2007 HCI Prelims Paper 1 FAQs

For Q4, see previous post by Mr WTY.

Explanation for Q12
O2 (i.e. gas K) is formed at anode.
4OH- --> O2 + 4H+ + 4e-
For 1 mol of O2 formed, 4 mol of e- are released.
These e- must be used to discharge the cations, Jn+.
1 mol of O2 is formed when 2 mol of Jn+ are discharged.
2Jn+ + 4e- --> 2J
Thus, n = 2.

Gas L is Cl2.
2Cl- --> Cl2 + 2e-
Note that the same quantity of electricity is used, i.e. 4 mol of e- must be involved.
Hence, the equation becomes 4Cl- --> 2Cl2 + 4e-.

Thus, ratio of J:K:L = ratio of O2:J:Cl2 = 2:1:2.

Explanation for Q14
[Br-] = [Cl-] = [I-] = [SO42-] = 1 x 10^-3 mol dm-3
Use the concentrations and the Ksp of the lead(II) compounds to find out the concentrations of lead(II) ions required for precipitation.

Explanation for Q15
[Ba(OH)2] = [HCl] = 1 mol/dm3
1 mol of Ba(OH)2 reacts with 2 mol of HCl.
20 cm3 of Ba(OH)2 is used. Thus, 40 cm3 of HCl is needed to reach equivalence point. No such titration curve is given.

Explanation for Q21
The "OH" group at the bottom can only give the cis-alkene after dehydration.

The "OH" on top lead to 2 possibilities. If the double bond forms at the carbon of the cyclohexyl ring, there are 2 isomers, i.e. cis and trans. If the double bond forms between carbon-2 and carbon-3 of the side chain (carbon-1 is bonded to Cl), then there is a chiral centre (carbon-4, counting from carbon-1 bonded to Cl). Hence, there are 2 optical isomers and each optical isomer can have cis or trans configuration at the double bond. Total of 4 isomers.

Hence, grand total = 2 + 4 = 6 isomers.

A Note for Q37
The concentrations of SO2, O2 and SO3 at equilibrium are 0.08, 0.04 and 0.02 (in mol/dm3) respectively. Kc = [SO3]2/{[SO2]2 x [O2]}.