Cambridge A Level Chemistry 9701 — 2010 May/June Paper 4 · Variant 3

This document consists of 17 printed pages and 3 blank pages. DC (SM/CGW) 27852 © UCLES 2010 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer all questions. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. * 9 1 2 5 8 5 5 4 3 1 * CHEMISTRY 9701/43 Paper 4 Structured Questions May/June 2010 1 hour 45 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet For Examiner’s Use 1 2 3 4 5 6 7 8 Total

2 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use Section A Answer all questions in the spaces provided. 1 Phenacyl chloride has been used as a component of some tear gases. Its lachrymatory and irritant properties are due to it reacting with water inside body tissues to produce hydrochloric acid. It undergoes a nucleophilic substitution reaction with NaOH(aq). O + OH– phenacyl chloride + Cl (a) Write the formulae of the products of this reaction in the two boxes above. [2] When the rate of this reaction was measured at various concentrations of the two reagents, the following results were obtained. experiment number [phenacyl chloride] [NaOH] relative rate 1 0.020 0.10 1.0 2 0.030 0.10 1.5 3 0.025 0.20 2.5 (b) (i) What is meant by the term order of reaction? … (ii) Use the above data to deduce the order with respect to each reactant. Explain your reasoning. … … … … … (iii) Write the overall rate equation for the reaction. …

3 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (iv) Describe the mechanism for this reaction that is consistent with your overall rate equation. You should show all intermediates and/or transition states and partial charges, and you should represent the movements of electron pairs by curly arrows. [7] (c) (i) Describe an experiment that would show that CH3COCl reacts with water at a much faster rate than phenacyl chloride. Include the reagents you would use, and the observations you would make with each chloride. … … … … … … … (ii) Suggest an explanation for this difference in reactivity. … … [4] [Total: 13]

4 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 2 (a) Describe and explain how the solubilities of the sulfates of the Group II elements vary down the group. … … … … … …[3] (b) The following table lists some enthalpy changes for magnesium and strontium compounds. enthalpy change value for magnesium / kJ mol–1 value for strontium / kJ mol–1 lattice enthalpy of M (OH)2 –2993 –2467 enthalpy change of hydration of M 2+(g) –1890 –1414 enthalpy change of hydration of OH–(g) –550 –550 (i) Use the above data to calculate values of ΔH o solution for Mg(OH)2 and for Sr(OH)2. Mg(OH)2 … … ΔH o solution = … kJ mol–1 Sr(OH)2 … … ΔH o solution = … kJ mol–1 (ii) Use your results in (i) to suggest whether Sr(OH)2 is more or less soluble in water than is Mg(OH)2. State any assumptions you make. … … (iii) Suggest whether Sr(OH)2 would be more or less soluble in hot water than in cold. Explain your reasoning. … … [5]

5 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (c) Calcium hydroxide, Ca(OH)2, is slightly soluble in water. (i) Write an expression for Ksp for calcium hydroxide, and state its units. Ksp = units … (ii) 25.0 cm3 of a saturated solution of Ca(OH)2 required 21.0 cm3 of 0.0500 mol dm–3 HCl for complete neutralisation. Calculate the [OH–(aq)] and the [Ca2+(aq)] in the saturated solution, and hence calculate a value for Ksp. [OH–(aq)] = … [Ca2+(aq)] = … Ksp = … (iii) How would the solubility of Ca(OH)2 in 0.1 mol dm–3 NaOH compare with that in water? Explain your answer. … … [6] [Total: 14]

6 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 3 (a) Fluorine is much more electronegative than both silicon and sulfur, but whereas the molecule of SF4 has an overall dipole, that of SiF4 has none. Suggest a reason for this difference. … …[1] (b) Predict whether or not the following molecules will have an overall dipole. Place a tick in the appropriate column. compound molecule has an overall dipole molecule does not have an overall dipole BCl 3 PCl 3 CCl 4 SF6 [2] (c) Boron and silicon are two elements adjacent to carbon in the periodic table. CCl4 does not react with water, whereas BCl 3 and SiCl4 do react. (i) Suggest a reason for this difference in reactivity. … … (ii) Construct equations showing the reaction of these two chlorides with an excess of water. BCl 3 … SiCl4 … [3]

7 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (d) When reacted with a small quantity of water, SiCl 4 produces an oxychloride X, SixCl yOz. The mass spectrum of X shows peaks at mass numbers of 133, 149, 247, 263 and 396. (You should assume that the species responsible for all these peaks contain the 16O, the 35Cl and the 28Si isotopes only.) (i) Use these data to deduce the molecular formula of X. molecular formula … (ii) Suggest the structures of the fragments responsible for the peaks at the following mass numbers. mass number structure 133 247 263 (iii) Hence suggest the displayed formula of X. [5] [Total: 11]

8 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 4 (a) Complete the electronic structures of the Cr3+ and Mn2+ ions. Cr3+ 1s22s22p6 … Mn2+ 1s22s22p6 … [2] (b) (i) Describe what observations you would make when dilute KMnO4(aq) is added slowly and with shaking to an acidified solution of FeSO4(aq) until the KMnO4 is in a large excess. … … … … … … (ii) Construct an ionic equation for the reaction that occurs. … [4] (c) By selecting relevant E o data from the Data Booklet explain why acidified solutions of Fe2+(aq) are relatively stable to oxidation by air, whereas a freshly prepared precipitate of Fe(OH)2 is readily oxidised to Fe(OH)3 under alkaline conditions. relevant E o values and half equations … … … … explanation … … [4]

9 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (d) Predict the organic products of the following reactions and draw their structures in the boxes below. You may use structural or skeletal formulae as you wish. hot conc. MnO4 – + H+ hot conc. MnO4 – + H+ hot Cr2O7 2– + H+ OH OH [4] (e) KMnO4 and K2Cr2O7 are the reagents that can be used to carry out the following transformation. E I II CHO OH (i) Draw the structure of intermediate E in the box above. (ii) Suggest reagents and conditions for the following. reaction I … reaction II … [3] [Total: 17]

10 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 5 (a) (i) Briefly explain why the benzene molecule is planar. … … … (ii) Briefly explain why all the carbon-carbon bonds in benzene are the same length. … … … [2] (b) Benzene can be nitrated by warming it with a mixture of concentrated sulfuric and nitric acids. (i) By means of an equation, illustrate the initial role of the sulfuric acid in this reaction. … (ii) Name the type of reaction and describe the mechanism for the nitration reaction, including curly arrows showing the movement of electrons and all charges. type of reaction … mechanism [4] (c) State the reagents and conditions needed to convert benzene into chlorobenzene. …[1]

11 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (d) Nitrobenzene undergoes further substitution considerably more slowly than chlorobenzene. In nitrobenzene the incoming group joins to the benzene ring in the 3-position, whereas in chlorobenzene the incoming group joins to the benzene ring in the 4-position. (i) Use these ideas to suggest the structures of the intermediate compounds Y and Z in the following synthesis of 4-chlorophenylamine. I II Z Y III NH2 Cl (ii) Suggest the reagents and conditions needed for reaction III in the above synthesis. … … (iii) Suggest the structural formulae of the products A, B, C and D of the following reactions. If no reaction occurs write “no reaction” in the relevant box. B A D C NH2 Br2(aq) OH –(aq) NaNO2 + HCl CH3COCl Cl [8] [Total: 15]

12 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use Section B Answer all questions in the spaces provided. 6 Human hair and silk both consist of proteins. Proteins are described as having three major levels of structure: primary, secondary and tertiary. (a) Outline what is meant by the terms primary structure and tertiary structure of a protein. primary structure … … … tertiary structure … … … [2] (b) In hair, the secondary structure consists of α-helices which are cross-linked by disulfide bonds. The amino acid responsible for this cross-linking is cysteine, H2NCH(CH2SH)CO2H. (i) Show by means of a diagram how the disulfide cross-links are formed. (ii) What type of reaction is this? …

13 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (iii) State three other interactions that stabilise the tertiary structure of proteins. … … … [4] (c) The β-pleated sheet is a different form of secondary structure found in proteins, such as those in silk. (i) What type of bonding is responsible for stabilising the β-pleated sheet in silk? … (ii) On the diagram below, draw a second polypeptide strand and show how bonds would be formed that stabilise this β-pleated sheet. N H CH CH CH CH R O R H O R C N H O R H O C N C N C [3] (d) The cysteine-containing protein in hair is called α-keratin. A similar sequence of amino acids can produce β-keratin proteins found in the scales, claws and shells of reptiles such as tortoises. In β-keratin the secondary structure of the protein is in the form of a β-pleated sheet. Suggest what makes the β-pleated sheet in β-keratin so much less flexible than the β-pleated sheet in silk. … … …[1] [Total: 10]

14 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 7 A mixture of amino acids may be separated using electrophoresis. A typical practical set-up is shown in the diagram. d.c. power supply + – glass slides filter paper soaked in buffer solution amino acid mixture placed here electrolyte (a) When the power supply is switched on, some amino acids may not move, but remain stationary. Suggest an explanation for this observation. … … …[2] (b) The amino acid glycine has the formula H2NCH2CO2H. Identify the species formed on the filter paper if glycine moves to the left (positive) end of the filter paper. … [1] (c) The following result was obtained from another electrophoresis. What can be deduced about the relative sizes of, and charges on, the amino acid species A, B and C? A + – B mixture placed here C amino acid relative size charge A B C [3]

15 © UCLES 2010 [Turn over 9701/43/M/J/10 For Examiner’s Use (d) The sequence of amino acids in a polypeptide may be determined by partial hydrolysis of the chain into smaller pieces, often tripeptides. (i) Following such a partial hydrolysis, the following tripeptides were obtained from a given polypeptide. ala-gly-asp gly-ala-gly lys-val-ser ser-ala-gly val-ser-ala Given that the N-terminal amino acid is lysine (lys) suggest the amino acid sequence of the shortest polypeptide that would give the above tripeptides. … The structural formulae of the amino acids in the polypeptide are given below. abbreviation amino acid structural formula ala alanine H2NCH(CH3)CO2H asp aspartic acid H2NCH(CH2CO2H)CO2H gly glycine H2NCH2CO2H lys lysine H2NCH(CH2CH2CH2CH2NH2)CO2H ser serine H2NCH(CH2OH)CO2H val valine H2NCH(CH(CH3)2)CO2H (ii) Which of the tripeptides in (i) has the lowest Mr? … (iii) Select one amino acid from those listed in the table which contains an ionic side-chain at pH 8. … [4] [Total: 10]

16 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use 8 The design and development of batteries has been a major research area in recent years. (a) Lead-acid batteries, used in cars, are made up of a number of rechargeable cells in series, and were first developed in 1860. They have the disadvantage of a relatively high mass compared to the energy stored. During discharge, the electrode reactions in the cells of these batteries are as follows. I Pb + SO4 2– J PbSO4 + 2e– II PbO2 + 4H+ + SO4 2– + 2e– J PbSO4 + 2H2O State which of these reactions occurs at the positive electrode in a lead-acid cell during discharge, explaining your answer. … …[1] (b) Use the Data Booklet and the equations I and II above to calculate the voltage produced by a lead-acid cell under standard conditions. [2] (c) Nickel-metal hydride batteries were developed in the 1980s and have become increasingly common particularly for small devices such as mobile phones and digital cameras that need near-constant sources of electrical energy. These cells use nickel oxohydroxide (NiO(OH)) as one electrode and a hydrogen-absorbing alloy such as LiNi5 as the other electrode. One reaction that takes place in these batteries is NiO(OH) + H2O + e– Ni(OH)2 + OH– (i) State the oxidation state of nickel in NiO(OH). … (ii) Suggest a likely advantage of these batteries compared with lead-acid batteries. … … [2]

17 © UCLES 2010 9701/43/M/J/10 For Examiner’s Use (d) Hydrogen fuel cells have been suggested as the next major advance in electrically powered vehicles. In these fuel cells hydrogen is oxidized to produce water, using a catalyst and inert electrodes. (i) Suggest a material for the electrodes. … (ii) Use your knowledge of hydrogen to suggest a disadvantage of these fuel cells in powering vehicles. … … [2] (e) Many of the world’s countries are developing ways of recycling materials which are valuable or which require large amounts of energy to produce. For each of the following recyclable materials, state whether recycling of this material is important in saving energy or in saving resources. Use your knowledge of chemistry to explain each choice. glass … … … steel … … … plastics … … … [3] [Total: 10]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9701 CHEMISTRY 9701/43 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 1 (a) C6H5-COCH2OH or C8H8O2 and NaCl or Cl – (1) + (1) [2] (b) (i) the exponent / power to which a concentration is raised in the rate equation (or in an equation, e.g. “a” in the equ: rate = k[A]a) (1) (ii) from 1 and 2: rate increases by 50% as does [RCl ], so rate ∝ [RCl ]1 (1) from 1 and 3: rate ∝ [NaOH]1 (1) (iii) (rate =) k[RCl ][OH– ] (1) (iv) R Cl C (H) (H) (+) (-) HO R C (H) (H) Cl HO HO R C (H)(H) + Cl (can be a solid line) marking points: • (+) or δ+ on C and (–) or δ– on Cl (1) • lone pair and charge on: OH– (1) • curly arrow from OH (lone pair) to (δ+)C, and either a curly arrow breaking C-Cl bond or 5-valent transition state (ignore charge) (1) • SN1 alternative for last mark (only award mark if candidate’s rate equation shows first order reaction): curly arrow breaking C-Cl bond and carbocation intermediate. [7] (c) (i) (add RCl / RCOCl to) (aq) Ag+ / AgNO3 or named indicator (e.g. MeOr) or use pH probe (1) White ppt appears (faster with RCOCl) or turns acidic colour (e.g. red) or shows pH decrease (1) if water is the only reagent, and no pH meter used: award only the second mark, for “steamy / white fumes” (ii) (C=O is polarised /) carbon is more δ+ than in R-Cl or carbon is positive or RCOCl can react via addition-elimination (mention of electronegativity on its own is not enough for the mark) (1) [3] [Total: 12]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 2 (a) less soluble down group (1) lattice energy and hydration energies both decrease (i.e. become less negative) (1) but H.E. decreases more (than L.E.) or change in H.E. outweighs L.E. (1) so ∆Hsol becomes more endothermic / less exothermic (1) [4] (b) (i) for Mg: ∆H = 2993 – 1890 – (2 × 550) = (+)3 (kJ mol–1) (1) for Sr: ∆H = 2467 – 1414 – (2 × 550) = –47 (kJ mol–1) (1) (ii) Sr(OH)2 should be more soluble in water, and ∆H is more exothermic / negative (1) Assuming “other factors” (e.g. ∆S, or temperature etc.) are the same (1) (iii) Sr(OH)2 should be less soluble in hot water, because ∆H is negative / exothermic (1) [5] (c) (i) Ksp = [Ca2+][OH– ]2 (needs the charges) units: mol3dm–9 (1) + (1) (ii) n(H+) = n(OH–) = 0.05 × 21/1000 = 1.05 × 10–3 mol in 25 cm3 [OH– ] = 1.05 × 1000/25 = 4.2 × 10–2 (mol dm–3) (1) [Ca2+] = 2.1 × 10–2 (mol dm–3) (1) Ksp = 2.1 × 10–2 × (4.2 × 10–2)2 = 3.7 × 10–5 (1) (iii) less soluble in NaOH due to the common ion effect or equilibrium is shifted to the l.h.s. by high [OH– ] (NOT just a mention of Le Chatr on its own) (1) [6] [Total: 15]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 3 (a) SiF4 is symmetrical or tetrahedral or bonds are at 109° or has no lone pair or 4 electron pairs shared equally or all Si-F dipoles cancel out, or SF4 has a lone pair (on S). (1) [1] (b) compound molecule has an overall dipole molecule does not have an overall dipole BCl 3
PCl 3
CCl 4
SF6
mark row-by-row, (2) [2] (c) (i) Si and B have empty / available / low-lying orbitals or C does not have available orbitals (allow “B is electron deficient” but not mention or implication of d-orbital on B) (1) (ii) BCl 3 + 3H2O → H3BO3 + 3HCl or 2BCl 3 + 3H2O → B2O3 + 6HCl (1) SiCl 4 + 2H2O → SiO2 + 4HCl etc., e.g. → Si(OH)4, H2SiO3 (1) [3] (d) (i) Si3Cl 8O2 (this has Mr = 84 + 280 + 32 = 396) or Si4Cl 4O9 or Si8Cl 4O2 (1) (ii) mass number structure 133 Cl 3Si 247 Cl 3Si-O-SiCl 2 263 Cl 3Si-O-SiCl 2-O (3) (if correct structures are not given for last 2 rows, you can award (1) mark for two correct molecular formulae: either Si2Cl 5O + Si2Cl 5O2 or Si3ClO8 + Si3ClO9 or Si7ClO + Si7ClO2) (iii) Cl Si O Si O Si Cl Cl Cl Cl Cl Cl Cl allow ecf on the structure drawn in the third row of the table in (ii) but any credited structure must show correct valencies for Si, Cl and O. (1) [5] [Total: 11]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 4 (a) Cr3+: 1s22s22p6... 3s23p63d3 (1) Mn2+: 1s22s22p6... 3s23p63d5 (1) (allow (1) out of (2) for 3s23p64s23d1 and 3s23p64s23d3) [2] (b) (i) any three of the following points: • initial (pale) green (solution) • fades to (almost) colourless (allow yellow) • then (permanent faint) pink • finally (deep) purple (3) (ii) MnO4 – + 8H+ + 5Fe2+ (+ 5e–) → Mn2+ + 4H2O + 5Fe3+ (+ 5e–) (1) [4] (c) Eo values: O2 + 4H+/2H2O = +1.23V Fe3+/Fe2+ = +0.77 V O2 + 2H2O/4OH– = +0.40V Fe(OH)3/Fe(OH)2 = –0.56V (2) Eo cell = +0.46V (allow –0.37) in acid, but +0.96V in alkali or Eo (OH–) > Eo (H+) (1) If Ecell is more positive it means a greater likelihood of reaction (1) [4] (d) O and CH3CO2H HO2C CO2H CO2H H3C O CHO H3C O or [1] [1] [1] [1] [1] (or CO2H) [5] (e) (i) (CH3)2C(OH)–CH2OH (1) (ii) reaction I: (cold dilute) KMnO4 (“cold” not needed, but “hot” or “warm” negates) (1) reaction II: Cr2O7 2– + H+ + distil (1) [3] [Total: 18 max 17] (1) (1) (1) (1) (1)

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 5 (a) (i) because the carbons are sp2 / trigonal planar / bonded at 120° or are joined by π bonds / orbitals (1) (ii) because the π electrons / double bonds are delocalised / in resonance or electrons are evenly distributed / spread out (1) [2] (b) (i) HNO3 + 2H2SO4 → NO2 + + H3O+ + 2HSO4 – (1) or HNO3 + H2SO4 → H2NO3 + + HSO4 – or → H2O + NO2 + + HSO4 – (ii) electrophilic substitution (1) mechanism: H NO2 NO2 curly arrows from benzene to NO2 +, and showing loss of H+ (1) correct intermediate (with “+” in the ‘horse-shoe’) (1) [4] (c) Cl 2 + AlCl3 / FeCl3 / Fe / Al / I2 (aq or light negates this mark) (1) [1] (d) (i) Y is chlorobenzene (1) Z is 4-chloronitrobenzene (1) (2) (ii) Sn / Fe + (conc) HCl (1) HCl is conc, and second step is to add NaOH(aq) (1) (iii) NH2 Cl NHCOCH3 Cl N2 Cl A B D C no reaction Br Br (Cl only 2 x Br, but ignore orientation allow NHOCCH3, but not NHCH3CO or NHCH3OC ) or Cl OH (4) [8] [Total: 15]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 6 (a) (i) Primary – the amino acid sequence / order / chain or diag. e.g. NH-C-CO-NH-C-CO or amino acids bonded by covalent / amide / peptide bonds (1) (ii) Tertiary – the coiling / folding of the protein / polypeptide chain due to interactions between side-chains on the amino acids or the structure which gives the protein its 3-D / globular shape (1) [2] (b) (i) Diagram: Minimum is CH2S-SCH2 (1) (ii) Oxidation / dehydrogenation / redox (1) (iii) Hydrogen / H bonds; ionic interactions / bonds or ion-dipole or salt bridges; van der Waals’ or id-id or induced / instantaneous dipole forces (ignore hydrophobic interactions) (2) [4] (c) (i) Hydrogen bonds (1) (ii) Correct new strand present (see below) needed Diagram showing C=O bonding to N-H in new strand...
...and N-H bonding to C=O in new strand
e.g. N O H R N O H R New strand must contain a minimum of two amino acid residues in a single chain. Deduct a penalty of –(1) for any wrong H-bond only if (2) marks have already been scored. (2) [3] (d) There are bonds or S-S bridges / linkages between the layers / sheets (in β-keratin) (but only van der Waals interactions between the layers in silk) (1) [1] [Total: 10]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 7 (a) The amino acid is uncharged / neutral / a zwitterion or charges balance / are equal (NOT “is non-polar”) It is equally attracted by the anode / + and the cathode / – or attracted by neither The pH of the buffer is at the isoelectric point/IEP of the amino acid any two

(2) [2] (b) (at pH 10), H2NCH2CO2 – or NH2CH2COO– (1) [1] (c) amino acid relative size charge A small(est) (1) –ve B large(st) (3) –ve C middle (2) +ve (numbers are OK to show relative sizes) Mark each row (3) [3] (d) (i) lys – val – ser – ala – gly – ala – gly – asp (2) (ii) gly – ala – gly (1) (iii) aspartic acid (or lysine) (1) [4] [Total: 10]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9701 43 © UCLES 2010 8 (a) Reaction II – since electrons are used up / required / gained / received (from external circuit) (1) [1] (b) (Pb2+ + 2e– → Pb) Eo = –0.13V (PbO2 + 4H+ + 2e– → Pb2+ + 2H2O) Eo = +1.47V two correct Eo values (1) Cell voltage is 1.6(0) (V) (1) [2] (c) (i) 3(+) (1) (ii) They are less heavy / poisonous / toxic / polluting or are safer due to no (conc) H2SO4 within them (1) [2] (d) (i) Platinum or graphite / carbon (1) (ii) They need large quantities of compressed gases which take up space or the hydrogen would need to be liquefied or the reactant is (highly) flammable / explosive / combustible (1) [2] (e) Glass: saves energy – the raw materials are easily accessible / cheap or making glass is energy-intensive (1) Steel: saves energy – extracting iron from the ore or mining the ore is energy intensive or saves a resource – iron ore (NOT just “iron”) is becoming scarce either one (1) Plastics: saves a valuable / scarce resource: (crude) oil / petroleum (1) [3] [Total: 10]