AES Prelim Ans

1 Assumption English School Science Department – Preliminary Examination 2024 4E Chemistry 6092 Marking Scheme Paper 1: Multiple-Choice Questions [40 m] 1 2 3 4 5 6 7 8 9 10 B C C D C C A D B A 11 12 13 14 15 16 17 18 19 20 D D C B D B D D A B 21 22 23 24 25 26 27 28 29 30 C D D C B C D A B D 31 32 33 34 35 36 37 38 39 40 A B C C A C C D B B Paper 2: Section A: Structured Questions [70 m] 1 (a) Silicon dioxide [1] (b) Lead(II) oxide [1] (c) Zinc bromide [1] (d) Potassium iodide [1] (e) Copper(II) oxide [1] 2 (a) 2 H2O2 → 2 H2O + O2 (state symbols not required) [1] (b) Bonding electrons-[1]; Non-bonding electrons of Oxygen- [1] [2] (c) When it is warmed, particles gain energy and move faster. More particles have the minimum activation energy to react and collide [1], which leads to a higher frequency of effective collisions between the reacting molecules. [1] [2] (d) Oxygen in H2O2 is oxidised as the oxidation state of oxygen increases from -1 in H2O2 to 0 in O2. [1] [2]

2 Oxygen in H2O2 is also reduced as the oxidation state of oxygen decreases from -1 in H2O2 to -2 in H2O. [1] Hence, since oxygen simultaneously undergoes oxidation and reduction, decomposition of hydrogen peroxide is a disproportionation reaction. 3 (a)  Soft, can be cut easily with a knife  Low melting point/ low boiling point  Low density Any 1- [1] [1] (b) As the atomic radius increases increases down the group from 152 pm in Lithium to 244 pm in Rubidium, the valence electrons are further away from the nucleus [1] The weaker attraction between the nucleus and the valence electron makes it easier for the electron to be lost, increasing the metal's reactivity. [1] [2] (c) Upon losing one electron to form a positively charged ion, lithium ion has one less electron shell (2) compared to lithium atom (3), hence it has a smaller radii. [1] (d)(i) The rapid production of hydrogen gas from the reaction creates bubbles, thereby pushing the sodium around the water surface and causing it to skid. [1] (d)(ii) Potassium will react more vigorously with water and catch fire [1] as it is more reactive than sodium, hence causing a faster reaction. [1] [2] 4 (a) Copper conducts electricity through mobile electrons whereas copper(II) sulfate conducts electricity through mobile ions. [1] Copper conducts electricity in solid (and molten) whereas copper(II) sulfate conducts electricity in aqueous and molten states. [1] [2] (b)(i) Cu2+(aq) + 2e → Cu(s) [1] Copper is less reactive than hydrogen/ Cu2+ ions are below H+ ions in the reactivity series, hence Cu2+ ions are selectively discharged over H+ ions to form a coating/ copper atoms on the cathode, hence increasing its mass. [1] [1]- Half equation, [1]- Explanation [2] (b)(ii) 0.92g [1] (c)(i) 4OH-(aq) → O2(g) + 2H2O(l) + 4e- Need to include state symbols [1] (c)(ii) Copper solid is deposited at the cathode in both experiments. [1] (c)(iii) Universal indicator will turn red. [1] The net discharge of OH- ions at the anode causes the electrolyte to be acidic due to more H+ ions than OH- ions being left in solution. [1] [2]

3 5 (a) Isomers are molecules that have the same molecular formula but different structural formula. [1] Compounds E and F are isomers of each other as they have the same molecular formula of C4H10 but the carbons and hydrogens in their compounds are arranged differently. [1] [2] (b)(i) Compound B. [1] Reagent and conditions: Water (vapour), 300 °C, 60atm, phosphoric(V) acid [1] [2] (b)(ii) Insert a damp blue litmus paper into the test tube. [1] Damp blue litmus paper will turn red if D is oxidised. [1] OR Add a drop of universal indicator into the test tube. [1] Uindicator indicator will turn from green to orange/ yellow if D is oxidised. [1] OR Add a reactive metal (e.g. magnesium ribbon)/ metal carbonate (e.g. sodium carbonate) into the test tube. [1] Effervescence/ bubbles of gas produced if D is oxidised. [1] [2] (c)(i) Ethyl propanoate. [1] [1] [2] (c)(ii) Pour the mixture into a separating funnel. [1] The ester can be collected as the top fraction as ester is insoluble in water and forms a separate layer from water/ ester is immiscible with water. [1] [2] 6 (a) C H N O Mass in 100g/ g 63.6 5.9 9.3 21.2 Ar 12 1 14 16 No. of moles/ mol 63.6/12 = 5.3 5.9/1 = 5.9 9.3/14 = 0.664 21.2/16 1.325 Simplest ratio 8 9 1 2 The empirical formula is C8H9NO2. Correct calculation of no. of moles for all 4 elements- [2]; any mistake minus 1 mark Correct simplest ratio- [1] [3]

4 (b)(i) Only one of the two intermediate products (2-nitrophenol and 4-nitrophenol) produced in step 1 is used for the synthesis of paracetamol. [1] (b)(ii) Reducing agent. [1] 4-nitrophenol (C6H5NO3) is reduced as it loses oxygen/ gains hydrogen to form 4-aminophenol (C6H7NO). [1] [Accept: change in oxidation state of N] [2] (b)(iii) Mass of paracetamol in 10 tablets = 10 x 0.5 = 5g [1] No. of moles of paracetamol = 5/ [8(12) + 9 + 14 + 2(16)] = 5/ 151 = 0.033113 mol [1] No. of moles of 4-aminophenol = 0.033113 mol Mass of 4-aminophenol = 0.033113 x [6(12) + 7 + 14 +16] = 0.033113 x 109 = 3.6093g = 3.61g [1] [3] (b)(iv) Verify that melting point is fixed/ check that paracetamol sample produces only one spot using paper chromatography. [1] (c) Paracetamol has a larger molecular size than ethanoic acid so the intermolecular forces of attraction between paracetamol molecules are stronger than that between ethanoic acid molecules. [1] More energy is required to overcome the stronger intermolecular forces of attraction, hence paracetamol has a higher melting point than ethanoic acid and exist as a solid while ethanoic acid exist as a liquid. [1] [2] 7 (a)(i) Photosynthesis: 6 CO2 + 6 H2O → C6H12O6 + 6 O2 Respiration: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Each equation- [1] [2] (a)(ii) Rate of removal of atmospheric carbon dioxide by photosynthesis is balanced by/ equal to rate of carbon dioxide added to the atmosphere by respiration. [1] (b) Excessive combustion of fossil fuels in power generators and vehicles lead to an increase in the amount of atmospheric carbon dioxide. [1] OR Deforestation leads to a reduction in plants and trees which can absorb CO2 from the atmosphere during photosynthesis. [1] An increase in the amount of CO2 enhances the greenhouse effect by increasing the amount of heat trapped and reducing the amount of heat energy escaping into space, hence causing the earth’s surface to warm up and resulting in global warming. [1] [2]

5 (c)(i) 18% [1] (c)(ii) No. B100 released 26% carbon dioxide [1] and this will not all be offset by photosynthesis/ growing plants for biodiesel/ farming also uses machinery which require fossil fuels and release carbon dioxide. [1] [2] 8 (a) The general trend for majority of ionic compounds shown in Fig. 8.1 is that solubility increases as temperature increases. [1] NH3 and Ce2(SO4)3 are exceptions to the trend as they have solubility that decreases as temperature increases. [1] OR NaCl is an exception since its temperature dependence is fairly flat, meaning that the solubility of NaCl changes very little as the temperature changes. [1]- General trend; [1]- Exception [2] (b) Ce2(SO4)3 [1] (c) At 50°C, ratio of solute to solvent is 4:5. Mass of KNO3 crystals in a 200 g saturated solution at 50°C = 4/9 x 200 = 88.9 g (3 s.f.) [1] (d) As the solvent is water, temperatures above 100°C would have caused water to be boiled off. [Reject: evaporation of water] [1] (e) Lattice enthalpy is positive/ endothermic as energy is absorbed to break the solid crystal lattice into its separate constituent ions. [1] (f)(i) Enthalpy of solution of MgSO4= +2874 – 1920 – 995 = – 41 kJ/ mol [1] Enthalpy of solution of SrSO4= +2484 – 1480 – 995 = + 9 kJ/ mol [1] [2] (f)(ii) MgSO4