SAJC Prelim P1_Worked_Solution

1 2017 H2 Chemistry Paper 1 Worked Solution 1. C 2. A 3. B 4. C 5. C 6. A 7. A 8. B 9. C 10. A 11. B 12. A 13. D 14. D 15. C 16. C 17. D 18. A 19. D 20. B 21. D 22. A 23. C 24. B 25. C 26. B 27. B 28. A 29. C 30. D 1 Amt of KMnO4 used for first experiment = 0.1 x 15/1000 = 1.5 x 10–3 mol 5Fe2+ + MnO4– + 8H+  5Fe3+ + Mn2+ + 4H2O Amt of Fe2+ present = 1.5 x 10–3 x 5 = 7.5 x 10–3 mol [Fe2+] = 7.5 x 10–3 /(25/1000) = 0.3 mol dm–3 Amt of KMnO 4 used for second experiment =0.1 x 34.2/1000 = 3.42 x 10 –3 mol Amt of Fe2+ present = 0.0171 mol Amt of Fe3+ originally in 25 cm3 = 0.0171 – (7.5 x 10–3) = 9.6 x 10–3 mol [Fe3+] = 9.6 x 10–3 / (25/1000) = 0.384 mol dm–3 Ans: C 2 9 x 4 = 36 nuclides were lost, hence the mass number of element A is now 241 – 36 = 205. 9 x 2 = 18 protons were lost, hence the atomic number of element A before accounting for the electrons is 94 – 76. However, the implication of e–1 0 is that each electron lost increases the atomic number by 1, so 5 electrons means that the atomic number increases by 5. Hence 76 + 5 = 81. Ans: A 3 A is not correct. From NH3 to H2S, the shape changes from trigonal pyramidal to bent, hence the bond angle actually decreases. B is correct. PH 3 has a smaller bond angle than NH 3 as P is less electronegative than N, hence the bond pairs are further away from the central atom and experience less bond–pair–bond pair repulsion, hence leading to a

2 smaller bond angle in PH 3. AlCl3 has a larger bond angle than NH 3 as it is trigonal planar with a bond angle of 120 º. C is not correct. From PH3 to PF3, H is less electronegative than F, hence the bonding electron pairs are closer to the central atom in PH3, hence experience greater bond–pair–bond pair repulsion, hence leading to a greater bond angle in PH3. From PH3 to PF3, there is a decrease in bond angle. D is not correct. XeF 4 is square planar with bond angle of 90 º. SCl6 is also square planar with bond angle of 90 º. Hence there is no increase in bond angle. Ans: B 4 Glycine forms zwitterions and has an ionic lattice structure with strong electrostatic forces of attraction between the zwitterions, hence has a higher melting point than 2–hydroxyethanoic acid. Ans: C 5 pV = nRT. For fixed mass of gas at constant T, pV α k. Hence graph should be horizontal line with gradient k. At lower T, pV should be be lower than original. Ans: C 6 Ans: A -110 -393 + 1 2O2 (g) CO(g) C(s) + O2 (g) CO2(g) Energy

3 7 ∆G = ∆H – T∆S ∆H will be positive, because interactions in the protein are broken going to the unfolded state. ∆S will also be positive, because the unfolded state is more disordered with coil being more random. ∆G is negative as denaturation is spontaneous when egg is cooked. Ans: A 8 At a higher temperature, the graph is skewed towards the right. Area under the graph represents the number of molecules with energy greater than or equal to EA. Ans: B