers 2021 Sec 3 End-of-Year Exam Chemistry Answers Paper 1 1 B 2 C 3 A 4 C 5 B 6 B 7 D 8 D 9 C 10 D 11 A 12 C 13 A 14 C 15 A 16 B 17 C 18 B 19 A 20 B 21 B 22 D 23 A 24 A 25 A 26 A 27 B 28 C 29 A 30 B Qn Ans Explanation 1 B At ‒46°C, M is a solid. At 110°C, M is a gas. Melting and boiling points of M must both be between ‒46°C and 110°C. 2 C For the water level at X to decrease, the rate of diffusion of gas P into the porous pot must be faster than the rate of diffusion of air out of the porous pot. Hence, gas P needs to have a lower Mr than air (< 28.96). The rate of diffusion of gas Q into the porous pot must be slower than the rate of diffusion of air out of the porous pot. Hence, gas Q needs to have a higher Mr than air (> 28.96). For option B, both carbon monoxide and nitrogen have the same Mr. Hence, there will be no change in the level of X. 3 A The independent variable is the one that is changed. This is the volume of acid used. A burette is used because the volume of acid that can be measured can vary but not when a pipette is used because the volume measured is fixed. The dependent variable is the one that is measured. This is the temperature recorded and hence, a thermometer is used. 4 C A is not possible as sulfur dioxide is highly soluble in water. B is wrong because sulfur dioxide is more densed than air and downward delivery should be used. D is wrong because the acidic gas of sulfur dioxide will react with concentrated potassium hydroxide, which is an alkali. 5 B Apparatus is a separating funnel that separates immiscible liquids (eg. oil and water). The bottom layer would have a higher density and would be collected first when the tap is opened. Since organic solvents like carbon disulfide and water are immiscible, they can be separated using the separating funnel. 6 B The Rf value of the sample as shown in Y is 0.5. Hence, the spot should appear at the 0.5 mark in a longer piece of paper.

2 7 D B and C are wrong because the number of protons should follow the proton numbers. A is wrong because an ion with 2+ charge should have 2 more protons than electrons. 8 D If X has 7 valence electrons (group 17), Y would have 8 valence electrons (group 18) and Z would have 1 valence electron (group 1). From statement 2, we can identify X and Y as period 3 elements chlorine, Cl, and argon, Ar. Z is potassium, K. A is false as X- ion and Z+ ion would form a compound ZX. B is false as relative atomic mass of Y (Ar) is 39.9 while that of Z (K) is 39.1. C is false as the next electron should be found in the 4s subshell. The 3d subshells are not filled. D is true as Z is a metal and X is a non-metal. 9 C Ar of potassium = 39(93.258/100) + 40(0.012/100) + 41(6.730/100) = 39.13 10 D H2‒ ion would have 3 electrons and the spdf notation would be 1s2 2s1, which does not follow the electronic configuration of any of the noble gases.