YJC_H1_CHEM_P1_Worked_Solution

2017 JC2 H1 Chemistry 8872 Preliminary Examinations Paper 1 Worked Solutions Page 1 of 4 Paper 1 Worked Solutions 1 A 2 B 3 D 4 D 5 B 6 B 7 A 8 A 9 B 10 A 11 C 12 B 13 C 14 D 15 D 16 B 17 B 18 C 19 A 20 C 21 D 22 C 23 B 24 C 25 A 26 D 27 B 28 B 29 D 30 A 1 Bearing in mind the mole ratio of the three gases when writing the balanced equation, you should be able to conclude that 1 mole of Ca(NO3)2 requires 3 moles of C: Ca(NO3)2 + 3C  CaO + 2CO2 + CO + Z It should also be clear that all the number of O atoms on both sides of the equation are already equal (i.e. balanced), and thus Z have to be N 2 so that the number of N atoms (the only remaining element to be balanced) on both sides are equal: Ca(NO3)2 + 3C  CaO + 2CO2 + CO + N2 Answer: A 2 A: 91.1(28) 7.9(29) 1.0(30) 28.099100 rA  B: 92.2(28) 4.7(29) 3.1(30) 28.109100 rA  C: 95.0(28) 3.2(29) 1.8(30) 28.068100 rA  D: 96.3(28) 0.3(29) 3.4(30) 28.071100 rA  Answer: B 3 To get the balanced equation: H2S(aq)  S(s) + 2H+(aq) + 2e− (2) SO2(aq) + 4H+ + 4e−  S(s) + 2H2O(l) 2H2S(aq) + SO2(aq)  3S(s)+ 2H2O(l) Dividing throughout by 3, we get: 2 3 H2S(aq) + 1 3 SO2(aq)  S(s)+ 2 3 H2O(l) Answer: D 4 For 10B: number of protons = 5; number of neutrons = 10 − 5 = 5  ratio of proton : neutron = 1 : 1 A: 40Ar: number of protons = 18; number of neutrons = 40 − 18 = 22  ratio of proton : neutron = 9 : 11 B: 40K: number of protons = 19; number of neutrons = 40 − 19 = 21  ratio of proton : neutron = 19 : 21 C: 32P: number of protons = 15; number of neutrons = 32 − 15 = 17  ratio of proton : neutron = 15 : 17 D: 32S: number of protons = 16; number of neutrons = 32 − 16 = 16  ratio of proton : neutron = 1 : 1 Answer: D 5 Around the S-atom, there are 3 bond pairs and 1 lone pair of electrons (a double bond is considered one bond pair). By VSEPR theory, the four electron pairs will space themselves as far apart as possible to minimise repulsion, leading to an electronic geometry of tetrahedral, and a bond angle of 109°. However, as the lone pair-bond pair repulsion are stronger than bond pair-bond pair repulsion, the bond angle will be smaller than 109°,  approximately 107°. Answer: B 6 In order for the anion to have a square pyramidal shape, it must have 5 bond pairs and 1 lone pair of electrons. Since Sb is from group 15, it has 5 valence electrons, which are used to form normal single bond with the five F- atoms (i.e. 5 bond pairs). Hence Sb must receive two electrons from external sources, in order to have one lone pairs of electrons.  n = 2, i.e. the anion is SnF 52−. Answer: B 7 Since both butane and methane are made up of non-polar molecules, only id-id interactions exist b