NYGH 2019-S3EOY-IP Chem P1 and P2 Ans

Secondary 3 (IP) EOY Answers 2019 Sec 3 End-of-Year Exam Chemistry Answers Paper 1 1 B 2 A 3 B 4 A 5 D 6 B 7 D 8 C 9 D 10 C 11 C 12 A 13 A 14 C 15 B 16 A 17 C 18 A / C 19 B 20 D 21 D 22 D 23 B 24 C 25 B 26 B 27 C 28 A 29 B 30 C Qn Ans Explanation 1 B Mixture of solid (ice) and liquid (water) will be at regions with constant temperature. Q-R is at a lower temperature so that would be the solid-liquid mixture. 2 A Neon has the lowest mass (20.2) compared to F2 (38.0), HCl (36.5) and SO2 (64.1). Hence, the one with the lowest mass will diffuse out of the balloon the fastest. 3 B The melting point of a pure substance would be lowered when impurities are added. The melting range for BOTH tin and lead (where both are liquids) would thus be from 80 to 220 °C. 4 A The drying agent is a basic oxide that can only dry alkaline or neutral gases. Acidic gases like carbon dioxide and chlorine cannot be dried using calcium oxide. Between ammonia and oxygen, ammonia (NH 3) has a molecular mass of 17 while oxygen (O2) is 32. Hence, ammonia, being less dense than air, can be collected by upward delivery method. 5 D In a school laboratory (i.e. at room temperature), substances X and Y are both liquids. Since they have different solubilities in water, they are considered immiscible liquids that need to be separated using a separating funnel. 6 B amino acid Rf value in solvent X Rf value in solvent Y Identity I 5.5/8=0.69 2/10=0.20 arginine II 3/8=0.38 3/10=0.30 glutamic acid III 4/8=0.50 8/10=0.80 cannot be identified 7 D Ozone makes up the ozone layer that cuts down on the harmful UV rays entering the Earth’s surface. Hence, it should not increase the chance of skin cancer. 8 C Combustion, flue gas desulfurisation and respiration all release carbon dioxide into the atmosphere. 9 D In acid, bromophenol blue is yellow and congo red is violet. Initial colours would be yellow and violet, giving D as the only option. 10 C As more copper(II) sulfate appears as crystals, the concentration of the solution decreases because there is less dissolved copper(II) sulfate in solution.

11 C HnX  nH+ + Xn- Let concentration of acid be 1 mol/dm3 and volume of acid be 2 dm3. no. of mol of acid = 2 mol Let concentration of alkali be 2 mol/dm3 and volume of alkali be 3 dm3. no. of mol of alkali = 6 mol Hence mole ratio of acid:alkali = 2:6 = 1:3 1 mole of acid requires 3 moles of OH - for complete neutralisation. Therefore, acid must be a tribasic acid and n=3. 12 A C6H5NH2 forms OH– when dissolved in water, showing that it is an alkaline. Hence, it should turn litmus paper blue and form a blue ppt with copper(II) sulfate due to the presence of hydroxide ions. It should also react with sulfuric acid to form the salt (C6H5NH3)2SO4. 13 A Presence of Al3+ ions forms white ppt in NaOH that is soluble in excess NaOH. Hence, a colourless solution would be obtained. Presence of SO4 2- ions forms a white ppt in BaCl2 that does not dissolve in excess. 14 C The following options show no observable change. Option A: Copper is unreactive in dilute acids. Option B: Magnesium does not react with carbonates. Option D: Both silver chloride and barium nitrate are salts. One is insoluble and the other is soluble in water. They will not react with each other. 15 B Ammonium chloride produces ammonia gas upon warming with alkalis. Iron(III) chloride gives a red-brown ppt, insoluble in excess in both reagents. Zinc sulfate gives a white ppt that is soluble in excess to give a colourless solution in both reagents. Only copper(II) ions will show a blue ppt, soluble in excess to give a dark blue solution in aqueous ammonia. A blue ppt, insoluble in excess will form in sodium hydroxide. 16 A Green ppt shows presence of Fe2+ ion. Since no aluminium metal was added and ammonia gas is produced, the solution should contain ammonium ions. 17 C Down a group, atomic radius increases. Across a period from left to right, atomic radius decreases. Atomic radius increases from Cl to Br and I. Hence bond strengths should be highest for H-Cl, followed by HBr and Hi. Atomic radius across a period from N to O to F decreases. Hence, bond strength should increase from H-N to H-O to H-F.

18 A/C There are 12 bonding electrons present. This is an ion that was formed by a transfer of electrons. All electrons on sulfur are already involved in covalent bonding. Hence, there are no lone pair electrons around sulfur atom. There is only covalent bonding within this ion. Hence C is also accepted. 19 B A triple bond has 1 σ and 2 π bonds. 20 D N is more electronegative and will pull electrons towards itself. Hence, electron density is said to have shifted towards N. There is a net dipole moment upwards due to the trigonal pyramidal shape. As N is bonded to H, hydrogen bonding would exist. 21 D Having a giant covalent structure causes graphene to have a high melting point. As each carbon atom is bonded to 3 other carbon atoms, leaving one electron delocalised and not involved in bonding, the delocalised electron is able to carry charge and conduct electricity. 22 D Sodium chloride is an ionic compound that does not conduct electricity in solid state but conducts electricity in aqueous or molten state. The sodium chloride crystals dissolve in the water and dissociate to form mobile ions that can act as charge carriers. 23 B 1 mole of NO reacts with oxygen to form 1 mole of NO2, which in turn reacts with water and oxygen to form 1 mole of HNO3 due to the mole ratios. 24 C no. of mol of LiOH = 1.50 x 0.025 = 0.0375 mol no. of mol of XCln = 0.250 x 0.050 = 0.0125 mol Since LiOH : XCln = n : 1, n = 0.0375/0.0125 = 3 25 B Na2CO3 + 2HCl  2NaCl + H2O + CO2 no. of mol of HCl = 0.100 x 0.020 = 0.002 mol By mole ratio, no. of mol of Na2CO3 = 0.001 mol Concentration of Na2CO3 solution = 0.001 / 0.025 = 0.04 mol/dm3 no. of mol in 0.5 dm3 of solution = 0.04 x 0.5 = 0.02 mol mass of Na2CO3 = 0.02mol x [2(23)+12+3(16)] = 2.12g % purity = 2.12/5 x 100% = 42.4% 26 B no. of mol of sulfuric acid used = 294/98 = 3 mol mass of 3 mol of (NH4)2SO4 = 3 x 132.1 = 396 g mass of 1.5 mol of Ca(H2PO4)2 = 1.5 x 234.1 = 351 g

27 C 70% = 0.325g 100% = 0.325/7 x 10 = 0.464g no. of mol of C9H8O4 = 0.464/180 = 0.002579 mol By mole ratio. no. of mol of C7H6O3 = 0.002579 mol mass of C7H6O3 needed = 0.002579 mol x 138 g/mol = 0.356 g 28 A no. of mol of O2 = 58.9 / 24000 = 0.00245 mol no. of mol of NaBrO3 = 0.00245/3 x 2 = 0.00164 mol 29 B Marble chips are in excess, so increasing the mass will not lead to a bigger decrease in contents. Moreover, the initial mass would be higher than curve 1. An increase in temperature will only lead to an increase in rate but not a further decrease in mass of beaker and contents. 30 C Only reactions that produce a gas will be suitable for this method of rate determination. Paper 2 Question Answers Marks A1 (a) Barium sulfate is insoluble in water (and will not be absorbed by the patient.) Accept: “sparingly soluble” Reject: “solid”, “precipitate”, ”toxic”, any other similar descriptions [1] (b)(i) 8cm3 [1] (b)(ii) Conc. of barium hydroxide = 0.35 x 8 / 20 = 0.140 moldm–3 [1] Conc in g/dm3 = 0.140 x 171.3 = 23.982 [1] Percentage purity = 23.982 / 25 = 95.9% [1] Reject: using impure quantity to get no of moles or concentration [3] (b)(iii) Yes (volume used is greater). The concentration of sulfuric acid is less/lower (due to dilution by water spilled) [1] [1] (b)(iv) Barium hydroxide contains m obile (Ba2+(aq) and OH –(aq)) ions/charged particles.[1] [1] (b)(v) Barium hydroxide reacted with sulfuric acid added to form barium sulfate which is insoluble and does not dissociate to give ions [1] Reject: “solid barium sulfate” Hence, electrical conductivity decreases as no./concentration of [3