C H E M I S T R Y N A T I O N A L Q U A L I F Y I N G E X A M I N A T I O N SOLUTIONS GUIDE
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1 C H E M I S T R Y A T I A L Q U A L I F Y I G E X A M I A T I SLUTIS GUIDE Answers are a guide only and do not represent a preferred method of solving problems.
2 Section A 1B, 2A, 3C, 4C, 5D, 6D, 7A, 8B, 9E, 10B, 11A, 12D, 13E, 14C, 15E Section B Q16 (a) 2PbS(s) 3 2 (g) 2Pb(s) 2S 2 (g) (b) It is important to limit the supply of air to avoid further oxidation of Pb to Pb 2 and S 2 to S 3. (c) The elemental lead must also be oxidised under these conditions, and so will react to form Pb as well: (d) PV S2 2Pb(s) 2 (g) 2Pb(s) n S2 RT n S2 PV S 2 RT but n PbS n S2 so n PbS PV S 2 RT 101.3kPa.V S JK 1 mol 1.298K molL 1 V S2 (epb(s) PbS(s) 3Pb(s) S 2 (g) (f) (g) Pb 2 oxidises S(-II) to S(IV) and so is the oxidising agent in this reaction From the equation given in part (d) we have n PbS molL 1 V S2 so n PbS molL L mol but this isonly for twothirds of the sample sototal n PbS mol (h) mol Purity m PbS m galena (i) mol gmol g 0.929g/ g of galena or 93.0% m Pb n PbS FW Pb mol 223.2gmol g 2
3 (j) so%elemental Lead n Pb PbS m Pb FW Pb g 207.2gmol mol n Pb n Pb PbS mol mol mol 2.9% (k) In the solution we have 4.18x10-3 mol of Pb 2 to 1.67x10-2 mol of H, which is in the ratio 1:4. Hence we would suspect [Pb(H) 4 ] 2- to be present in solution. ote that the oxidation state of Pb is unchanged in this species. Q17 (a) 9 Ser-Gly-Cys-Lys-Ile-Ile-Ser-Ala-Ser-Thr-Cys-Pro-Ser-Tyr-Pro-Asp-Lys (b Lys-Ser-Cys-Cys-Pro-Asn-Thr-Thr-Gly-Arg (c) 6 Asn-Thr-Cys-Arg-Phe 7 Gly-Gly-Gly-Ser-Arg-Glu-Val-Cys-Ala-Ser-Leu (d) Peptides from trypsin cleavage: 1 Asn-Ile-Tyr-Asn-Thr-Cys-Arg 2 Lys-Ser-Cys-Cys-Pro-Asn-Thr-Thr-Gly-Arg 3 Ile-Ile-Ser-Ala-Ser-Thr-Cys-Pro-Ser-Tyr-Pro-Asp-Lys 4 Phe-Gly-Gly-Gly-Ser-Arg 5 Ser-Cys-Cys-Pro-Asn-Thr-Thr-Gly-Arg Peptides from chymotrypsin cleavage: 6 Asn-Thr-Cys-Arg-Phe 7 Gly-Gly-Gly-Ser-Arg-Glu-Val-Cys-Ala-Ser-Leu 8 Lys-Ser-Cys-Cys-Pro-Asn-Thr-Thr-Gly-Arg-Asn-Ile-Tyr 9 Ser-Gly-Cys-Lys-Ile-Ile-Ser-Ala-Ser-Thr-Cys-Pro-Ser-Tyr-Pro-Asp-Lys (e) 7 Gly-Gly-Gly-Ser-Arg-Glu-Val-Cys-Ala-Ser-Leu Leu is a C-terminus resulting from a chymotrypsin cleavage and this is unexpected as chymotrypsin normally cleaves an amide bond to leave a C-terminus of either Phe, Tyr or Trp. 3
4 (f) Lys Ser Cys Cys Pro Ans Thr Thr Gly Arg Asn Ile Tyr Asn Thr Cys Arg Phe Gly Gly Gly Ser Arg Glu Val Cys Ala Ser Leu Ser Gly Cys Lys Ile Ile Ser Ala Ser Thr Cys Pro Ser Tyr Pro Asp Lys Q18 (a) (i geometrical isomers (ii geometrical isomers H 3 H 3 Br H 3 Br H 3 H 3 H 3 h 3 H 3 H 3 H 3 h 3 H 3 (iii) 3 geometrical isomers (iv geometrical isomers H 3 CS H 3 Br H 3 H 3 SC H 3 SC H 3 SC H 3 H 3 H 3 H 3 C H 3 C Br Br H 3 CS CS (b) (i) cis-[( 2 )(H 3 ) 4 ] has the structure 2 H 3 1 geometrical isomer 1 linkage isomer H 3 h 3 H 3 H 3 H 3 H 3 H 3 h 3 2 H 3 h 3 H 3 4
5 (ii) trans-[(sc (H 3 ]. Has the structure H 3 SC CS H 3 1 geometric isomer 2 linkage isomers H 3 SC H 3 SC H 3 CS H 3 SC SC H 3 SC H 3 (c) (i) square planar [(SC (dmen)] has the structure SC SC As there is only one structure possible there is therefore no geometrical isomer. 3 linkage isomers CS SC SC CS CS CS (ii) cis-[( (H 3 (en)].has the structure H 3 H 3 2 geometrical isomers H 3 H 3 H 3 H 3 3 linkage isomers 2 H 3 2 H 3 H 3 2 H 3 2 H 3 H 3 H 3 H 3 5
6 If you have been alert and careful you will have seen that there is another structure for the cis isomer and each of the three linkage isomers, these are shown and explained in (d). (d) The complex ion cis-[( (H 3 (en)] (c) (ii) and its three linkage isomers all have nonsuperimposable mirror images and hence exhibit optical isomerism. All the rest have superimposable mirror images and hence do not exhibit optical isomerism. The alternative structures alluded to in (c) (ii) are cis-[( (H 3 (en)] 3 linkage isomers H 3 H H 3 H 3 2 H 3 H 3 2 H 3 H 3 H 3 H 3 Rotating these four structures 180 around the vertical axis produces the structures drawn below and we can see that they are the mirror images of the cis isomer and its three linkage isomers in (c) (ii) H 3 H 3 H 3 H 3 H 3 H 3 H 3 H 3 H 3 H 3 (e) (i) There are two stereoisomers. mirror n n Al Al (where is C and n 3) (ii) There are three stereoisomers. mirror (where is en 6
7 (iii) There are two stereoisomers. ' ' 2 ' ' 2 (where ' is dmen (iv) There are two stereoisomers. ' ' Zn n mirror n ' Zn ' (where ' is dmen and n is 2 Q19 (a) (b) Person A has a dormant tumor in the left eye. Person B has an active tumor in the right eye. average count unt rate for around 3 hours normal eye Activity after 3 hours 30 photons / Bq Proportion of original injection in eye Bq % The β particle is an electron emitted from the nucleus when a neutron transforms to a proton. The 32 P becomes 32 S 1 0 n 1 p 0 β P S β (c) t 12 ln 2 k so k ln 2 t If A is final activity, A 0 is initial activity and t is time then A A 0 e kt e ( ) Bq 7
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