Copy of VS 2024 Sec 4 Chem Prelim Paper 1 and 2 Answers

1 Victoria School 2024 Secondary 4 Chemistry Prelim Answer Scheme Paper 1 1 2 3 4 5 6 7 8 9 10 C D B B B A C A C D 11 12 13 14 15 16 17 18 19 20 D B A B C C C D D D 21 22 23 24 25 26 27 28 29 30 B A D C D B C C D B 31 32 33 34 35 36 37 38 39 40 A A D D A C B C D C Paper 2 Section A Qn Suggested answers Mark A1 2 Across: fluoride 1 3 Down: addition 1 4 Down: alloy 1 5 Down: sublimation 1 6 Down: fixed 1 A2a Sodium: The indicator would turn from green to purple REJECT: Blue Carbon: The indicator would turn from green to orange REJECT: Red 1 A2b [1m] – cation [1m] – anion 2 A2c 4Na + O2  2Na2O No of moles of sodium= 2.30 / 23 = 0.100 mol 4mol Na ≡ 2mol Na2O 0.100mol Na ≡ 0.05 mol Na2O Theoretical yield of Na2O = 0.0500 mol x 62 = 3.10 g 4Na + O2  2Na2O 1

2 Qn Suggested answers Mark Mass gained = 94.82 – 94.50 = 0.32g No of moles of oxygen = 0.32 / 16+16 = 0.01 mol 1mol O2 ≡ 2mol Na2O 0.0100mol O2 ≡ 0.0200 mol Na2O Actual Yield = 0.0200 x 62 = 1.24 g Percentage Yield = 1.24 / 3.10 x 100% = 40.0 % 1 1 A3ai Bond breaking absorbs energy, hence is endothermic. Bond forming releases energy, hence is exothermic. Since more energy is released than absorbed, the reaction is exothermic. OR The energy absorbed when breaking the bonds in 1 mole of carbon monoxide and 1 mole of chlorine is less than the energy released when forming the bonds in 1 mole of phosgene, hence making it an exothermic reaction. 1 1 1 1 A3aii The oxidation state of chlorine decreased from 0 in Cl2 to -1 in COCl2. This is reduction. The oxidation state of carbon increased from +2 in CO to +4 in COCl2. This is oxidation. Since oxidation and reduction occurs at the same time, this is a redox reaction. (Must state but no marks) 1 1 A3b At –128 oC, molecules are packed tightly and very close together in a regular/orderly manner and they can only vibrate about their fixed positions. (As temperature decreases, the molecules move faster and faster) and begin to settle in fixed positions. At –108 oC, the phosgene molecules are packed closely together in a disorderly manner and they are able to slide over each other. 1 1 A4a • Catalysts can be regenerated/are not consumed at the end of a reaction. • Thus they can be (reused), so they only need to be purchased once • small amount need to be used. • 1 1 A4b In pellet form, there is a higher surface area (per unit volume). Higher frequency of effective collisions between reacting particles (and the catalyst), leading to a faster reaction rate. 1 1

3 Qn Suggested answers Mark A4ci 1 A4di Methyl ethanoate 1 A4dii [1m] – balanced equation with irreversible arrow [1m] – full structural formula 2 A5a 1 A5bi 1 A5bii Reaction in (a)(ii) requires a carboxylic acid and an amine. and (b)(i) requires C=C OR Reaction in (a)(ii) results in a loss of small molecules, reaction in (b)(i) has no loss of small molecules OR Water is a byproduct of the reaction in (a)(ii) while there are no byproducts in (b)(i) OR Breaking of C-C double bond in (a)(ii) while no breaking of double bonds in (b)(i) 1 A5c Any 4 carbon branched chain with 2 -NH2 groups OR 1

4 Qn Suggested answers Mark Any 4 carbon straight chain with 2 -NH2 groups that is not on the terminal carbons. A6a C H N Mass in 100 g/g 74.1 8.60 17.3 No of moles/mol 74.1 / 12 = 6.175 8.60 / 1 = 8.60 17.3 / 14 = 1.236 Mole Ratio 6.175 / 1.236 ≈ 5.00 8.60 / 1.236 ≈ 7 1.236 / 1.236 = 1 Empirical Formula C5H7N Molecular Formula = nC5H7N n = Relative molecular mass / relative formula mass = 162 / 81 = 2 Molecular Formula = C10H14N2 1 1 1 A6b Concentration = 0.123 mol/dm³ Volume = 2 cm3 = 2/1000 dm³ = 0.002 dm³ Number of moles = Concentration x Volume = 0.123 mol/dm³ x 0.002 dm³ = 0.000246 mol Molar mass of nicotine = 162 g/mol Mass = number of moles x molar mass = 0.000246 mol x 162 g/mol = 0.03985 g = 39.9 mg (to 3sf) (convert g to mg) 1 1 A6c When a cigarette is smoked, carbon monoxide is inhaled by the smoker and the people around them. Carbon monoxide bonds readily and irreversibly with haemoglobin to form carboxyhaemoglobin/stable compound. This reduces the amount of haemoglobin available to transport oxygen/results in less O2 transported around the body, leading to fatigue/dizziness, and eventually death. 1 1 A6(d)(i) step number description 2 Soak the ground tobacco in the ethanol solvent. 4 Separate the nicotine-rich solution from the solid plant material using process A. 1

5 Qn Suggested answers Mark 5 Obtain nicotine from the mixture of ethanol and nicotine using process B. 3 Stir the mixture to enhance the extraction of nicotine into the solvent. 1 Grind the dried tobacco leaves into a fine powder. A6(d)(ii) Process A: filtration Process B: Fractional Distillation 1 A7a T, F, T, T Any one wrong minus 1 A7bi Immiscible in water and less dense than water, preventing entry of air/oxygen for the respiration by larvae. 1 1 A7bii Kerosene has a higher Mr, hence there are Thus more energy is needed to overcome the stronger intermolecular forces of attraction. It is less volatile, so it does not evaporate away easily compared to petrol 1 1 A8ai To prevent the hot reactive sodium metal and hot chlorine gas from reacting and reforming back sodium chloride. 1 A8aii No. Hydrogen gas will be produced instead of sodium metal. The electrolyte produced will be sodium hydroxide instead of sodium chloride. 1 1 A8b Cu → Cu2+ + 2e- Copper loses electrons more readily than chloride ions, hence it will be preferentially discharged, forming copper(II) ions instead of chlorine gas 1 1 A8c No. The litmus paper must be bleached in order to confirm the presence of chlorine. 1 A8di colourless solution turns brown 1 A8dii chlorine is more reactive than iodine and it can displace iodine from potassium iodide 1 A9a Metals X and Y are more reactive than metal W. Metal W is more reactive than metal Z. 1 1 A9b 1. Weigh a piece of metal X. 2. Half fill a test tube with water and stopper with a delivery tube. 3. Place metal X in another test tube. Connect this test-tube to the one with water. 1

6 Qn Suggested answers Mark 4. Heat the test tube of water until steam is formed and allow the steam to pass over the heated metal X. 5. Stop heating when there is colour change observed/after 5 minutes. 6. Weigh the resulting solid after heating [method] when it has cooled down. 7. Repeat Steps 1 to 6 with metal Y. 8. Compare the change in mass for both metals. 9. The more reactive metal will have a higher change in mass after 5 mins. method: measure volume of gas after 10 minutes/ observe colour change of the solid before and after heating 1 1 1 A10a Across period 2, the melting points of the fluoride compounds decreases. 1 A10bi GeF4 1 A10bii No. The number of F atoms that bond to the halogens should be 1, however, for Cl, Br and I, they can form more than one compound, which have varying number of F atoms that are bonded to them. Or give any counter example: i.e. The number of F atoms that bond to the halogens should be 1, but C lF 3 has 3 F atoms bonded to Cl. 1 1 A10biii The number of compounds formed between a halogen and fluorine is equal to the period number of the halogen minus 1. Or The number of compounds formed between a halogen and fluorine is always an odd number