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. H‒ ion is a possible ion that has an spdf notation of 1s2. 11 A There are no negative values for options B, C and D. 12 C Calcium hydroxide reacts with ammonium nitrate to form calcium nitrate, ammonia and water. Nitrogen is lost through ammonia gas being released to the surroundings. 13 A chemical equation: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) ionic equation: 2Na+(aq) + CO32‒(aq) + 2H+(aq) + 2Cl‒(aq) → 2Na+(aq) + 2Cl‒(aq) + CO2(g) + H2O(l) 14 C The difference between the two test tubes is the solvent that is used. With water being used as a solvent in tube B, only tube B shows effervescence, indicating that the citric acid is able to react with magnesium in an acid + metal reaction. This shows citric acid only produces hydrogen ions when dissolved in water, showing its acidic properties.

3 15 A no. of mol of HX required = 0.100 mol dm‒3 x 0.020 dm3 = 0.002 mol Hence, A is incorrect. HX is a strong acid as the final pH is about 1. Concentration of NaOH = 0.002 mol / 0.025 dm3 = 0.08 mol dm‒3 16 B Candidates are expected to know :  metallic oxides are basic  non-metallic oxides are acidic in nature  aluminium oxide is amphoteric  carbon monoxide is neutral Hence, to neutralize NaOH, only Al2O3 , P4O10 and CO2 are able to display acidic properties. 17 C The common uses of SO2 are in the use of the manufacture of bleaching agents and food preservatives. Noble gases are used in the manufacture of low energy light bulbs. 18 B PbCl2 is an insoluble salt and should be prepared via precipitation method – mixing of two aqueous solution containing Pb2+ and Cl . Hence, it is important to first prepare a solution containing aqueous Pb2+. Option B is the only correct answer because : Adding lead to excess nitric acid will ensure that all the solid lead will completely react and dissolve to produce aqueous Pb(NO3)2. Subsequently, Pb(NO3)2 (aq) is added to dilute HCl to produce solid PbCl2 which can be easily collected as residue via filtration. Option C is wrong as excess lead is added but not removed which will affect the purity of the PbCl2 collected. 19 A Statement 1 is correct. A lower pH signifies a higher concentration of H+. Statement 2 is correct. pH = - log [H+] Statement 3 is incorrect. pH 5 is weakly acidic. Hence it should be orange / yellow. Statement 4 is correct. For example, the ionic equation for the reaction between hydrochloric acid and sodium metal: 2H+ (aq) + Na (s)  Na+ (aq) + H2 (g) It is evident that the oxidation state of Na increases from 0 to +1 in Na+. Hence it lose electrons and is thus oxidized.

4 20 B Assuming all the solutions have a concentration of 1 mol dm-3 and exist in a volume of 1 dm3. Option A: 1 mol of (NH4)2SO4 = 3 mol of ions (2 x NH4+ + 1 x SO42) Option B: 1 mol of Al(NO3)3 = 4 mol of ions (1 x Al3+ + 3 x NO3) Option C: 1 mol of Fe(NO3)2 = 3 mol of ions ( 1 x Fe2+ + 2 x NO3) Option D: 1 mol of NaCl = 2 mol of ions ( 1 x Na+ + 1 x Cl) 21 B Percentage by mass = 100% x compound of Mr element of x Arelement of No. Element No. of element Total Mass Contributed C 5 5 x 12.0 = 60.0 Cl 2 2 x 35.5 = 71.0 H 11 11 x 1.0 = 11.0 N 1 1 x 14.0 = 14.0 Hence, the element with highest percentage by mass is Chlorine. 22 D Mass of O3 permitted in 1 cm3 of air = 8 x 10-9 g Amount of O3 = 8 x 10-9 / 48 No. of molecules of O3 = 8 x 10-9 x L / 48 23 A Amount of gas produced per day = 480 / 24 = 20 mol Since CO2 : CH4 = 1:1, Amount of CO2 = Amount of CH4 = 10 mol 24 A Since the reaction vessel is enclosed, there is no net gain or loss of reactants and products (By law of conservation of mass) , hence statement 1 is correct. By mole ratio CH4 : O2 : CO2 1 : 2 : 1 Volume ratio 50 : 100 : 50 Hence, the volume of oxygen required to react with 50 cm3 of methane is 100 cm3 and there are 50 cm3 of O2 that is unreacted and in excess. Hence, at the end of the reaction the total volume of gas comprise of O2 (unreacted) + CO2 produced: 50 + 50 = 100 cm3. The 50 cm3 of CO2 produced is acidic and will be absorbed by aqueous potassium hydroxide, which is an alkaline.

5 25 A SO2 is a pungent and acidic gas. It is able to decolourise acidic potassium manganate. 26 A Green solution and green precipitate upon the addition of aqueous ammonia is characteristic of the Fe2+ ion. Fe2+ can be easily oxidized to Fe3+ which the reddish brown precipitate (Fe(OH)3). Nitric acid is added to test for carbonate anion which is not present as no effervescence is observed. Hence, the yellow precipitate formed upon the addition of silver nitrate is due to the presence of I 27 B Statement 3 is incorrect. Across the period, atomic radius decreases. This is so as nuclear charge increases as number of protons increases. Shiel