9701 chemistry 9701/22 paper 2 (as structured questions), maximum raw mark 60 this mark scheme is...

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

9701/11 Paper 11 (Multiple Choice), maximum raw mark 40

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.

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Mark Scheme: Teachers’ version Syllabus Paper

GCE A/AS LEVEL – May/June 2010 9701 11

© UCLES 2010

Question Number

Key Question Number

Key

1 C 21 C

2 A 22 B

3 C 23 D

4 C 24 C

5 B 25 B

6 D 26 D

7 A 27 A

8 D 28 D

9 C 29 B

10 D 30 D

11 A 31 B

12 D 32 D

13 A 33 B

14 A 34 B

15 A 35 C

16 C 36 B

17 B 37 A

18 D 38 C

19 A 39 A

20 C 40 B

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9701 CHEMISTRY




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Question Number

Key Question Number

Key

1 C 21 B

2 C 22 D

3 A 23 C

4 C 24 B

5 B 25 C

6 D 26 D

7 C 27 B

8 D 28 D

9 D 29 D

10 A 30 A

11 A 31 B

12 A 32 B

13 D 33 D

14 A 34 B

15 A 35 C

16 D 36 B

17 A 37 A

18 C 38 A

19 B 39 C

20 C 40 B

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9701 CHEMISTRY




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Question Number

Key Question Number

Key

1 C 21 B

2 A 22 C

3 C 23 C

4 B 24 D

5 C 25 B

6 C 26 D

7 D 27 D

8 A 28 A

9 D 29 D

10 D 30 B

11 A 31 D

12 A 32 B

13 D 33 B

14 A 34 C

15 A 35 B

16 A 36 B

17 C 37 C

18 B 38 A

19 D 39 A

20 C 40 B

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9701 CHEMISTRY

9701/21 Paper 2 (AS Structured Questions), maximum raw mark 60

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.



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GCE AS/A LEVEL – May/June 2010 9701 21

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1 (a) fewer electrons in Cl 2 than in Br2 (1)

smaller van der Waals’ forces in Cl 2 or stronger van der Waals’ forces in Br2 (1) [2]

(b) CO has a permanent dipole or N2 does not (1)

permanent dipole-permanent dipole interactions are stronger than those from induced

dipoles (1) [2]

(c) (i) a co-ordinate bond (1)

(ii) a covalent bond (1)

or

(iii) a lone pair (1)

or

penalise any groups of 3 or 4 electrons that are circled [3]

(d) CO and HCN both have a dipole or N2 does not have a dipole (1) [1]

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(e) (i)

H H

H C C O H

H C≡N

C≡N must be shown (1)

(ii) nucleophilic addition (1)

(iii)

H H

δ+ δ |

CH3 C O CH3 C O HCN CH3 C OH + CN

| H CN CN CN

C=O dipole correctly shown or correct curly arrow on C=O (1)

attack on Cδ+

by C of CN (1)

correct intermediate (1)

CN regenerated (1) [5 max]

[Total: 13]

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2 (a) (i) new graph has lower maximum (1)

maximum is to the right of previous maximum (1)

(ii) H is at Ea (1) [3]

(b) the minimum amount of energy molecules must have or energy required (1)

in order for the reaction to take place (1) [2]

(c) (i) iron or iron oxide (1)

100 to 500 atm and 400–550°C

units necessary – allow other correct values and units (1)

(ii) C is placed to the left of H (1)

(iii) more molecules now have energy >Ea (1) [4]

(d) reaction 1

has greater Ea (1)

because energy is needed to break covalent bonds (1)

reaction 2 has lower Ea

or actual reaction is H+ + OH → H2O

or reaction involves ions (1)

opposite charges attract (1) [4]

[Total: max 12]

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3 (a) Accept only symbols.

(i) S or S8 (1)

(ii) K or K+ (1)

(iii) Na – allow K or Li (1)

(iv) Cl or Br or F (1)

(v) Mg or Ca or Li

allow Ni, Cu, or Zn (1) [5]

(b) Accept only formulae.

(i) F2O (1)

(ii) SO2 and SO3

or P2O3/P4O6 and P2O5/P4O10

or any two from N2O3, NO2/N2O4, N2O5

or any two from Cl 2O, ClO2, ClO3, Cl 2O7 (1+1) [3]

(c) (i) NaF, MgF2, AlF3 – any two (1)

(ii) octahedral (1)

(iii) I atom is larger than Cl atom (1)

(iv) cannot pack 7 F atoms around Cl atom

or can pack 7 F atoms around I atom (1) [4]

[Total: 12]

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4 (a)

Br2 KMnO4/H

+ cold, dilute HBr KCN in NH3 aqueous in an ethanol excess H2SO4(aq) NaOH(in ethanol) heat under heat under reflux reflux

give 1 for each correct structure (7 × 1) [7]

(b) (i) ester (1)

(ii) heat under reflux (1)

trace of conc. H2SO4 or presence of HCl (g) (1) [3]

[Total: 10]

CH3CH=CH2

CH3CHBrCH2Br

A

CH3CH(OH)CH2OH

B

CH3CH2CH2CN

or

CH3CH(CN)CH3

D

CH3CH2CH2Br

or

CH3CHBrCH3

C

CH3CH2CH2CO2H

or

CH3CH(CO2H)CH3

E

CH3CH2CH2NH2

or

CH3CH(NH2)CH3

F

CH3CH=CH2

G

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5 (a) (i) same molecular formula

but different structural formula/structure (1)

(ii) asymmetric C atom/chiral centre present (1)

>C=C< bond present (1) [3]

(b) NaO2CCH(OH)CH(OH)CO2Na (1) [1]

(c) no because there is no chiral carbon atom present (1) [1]

(d) (i) C : H : O =

12

35.8 :

1

4.5 :

16

59.7 this mark is for correct use of Ar values (1)

C : H : O = 2.98 : 4.5 : 3.73

C : H : O = 1 : 1.5 : 1.25 this mark is for evidence of correct calculation (1)

gives empirical formula of W is C4H6O5

(ii) C4H6O5 = 12 × 4 + 1 × 6 + 16 × 5 = 134

molecular formula of W is C4H6O5 (1) [3]

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(e) (i) n(OH ) =

1000

10029.4 × = 0.0294 (1)

n(W) =

134

1.97 = 0.0147 (1)

no. of –CO2H groups present

in one molecule of W =

0.0147

0.0294 = 2 (1)

or n(OH ) =

1000

1.0029.4× = 0.0294 (1)

1.97 g W ≡ 0.0294 mol NaOH

134 g W ≡

1.97

1340.0294 × = 1.999 ≈ 2 mol NaOH (1)

no. of –CO2H groups present in 1 molecule of W = 2 (1) [3]

(ii)

H H

H—O O—H

C—C—C—C

O O H O—H or

CH3

H—O O—H

C —C—C

O O O—H or

OH

H—C—H

H—O O—H

C—C—C

O O H

one correct structure (1)

correctly displayed (1)

allow any correct ether [2]

[Total: 13]

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9701 CHEMISTRY





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1 (a) (i)

correct 1s and 2s (1)

correct 2px, 2py and 2pz (1)

(ii)

spherical s orbital (1)

double lobed p orbital along one axis (1)

both orbitals correctly labelled (1)

(iii)

both correct (1) [6]

(b) (i) N 1400 kJ mol 1 O 1310 kJ mol 1 both (1)

(ii) N is all singly filled 2p orbitals or O has one filled/paired 2p orbital (1) these paired 2p electrons in the O atom repel one another (1) [3]

[Total: 9]

2px 2py 2pz

2s

1s

nucleus

nitrogen oxygen

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2 (a)

element particle formula

copper cation Cu2+

allow Cu+

argon atom or molecule Ar

one mark for each correct row or column (2 × 1) [2] (b) Cu cations held in ‘sea’ of delocalised electrons (1) by strong metallic bonds (1) Ar van der Waals’ forces between molecules (1) which are weak (1) [4] (c) (i) oxidising agent or electron acceptor (1) Ar has very high first I.E or Ea for reaction is very high or Ar has full valency shell/complete octet (1) [2] (d) from Ne to Xe more electrons in atom (1) hence more induced dipoles/van der Waals’ forces (1) [2] [Total: 10]

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3 (a)

oxide Na2O MgO Al 2O3 SiO2 P4O6 SO2

bonding ionic ionic ionic/covalent covalent covalent covalent

structure giant giant giant giant simple simple

(i) fully correct ‘bonding’ row (1) (ii) fully correct ‘structure’ row (1) [2] (b) Al 2O3 or SiO2 (1) [1]

(c) (i) Na2O Na2O + H2O → 2NaOH (1) pH 10–14 (1)

SO2 SO2 + H2O → H2SO3 (1) pH 2–5 (1)

(ii) NaOH + H2SO3 → NaHSO3 + H2O

or 2NaOH + H2SO3 → Na2SO3 + 2H2O (1) [5] (d) MgO(l) conducts (1) MgO(l) contains free/mobile ions (1) SiO2(l) does not conduct (1) SiO2(l) has no free ions (1) [4] [Total: 12]

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4 (a) C : H : O = 12

48.7 :

1

8.1 :

16

43.2 (1)

= 4.06 : 8.1 : 2.70 = 1.5 : 3 : 1 = 3 : 6 : 2 empirical formula is C3H6O2 (1) [2]

(b) (i) Mr = 65

1058.0101.00

4008.310.13

×××

××

pV

mRT (1)

= 74.5 (1) (ii) C3H6O2 = 36 + 6 + 32 = 74 (1) n(C3H6O2) = 74.5 hence molecular formula of E is C3H6O2 (1) [4] (c) structures of F are

HCO2CH(CH3)2

S

HCO2CH2CH2CH3

T

CH3CO2CH2CH3

U

CH3CH2CO2CH3

V

each correct structure is worth one mark (3 × 1) [3] (d) (i) H2SO4/HCl/mineral acid or NaOH/KOH (1) (ii) carboxylic acid not ‘acid’ (1) [2] (e) (i) aldehyde (1) (ii) must be a primary alcohol (1) (iii) CH3OH or CH3CH2OH or CH3CH2CH2OH (1) [3] (f) (i) S (1) (ii) only S is not the ester of a primary alcohol or only S is the ester of a secondary alcohol (1) [2] [Total: 16]

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5 (a) (i) propan-1-ol or propan-2-ol (1) (ii)

OH

or OH (1) (iii) dehydration or elimination (1) [3] (b) (i) carbon (1) by decomposition/cracking of the alcohol (1) (ii) to avoid ‘sucking back’ of water into the hot tube (1) (iii) SiO2 (1) (iv) conc. H2SO4 or P4O10 or Al 2O3 or H3PO4 (1) [5] (c) (i) CH3CHBrCH2Br (1) (ii) CH3CH(OH)CH2OH (1) (iii) CH3CO2H (1) [3] (d) (i) (very) high pressure or Ziegler-Natta catalyst (1) (ii) does not biodegrade or gives harmful combustion products (1) [2] [Total: 13]

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9701 CHEMISTRY





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1 (a) enthalpy change when 1 mol of a compound is formed (1) from its elements (1) in their standard states under standard conditions (1) [3]

(b) (i) N2H4(l) + O2(g) → N2(g) + 2H2O(g) ∆Hf

o/kJ mol 1 +50.6 –241.8 ∆H

o

reaction = 2(–241.8) – (+50.6) (1) = –534.2 kJ mol 1 (1) (ii) Ea is too high (1) (iii) products are H2O and N2 which are harmless/non toxic or are already present in the atmosphere (1) [4] (c) (i) ‘dot-and-cross’ diagram (1) (ii) (1) (iii) minimum is allow bond angle around N atom between 109o and 104o (1) [4] (d) –2 (1) [1]

[Total: 12]

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2 (a) the energy required to remove one electron from each atom (1) in one mole of gaseous atoms (1) or the enthalpy change in kJ mol 1 for (1)

M(g) → M+(g) + e (1) [2] (b) (i) first ionisation energy decreases down Group 1 (1) outermost electron is further from nucleus or has greater shielding (1) (ii) outermost electron experiences less attraction or formation of M+ cation becomes easier down Group 1 (1) [3]

(c) (i) n(Li) = 6.9

0.83 = 0.12 (1)

(ii) 2 mol Li → 1 mol H2

0.12 mol Li → 2

0.12 1× = 0.06 mol H2 (1)

volume of H2 = 0.06 × 24.0 = 1.44dm3 (1)

(iii) 2 mol Li → 2 mol LiOH

0.12 mol Li → 0.12 mol LiOH in 0.50 dm3 (1)

[LiOH] = 0.50

1 0.12×

= 0.24 mol dm 3 (1) [5]

(d) sodium burns with a yellow flame or white solid formed or colour of chlorine disappears (1)

2Na + Cl2 → 2NaCl (1) [2]

[Total: 12]

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3 (a) (i) Ca (1) (ii) S or C [allow H (H2O2) or N (NO, NO2)] (1) (iii) He (1) (iv) Al (1) (v) Si or Ge (1) (vi) Al (1) [6] (b) any two from N or O or F (1) [1] (c) (i) Al2O3 or SiO2 (1) (ii) SO2 or P2O3/P4O6 (1) and and SO3 or P2O5/P4O10 (1) (iii) Na2O (1) (iv) Al2O3 (1) [5]

[Total: 12] 4 (a) reaction 1 free radical substitution (1) reaction 2 elimination (1) [2] (b) (i) in reaction 4 CH3C(OH)(CN)CH3 (1) (ii) in reaction 3 I (1) (iii) in reaction 3 CH3I or in reaction 4 CH3COCH3 (1) [3] (c) a species which has a lone pair of electrons

or which reacts with an electron deficient (δ+) centre in a molecule (1) [1] (d) in reaction 3 OH (1) in reaction 4 CN (1) [2]

(e) π bonding is electron rich (1) [1]

[Total: 9]

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5 (a)

C*

HOH

C

OH

CH3

O

[3] (b)

reagent(s) condition(s)

step 1 Cr2O7

2 / H+

(1)

distil off aldehyde

(1)

step 2

HCN in presence of CN or

KCN + dil H2SO4 (1)

room temperature

(1)

step 3

aqueous mineral acid/ /H2SO4/HCl

not HNO3 (1)

heat under reflux

(1)

in each case, the reagent must be correct before the condition mark is awarded [6] (c) (i) a protein (1) (ii) 2,4-dinitrophenylhydrazine/Brady’s reagent (1) yellow-orange-red ppt. (1) (iii) acidified K2Cr2O7 or Lucas test or CH3CO2H/H+ (1) colour changes or cloudiness or fruity smell from orange to green (1) (iv) LiAlH4/NaBH4 or H2/Ni etc. (1) [6]

[Total: 15]

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9701 CHEMISTRY

9701/31 Paper 31 (Advanced Practical Skills), maximum raw mark 40




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© UCLES 2010

Question 1 Round all thermometer readings to the nearest 0.5°C.

Question Sections Indicative material Mark

1 (a) PDO Recording MMO Collection MMO Decisions MMO Quality MMO Quality

(i) Presents data in single table of results – to include volume of FA 2, initial and final temperatures and temperature change.

(ii) All columns correctly labelled with appropriate unit

shown. Must use solidus, brackets or describe unit fully in words. If units not included in column headings every entry must have the correct unit shown.

(iii) All thermometer readings recorded to 0.5°C (iv) Follows instructions – uses 10, 20, 30, 40, 50 cm3 of

FA 2 + two additional volumes (v) One extra volume of FA 2 on either side of the

maximum for the first five expts. or Two extra volumes between identical values for the first five expts. or Two extra volumes the same side as the next highest reading.

(vi) and (vii) Check and correct ∆T where necessary. (If multiple readings for max. T then apply hierarchy: take value of consistent readings; take average and correct to nearest 0.5°C) Compare temp rise with that obtained by the Supervisor (Expected value is 14.0°C) For 30 cm3 FA 2: Award (vi) and (vii) for a temp rise of 0.0°, 0.5°, 1.0°C Award (vi) only for a difference of 1.5°C

(viii) and (ix) Check and correct ∆T where necessary. Compare temp rise with that obtained by the Supervisor (Expected value is 13.5°C) For 40 cm3 FA 2: Award (viii) and (ix) for a temp rise of 0.0°, 0.5°, 1.0°C Award (viii) only for a difference of 1.5°C

1

1

1

1

1

2

2

[9]

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(b) PDO Layout

(i) Temperature (rise) plotted on y-axis against volume (of FA 2) or FA 2 added /cm3 on x-axis. Clearly labelled

axes (ignore units unless T, ∆T or V used as labels) (ii) Uniform and sensible scales that allow points to be

plotted in at least half of the squares on each axis. (6 × 4 big squares). (0,0) may be considered – as an additional point or with a line going through it

(iii) Visual check the “sweep” of all points, for all

experiments recorded. Check the plotting of points for 10, 30 and 50 cm3 of FA 2 (and any other “suspect” point) If any point is missing and that experiment was not carried out, check adjacent point Points should be within ½ of a small square, in the correct square

Do not award if T plotted instead of ∆T (iv) Appropriate lines drawn through the ascending and

descending points. (Ignore any deviation through rounding at the maximum temperature rise) Do not award if both straight lines and curves drawn or there is any forced change in gradient.

1

1

1

1

[4]

(c) ACE Interpretation

Reads from the graph (to within ½ small square) the volume of FA 2 at the intersection of two lines. Allow rounding to the closest cm3 Do not award this mark if the lines/curves have been

rounded at the maximum ∆T.

1

[1]

(d) PDO Layout

Explains that the temperature rise is the dependent variable or Volume of FA 2 is the independent variable/one that is controlled/one that you vary (or words to that effect)

1

[1]

(e) ACE Conclusion

Gives correct equation for the reaction (ignore state symbols)

2NaOH + H2SO4 → Na2SO4 + 2H2O or NaOH + H2SO4 → NaHSO4 + H2O

1

[1]

(f) PDO Display ACE Interpretation

Working is shown in (f)(i) (involves volumes and concentration, 2.0 mol dm 3) and (f)(ii) (any clear mole ratio)

Has correct expression for 2.01000

10.00×

or an answer of 0.02(00) in (f)(i) and 0.04(00) in (f)(ii) There is no ecf within (f)

1

1

[2]

(g) PDO Display

Expression given in the question paper is correctly evaluated to 2 or 3 significant figures. Allow a volume, read from rounded curves to be used in this expression. Normal rounding rules apply to the sig fig.

1

[1]

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(h) ACE Interpretation

Uses the expression:

(answer from (c) + 10) × 4.3 × ∆T read from graph Divides the answer above by answer to (f)(i) and gives answer in kJ mol 1 with –ve sign Do not award this second mark unless candidate has

calculated (a volume of soln × 4.3 × ∆T)

1

1

[2]

(i) ACE Improvements

Advantage of burette: Lower % error or more accurately calibrated (must refer to or infer scale/graduations/markings/divisions) Disadvantage of burette: Takes longer to add the FA 2

1

1 [2]

(j) ACE Interpretation

Candidate gives two of the following as significant sources of error. Heat loss (to the surroundings) Thermometer graduated at 1°C intervals Drying of cup/thermometer Initial temps of both solutions should be taken Other acceptable sources of error may be seen.

1

[1]

(k) ACE Interpretation

(i) Maximum error in reading a 1°C graduated thermometer is given as 0.5°C

(iii) Calculates answer in 100% in answer

2 in answer×

×

(k)(ii)

(k)(i)

1

1 [2]

Total [26]

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Question 2


FA 3 is BaCl2(aq); FA 4 is MgBr2(aq) [MgCl2 + NaBr]; FA 5 is CaI2(aq) [CaCl2 + NaI]; FA 6 is K2CrO4(aq)

2 (a) MMO Decisions

Chooses silver nitrate/Ag+(aq)/solution containing Ag+ ions followed by (aqueous) ammonia.

1 [1]

(b) PDO Recording MMO Collection

Results for three solutions and the two reagents from (a) (or three reagents if (a): ‘Ag++ NH3’, Pb2+) if recorded in a single table (no repetition of solutions or reagents) Give one mark for correct observations with FA 3, FA 4 and FA 5. FA 3 – white ppt with Ag+, soluble in NH3(aq) FA 4 – cream ppt with Ag+, partially soluble or insoluble in NH3(aq) (allow “creamy” not “creamy white”) FA 5 – yellow ppt with Ag+, insoluble in NH3(aq) If Ag+ and Pb2+ in (a), all observations must be correct (ignore any ‘extra’ NH3 if not in (a)) (Pb2+: white, white, yellow ppts respectively)

1

1

[2]

(c) ACE Conclusion

Mark consequentially on observations in (b) Expected conclusion Identifies FA 3 as solution containing Cl

from “white ppt with Ag+ (soluble in NH3(aq)) given as evidence. Mark consequentially – ecf allowed here. (No retrospective to observations)

1

[1]

(d) MMO Collection

Mark each of the boxes and see whether correct columns or rows give the better mark. Award the better mark. See table below for the expected observations

1 1 1 [3]

FA 3 FA 4 FA 5

+ NaOH(aq) ignore white ppt white ppt or “cloudiness”

+ NH3(aq) no ppt (allow reference to “cloudiness”/”slight white ppt”)

white ppt no ppt/no change/ no reaction

+ FA 6 yellow ppt no ppt/no change/ no reaction/yellow soln

no ppt/no change/ no reaction/yellow soln

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(e) MMO Collection

Records (yellow) solution turning orange (or wtte, e.g. orange solution forms)

1 [1]

(f) ACE Conclusions

Mark consequentially on observations in (d) and (e) Expected conclusions: Anion in FA 6 is chromate, CrO4

2 , from yellow soln turning orange in (e) or yellow ppt with FA 3 in (d) provided FA 3 not also identified from (d) and FA 3 contains Ba2+ from observations with NaOH and NH3 (or just NaOH if obs with FA 4 and FA 5 are correct with it) or FA 6 in (d)

1

[1]

FA 7 is a tertiary alcohol; FA 8 is an aldehyde; FA 9 is a ketone; FA 10 is a primary alcohol

(g) MMO Collection

One mark for two correct observations with FA 7 One mark for correct observations with FA 8 and FA 9 One mark for two correct observations with FA 10 See table below for expected observations

1

1

1 [3]

observations reagent

FA 7 FA 8 FA 9 FA 10

acidified dichromate

no reaction no reaction

(colour change to) green/blue-green/ cyan/turquoise (solution not ppt)

2,4-DNPH no reaction yellow ppt yellow ppt

Tollens’ reagent no reaction silver mirror or black/grey solution or ppt

no reaction

(h) ACE Conclusions

No ecf from (g) FA 7 contains the tertiary alcohol from no reaction with all three reagents or no reaction with dichromate and 2,4-DNPH provided there is no CON in the observation with Tollens’ FA 8 contains the aldehyde from the silver (mirror), black or grey precipitate or solution with ammoniacal silver nitrate Allow from brown ppt if it is the only positive result with Tollens’.

1

1

[2]

Total [14]

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9701 CHEMISTRY





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Question 1 Round all thermometer readings to the nearest 0.5°C. Supervisor’s Report

Calculate m/∆T correct to 3 s.f. for each experiment. Candidate’s scripts

Calculate m/∆T correct to 3 s.f. for each experiment.


1 (a) MMO Collection PDO Recording

(i) Follows instructions with regard to times and temperature readings 0–3 minutes at 1 minute intervals; 5–8 minutes at ½ minute intervals, and T1 recorded in box. (Ignore if also in table)


shown. Must use solidus, brackets or describe unit fully in words. If units not included in column headings every entry must have the correct unit shown Accept min, mins or minutes

(iii) Look at results here and in (d).

All balance readings consistent to at least 1 decimal place. and All thermometer readings (table and box) recorded to nearest 0.5 oC. There must be at least one at 0.5 in (a).

1

1

1

[3]

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(b) PDO Layout ACE Interpretation

(i) Temperature of water in the beaker plotted on y-axis against time on x-axis. Clearly labelled axes (ignore units) [temp/time are minimum acceptable labels] but accept T / oC and t / min as labels. The unit is necessary in this case

(ii) Uniform and sensible scales for candidate’s choice of

graph. Plotted points must be in at least 4 large squares on the temperature axis and 5 large squares on the time axis. Do not include any plotted value of T1.

(iii) There should be a minimum of 5 plotted points

between 5 and 8 minutes. Examiner then checks plotting of points at t0 min,

t5 min and t8 min and the plotting of any suspect point. If any of the t0 min, t5 min and t8 min points is missing check the adjacent point. Points should be within ½ of a small square of the correct position and in the correct small square

(iv) Acceptable straight lines drawn – an acceptable

straight line is one passing through the majority of points or has balanced points on either side of the line and correct values of T2 and T3 read (to within ½ small square) from the graph. Extrapolation need not be drawn on the graph

1

1

1

1

[4]

(c) ACE Interpretation

(i) and (ii) Award one mark if both of the following expressions are correctly evaluated. heat gained = 210 × candidate value of (T3 – T2)

heat lost = 210 × candidate value of (T1 – T3) Units should be consistent. Ignore any sign given. (iii) No mark.

1

[1]

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(d) MMO Quality

Compare the two m/∆T values (goC–1) for the candidate’s two experiments. Award three marks for a difference up to 0.2 Award two of these three marks for a difference of 0.2+ oC – 0.3 Award one of these three marks for a difference of 0.3+ oC – 0.4

Compare the standard m/∆T value of 1.70 g.oC–1 with the closer value from the candidate’s results. Award three marks for a difference up to 0.2 Award two of these three marks for a difference of 0.2+ oC – 0.3 Award one of these three marks for a difference of 0.3+ oC – 0.4

3

3

[6]

(e) ACE Interpretation ACE Conclusions ACE Interpretation ACE Conclusions PDO Display

(i) Give one mark for correct evaluation of 430 ×

candidate’s ∆T(expt 1)

(ii) The candidate explains that the beaker as well as the

solution has cooled Short term – beaker loses heat energy as it cools Long term – Heat energy is absorbed by beaker (and solution) Read in context – Beware of: Heat is absorbed by beaker and surroundings (con)

(iii) Give one mark if the candidate adds (∆T(expt 1) × answer in (c)(iii)) to the answer in (e)(i). Correct expression is sufficient, evaluation not required.

(iv) The candidate correctly calculates the moles of FA 1

(candidate’s mass / 53.5) used in expt 1. In (i), (iii) or (iv) withhold one mark for use of data from expt 2 the first time it is seen. Do not withhold more than one mark for this. (v) The candidate correctly divides the answer to (e)(iii) by

the answer to (e)(iv) and by 1000. Ignore errors in evaluation and sign

(vi) Award this mark if the candidate has given a +ve sign

and explains that: the reaction is endothermic or heat is absorbed in the reaction or the temperature falls during the reaction.

(vii) Award this mark if working is shown in sections (c)(i),

or (c)(ii) or (e)(i) and (e)(iv) and (e)(v).

1

1

1

1

1

1

1

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(e) contd

PDO Display

(viii) Award this mark if the final answer in section (iii) of (c) and section (iv) of (e) is given to 2 or 3 significant figures.

1

[8]

(f) ACE Interpretation

Correctly calculates the difference and the percentage error. Ignore significant figures. Beware mixed units

1

[1]

(g) ACE Conclusions ACE Improvements

Clearly described source of error (i) Heat loss / gain (ii) Use of glass beaker (iii) Precision of thermometer (iv) Small temperature fall Specific improvement given with some attempt at justification. (i) Lid – prevents convection or evaporation

Insulation – prevents conduction Use plastic beaker – provides insulation

(ii) Polystyrene cup – lower heat capacity (iii) Use thermometer at 0.5 oC or better, gives smaller %

error. (iv) Larger mass of NH4Cl or smaller water volume. Gives

greater temperature change Do not award either of these marks for answers referring to use of measurement of volume or measurement of mass.

1

1

[2]

Total [25]

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FA 2 is Na2SO4(aq); FA 3 is Na2CO3(aq); FA 4 is a mixture of Na2SO4(aq) and Na2CO3(aq)

2 (a) MMO Decisions MMO Collection

(i) reagent 1 – chooses any specified acid to detect the carbonate present (name or formula may be in results table) and reagent 2 – chooses BaCl2 or Ba(NO3)2. Accept Ba2+(aq) or soln containing Ba2+(aq) as reagent. Also accept incorrect formulae for a compound, e.g. BaCl, providing the identity of the reagent is obvious.

(ii) Explains significance of order in which reagents added.

acid first – to remove carbonate from solution or after Ba2+ – to dissolve any barium carbonate precipitated. Candidates must make clear the relationship of acid to barium carbonate. Do not award this mark if sulfuric acid has been used in (i)

In section (iii), assume reagents follow each other in the same test-tube unless otherwise stated. Allow lead(II) nitrate as the 2nd reagent providing it is used with nitric acid. (iii) Addition of acid

No reaction with FA 2, effervescence/bubbles/bubbling (or gas tested with limewater) for FA 3 and FA 4 Addition of Ba2+(aq) white ppt with all three solutions, if added as first reagent or to a separate sample or white ppt, insoluble in acid for FA 2, soluble in acid for FA 3 and insoluble/partially soluble in acid for FA 4 if added before the addition of acid or white ppt with FA 2 and FA 4 and no ppt with FA 3 if added after addition of acid Do not award this mark if sulfuric acid has been used unless it has been stated that Ba2+(aq) was added to a fresh sample

Allow deductions from lead nitrate (as for barium salt) ONLY if nitric acid has been specified. If a candidate selects limewater as a reagent, can get observation marks, CaSO4 soluble at this concentration.

1

1

1

1

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FA 5 is CuSO4(aq); FA 6 is NH4Cl(aq); FA 7 is CoCl2(aq); FA 8 is MnSO4(aq)

(a) contd

ACE Conclusions

No ecf to be applied in these conclusions (iv) Give one mark for identifying carbonate in FA 3 and

FA 4, with supporting evidence. Minimum acceptable evidence – gas with acid or white ppt (barium carbonate) soluble in acid

Give one mark for identifying sulfate in FA 2 and FA 4 only, with supporting evidence. Minimum acceptable evidence – white ppt with Ba2+ (if insoluble in acid) but con if soluble. If no mark has been awarded in (iii) or (iv), allow one mark if evidence given is consistent with the ions identified

1

1

[6]

(b) MMO Collection MMO Decisions MMO Collection

(i) For FA 5, records blue ppt, insoluble in an excess of NaOH blue ppt with ammonia; soluble or forming a deep blue colour with excess of the reagent

(ii) For FA 6, records

no precipitate with both NaOH and NH3(aq), and ammonia detected or red litmus turning blue on heating with NaOH (No reference to a gas is necessary)

(iii) Describes test carried out on gas evolved to identify

ammonia. An observation of no ppt with either reagent and a gas

turning red litmus blue on heating with NaOH would earn both the C3 and the De7 mark. If the mark is not awarded for ammonia test a retrospective mark can be given here for testing gas evolved with limewater in (a)

(iv) For FA 7 records

blue (only) ppt, on initial addition of NaOH blue/green ppt, insoluble in excess ammonia

(v) Any reference to pink in either precipitate on initial

formation, in excess or on standing.

1

1

1

1

1

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(b) contd

MMO Collection

(vi) For FA 8, records off-white/pale brown/buff/beige precipitate insoluble in an excess of NaOH off-white/pale brown/buff/beige precipitate insoluble in an excess of ammonia Do not accept a cream ppt.

(vii) Darkening of the initial precipitate

or appropriate coloured precipitate (allow white or cream ppt here as colour of initial ppt) turning brown at any stage with either reagent.

1

1

[7]

(c) ACE Conclusions

No ecf to be applied in these conclusions Identifies all cations correctly:

FA 5 is Cu2+ / copper(II) FA 6 is NH4

+ / ammonium FA 8 is Mn2+ / manganese(II)

Gives appropriate supporting evidence for two of the three ions Minimum evidence for each of the ions. Cu2+ (i) Blue ppt with both NaOH and NH3(aq), or (ii) Blue ppt with NaOH, insoluble in excess of the

reagent, or (iii) Blue ppt with NH3(aq), soluble in excess of the

reagent, or (iii) Dark blue colour formed at any stage with NH3(aq) NH4

+ (i) Ammonia, NH3, alkaline gas, or gas turning red litmus

blue with NaOH (hot or cold) Mn2+ (i) Initial off-white to beige ppt with NaOH and with

NH3(aq) or

(ii) Precipitate darkening / turning brown with either reagent – providing the colour of the initial precipitate is not completely inappropriate (e.g. blue or green).

1

1

[2]

Total [15]

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9701 CHEMISTRY





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Question 1 Round all thermometer readings to the nearest 0.5°C.


1 (a) PDO Recording MMO Collection MMO Decisions MMO Quality

(i) Presents data in single table of results – to include masses of tube with solid and residue, (mass of solid), initial and final temperatures and temperature change.


shown. Must use solidus, brackets or describe unit fully in

words. There must be at least one mass and one temperature.

If units not included in column headings every entry must have the correct unit shown.

(iii) All balance readings recorded to 1 or more decimal

places and showing same precision of measurement (iv) All thermometer readings recorded to 0.5°C (must be

at least one at 0.5°C) (v) Two additional masses of NaHCO3 evenly spaced

between the other readings or one or both extending the plot

(Not two between same pair) (Not within 0.5 g of any other) (Not >3.0 g away from any other)

(vi) and (vii) Check and correct ∆T where necessary. Compare temperature fall with that obtained by the Supervisor for FB 2 Award (vi) and (vii) for a temp fall difference of 0.0°, 0.5° or 1.0°C Award (vi) only for a difference of 1.5°C

(viii) and (ix) Check and correct ∆T where necessary. Compare temperature fall with that obtained by the Supervisor for FB 3 Award (viii) and (ix) for a temp fall difference of 0.0°, 0.5° or 1.0°C Award (viii) only for a difference of 1.5°C

1

1

1

1

1

2

2

[9]

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(b) PDO Layout

(i) Temperature (fall) plotted on y-axis against mass of sodium hydrogencarbonate on x-axis.

Clearly labelled axes (ignore units unless T/δT/θ or m as labels)

Award even if final temp plotted (ii) Uniform and sensible scales that allow points to be

plotted in at least half of the squares on each axis. (6 × 4 big squares)

(iii) check the “sweep” that points plotted for all

experiments recorded. Check the plotting of points for FB 1, FB 2 and FB 3

(and any other “suspect” point). Points should be within ½ of a small square, in the correct half of the small square.

Not awarded if final T is plotted Not awarded if only FB 1, FB 2 and FB 3 are recorded (iv) Straight line drawn, passing within ½ small square of

the origin and close to the majority of the points

1

1

1

1

[4]

(c) PDO Layout

Explains that the mass of sodium hydrogencarbonate is the independent (controlled) variable or Temperature change is the dependent variable (or words to that effect)

1

[1]

(d) ACE Conclusion

Gives correct equation for the reaction

NaHCO3 + HCl → NaCl + H2O + CO2 (Not H2CO3) (Ignore state symbols)

1

[1]

(e) PDO Display ACE Interpretation

Construction on graph for determining the gradient clearly shown (Must span at least 3 large squares in each direction) and working shown for calculation (could be found on graph page) Reads intercepts or selects two points on the line to within ½ small square in either direction.

Evaluates ∆y/∆x from candidate’s intercept figures to find gradient correctly to sf shown

1

1

1

[3]


Multiplies answer to (e) by 84 1 [1]

(g) ACE Interpretation PDO Display

Some use of 30 × 4.3 × (answer to (f)) Answer, in kJ mol 1, correct to 2 or 3 significant figures and showing +ve sign Correct answer is given by 0.129 × answer to (f) (No ecf from first part)

1

1

[2]

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(h) ACE Improvements

Suggests use of a lid and increased/thicker insulation round the sides. (Accept use of a vacuum flask for improved insulation.)

1

[1]

(i) ACE Improvements

Lower % error with burette or more accurately calibrated (must refer to or infer scale/graduations/markings/divisions)

1

[1]

(j) ACE Interpretation

Maximum error in reading a balance reading to 1 decimal place is 0.05 g and Maximum error in reading a 1°C graduated thermometer is given as ±0.5°C Expected % errors 0.89/0.9 or 1.79/1.8% (correct for sf shown) for the balance and 8.3% for the thermometer. (allow ecf from first part – error (× 2)/0.056 and error × 2/0.12) This section may be marked horizontally or vertically

1

1

[2]

(k) ACE Interpretation

Systematic error stated (or explained in other words) 1 [1]

Total [26]

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Question 2


FB 7 is (NH4)2SO4.FeSO4(aq); FB 8 is NH4Cl(aq)

2 (a) MMO Decisions

Chooses barium chloride (or nitrate)/ Ba2+(aq)/solution containing Ba2+ (ions) followed by a specified mineral acid other than sulfuric acid or vice versa. (Allow if acid name specified in (b)) Do not accept lead nitrate/lead ions

1

[1]

(b) PDO Recording MMO Collection

Results for two solutions and two reagents (ecf from (a)) recorded in a single table with distinct separation between reagents (no repetition of solutions or reagents) Give one mark for correct observations with FB 7 and FB 8 Expected results: FB 7 – white ppt with Ba2+, insoluble in acid (do not award if H2SO4 added) FB 8 – No reaction or no precipitate If Pb2+ used then white ppt with both or obs as for Ba2+ Do not award if any obs are dashes except for FB 8 adding acid after Ba2+

1

1

[2]

(c) ACE Conclusion

Mark consequentially on observations in (b) Expected conclusion Identifies FB 7 as solution containing SO4

2 from “white ppt with Ba2+, insoluble in acid given in evidence ecf allowed here. Allow deduction if H2SO4 has been added after Ba2+ (not with Pb2+)

1

[1]

(d) MMO Collection

One mark for two correct observations with NaOH(aq) – before heating FB 7 – green/dirty green/muddy green ppt FB 8 – no ppt/no change/no reaction The mark from (e) may be awarded here if the green ppt with FB 7 is recorded as turning brown One mark for correct observations with NaOH(aq) – after heating gas evolved turns red litmus paper blue for both FB 7 and FB 8 One mark for correct observations with NH3(aq) FB 7 – green ppt (as above) FB 8 – no ppt/no change/no reaction

1

1

1

[3]

(e) MMO Collection

Records brown precipitate/residue provided green ppt in first box of (d)

1 [1]

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(f) ACE Conclusions

Mark consequentially on observations in (d) and (e) Expected conclusions: (i) Common cation is ammonium/NH4

+, from evolution of ammonia or alkaline gas (minimum) or clear ref back to relevant obs

(ii) Second cation is Fe2+ in FB 7 from green ppt with NaOH and NH3 or green ppt, turning brown, with either.

If Cr3+ then grey-green ppt with both alkalis or grey-green ppt with NaOH soluble in excess giving (dark) green solution or grey-green ppt insol in excess NH3 (no con obs)

1 [1]

FB 9 is a ketone; FB 10 is a primary alcohol; FB 11 is a secondary alcohol; FB 12 is an aldehyde

(g) MMO Collection

One mark for correct observations with FB 9 and FB 12 One mark for two correct observations with FB 10 One mark for two correct observations with FB 11 See table below for expected observations

1

1

1 [3]

observations reagent

FA 9 FA 10 FA 11 FA 12

acidified dichromate

no reaction/no change/yellow or

orange soln (not no ppt*)

(colour change to) green/blue-green/

cyan/turquoise (solution not ppt*)

(colour change to) green/blue-green/

cyan/turquoise (solution not ppt*)

*penalise ppt error once in this row If FB9 colour wrong penalise 1st of additional ppts

2,4-DNPH yellow ppt no reaction/ no change

yellow ppt

Tollens’ reagent no reaction/

no change/no ppt no reaction/

no change/no ppt

silver mirror or black/grey

solution or ppt

(h) ACE Conclusions

Mark consequentially on observations in (g) FB 10 and FB 11 contain the alcohols from both oxidised by dichromate(VI) and give no ppt with Tollens’ FB 9 contains the ketone from No reaction with dichromate (Cr2O7

2 obs all correct) or no reaction with Cr2O7

2 and yellow ppt with 2,4-DNPH. (If FB 12 selected evidence must have yellow ppt with 2,4-DNPH and no reaction with Tollens’)

1

1

[2]

Total [14]

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9701 CHEMISTRY





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© UCLES 2010

Question 1


1 (a) MMO Collection PDO Recording

(i) Follows instructions with regard to weighings and mass of NaHCO3

2 balance readings and mass of NaHCO3 and times and temperature readings

0–3 minutes at 1 minute intervals; 5–8 minutes at ½ minute intervals (ii) All columns correctly labelled with appropriate unit

shown. Must use solidus, brackets or describe unit fully in

words. If units not included in column headings every entry

must have the correct unit shown Accept min, mins or minutes (iii) Look at results here and in (d). All balance readings consistent to at least 1 decimal

place. and All thermometer readings recorded to nearest 0.5°C. There must be at least one at 0.5 in (a).

1

1

1

[3]

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(b) PDO Layout ACE Interpretation

(i) Temperature of acid/solution in the beaker plotted on y-axis against time on x-axis.

Clearly labelled axes (ignore units) [temp/time are minimum acceptable labels] but accept T / °C and t / min as labels. The unit is

necessary in this case (ii) Uniform and sensible scales for candidate’s choice of

graph. Plotted points must be in at least 5 large squares on

the temperature axis and 5 large squares on the time axis.

(iii) There should be a minimum of 5 plotted points

between 5 and 8 minutes. Examiner then checks plotting of points at t0 min,

t5 min and t8 min and the plotting of any suspect point.

If any of the t0 min, t5 min and t8 min points is missing check the adjacent point.

Points should be within ½ of a small square of the correct position and in the correct small square

(iv) Acceptable straight lines drawn – an acceptable

straight line is one passing through the majority of points or has balanced points on either side of the line

and correct values of the minimum and maximum

temperatures at t = 4 minutes are read (to within ½ small square) from the graph.

Extrapolation need not be drawn on the graph

1

1

1

1

[4]

(c) ACE Interpretation ACE Conclusions

(i) No mark. (ii) The candidate correctly calculates (to sig figs

displayed) the moles of FA 1 used, (cand mass of FA 1/84)

(iii) The candidate correctly divides their answer to (i) by their answer to (ii) and by 1000.

Ignore errors in evaluation and sign. (iv) Award this mark if the candidate has given a +ve sign

and explains that: the reaction is endothermic or heat is absorbed in the reaction or the temperature falls during the reaction

1

1

1

[3]

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(d) MMO Quality

Compare the two ∆T/m values (°C g–1) for the candidate’s two experiments. Award three marks for a difference up to 0.2 Award two of these three marks for a difference of 0.2+ °C – 0.3 Award one of these three marks for a difference of 0.3+ °C – 0.4

Compare the standard m/∆T value of 1.55°C g–1 with the closer value from the candidate’s results. Award three marks for a difference up to 0.2 Award two of these three marks for a difference of 0.2+ °C – 0.3 Award one of these three marks for a difference of 0.3+ °C – 0.4

3

3

[6]

(e) ACE Interpretation ACE Conclusions PDO Display

(i) No mark is awarded for this section but check that temperature rise from expt 1 has been used

(ii) The candidate correctly calculates (to sig figs displayed) the moles of FA 2 used in expt 1.

Do not award this mark if data from expt 2 has been used in (i) or (ii).

(iii) The candidate correctly divides the answer to (e)(i) by the answer to (e)(ii) and by 1000. Ignore errors in evaluation and sign

(iii) Award this mark if the candidate has given a –ve sign and explains that: the reaction is exothermic or

heat is released in the reaction or the temperature rises during the reaction Award this mark if working is shown in sections (c)(ii), (c)(iii), (e)(ii) and (e)(iii) Award this mark if the final answer in section (c)(iii) and (e)(iii) is given to 2 or 3 sig fig.

1

1

1

1

1 [5]


Correctly calculates ∆Hdecomposition from candidate values in (c)(iii) and (e)(iii).

Allow (±1) on final significant figure given The answer given must include a mathematically correct sign

1

[1]

(g) ACE Interpretation

Correctly calculates the difference and the percentage error. Ignore significant figures.

1

[1]

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(h) ACE Conclusions ACE Improvements

Clearly described source of error (i) Heat loss / gain (ii) Precision of thermometer (iii) Acid spray (iv) Use of a glass beaker Specific improvement given with some attempt at justification. (i) Lid – prevents convection or evaporation Insulation – prevents conduction (ii) Use thermometer at 0.5°C or better, gives smaller %

error. (iii) Lid – acts as a physical barrier (iv) Use of plastic cup – better insulator or lower (specific)

heat capacity Do not credit either mark for answers referring to use of measurement of volume or measurement of mass.

1

1

[2]

Total [25]

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Question 2


2 (a) MMO Decisions MMO Collection

(i) reagent 1 – chooses any specified acid to detect the carbonate present

(name or formula may be in results table) and reagent 2 – chooses AgNO3. Accept Ag+(aq) or soln containing Ag+ as reagent. Also accept incorrect formulae for a compound,

e.g. Ag(NO3)2, providing the identity of the reagent is obvious.

(ii) Explains significance of order in which reagents

added. acid first – to remove carbonate from solution or after

Ag+ – to dissolve any silver carbonate precipitated. Candidates must make clear the relationship of

acid to silver carbonate. Do not award this mark if hydrochloric acid has been

used in (i) In section (iii), assume reagents follow each other in the same test-tube unless otherwise stated. Allow lead(II) nitrate as the 2nd reagent providing it is used with nitric acid. (iii) Addition of acid No reaction with FA 4, effervescence/bubbles/bubbling

(or gas tested with limewater) for FA 5 and FA 6 Addition of Ag+(aq) Ignore any addition of NH3(aq) after Ag+ white ppt with FA 4, if added as first reagent or to a

separate sample Allow off-white to brownish ppt with FA 6 or white ppt, insoluble in acid for FA 4, soluble in acid for

FA 6 and insoluble/partially soluble in acid for FA 5 if added before the addition of acid

or white ppt with FA 4 and FA 5 and no ppt with FA 6 if added after addition of acid

Do not award this mark if hydrochloric acid has been used unless it has been stated that Ag+(aq) was added to a fresh sample

Allow deductions from lead nitrate (as for silver salt) ONLY if nitric acid has been specified

1

1

1

1

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2 (a) contd

ACE Conclusions

No ecf to be applied in these conclusions (iv) Give one mark for identifying carbonate in FA 5 and

FA 6, with supporting evidence. Minimum acceptable evidence – gas with acid

or off-white, white, cream. brownish ppt (silver carbonate)

soluble in acid Do not award this mark from the colour of the precipitate alone. Give one mark for identifying chloride in FA 4 and FA

5 only, with supporting evidence. Minimum acceptable evidence – white ppt with Ag+ (if

insoluble in acid) but con if soluble. Do not award this mark from the colour of the precipitates with Ag+ alone If no mark has been awarded in (iii) or (iv), allow one

mark if evidence given is consistent with the ions identified

1

1

[6]

FA 7 is Al

K(SO4)2(aq); FA 8 is NiCl2(aq); FA 9 is CrCl3(aq); FA 10 is Pb(NO3)2(aq)

(b) MMO Collection

For FA 7, records (i) white ppt, soluble in an excess of NaOH and white ppt, insoluble in excess ammonia For FA 8, records (ii) green ppt, insoluble in an excess of NaOH and Dark or deep blue solution with (excess) ammonia or

Initial blue solution, darkening with excess of the reagent

For FA 9 records (iii) grey-green ppt, soluble in an excess of NaOH to give a

dark green solution (iv) grey-green ppt, insoluble in excess ammonia For FA 10, records (v) white ppt, soluble in an excess of NaOH and white ppt, insoluble in excess ammonia

1

1

1

1

1

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(b) contd

MMO Decisions MMO Collection

(vi) Selects hydrochloric acid, sulfuric acid potassium iodide, potassium chromate (or dichromate) as suitable reagent

Not barium chloride If no reagent is chosen, a retrospective De7 mark can be given here for testing gas evolved with limewater in (a) (vii) Observes appropriate ppt with chosen reagent and

FA 10 but not with FA 7. Do not award this mark if reagent is added to any solution recorded as giving a coloured ppt or no ppt with NaOH or with NH3 or barium chloride is used

1

1

[7]

(c) ACE Conclusions

No ecf to be applied in these conclusions Identifies all cations correctly: FA 7 is Al

3+ / aluminium FA 9 is Cr3+ / chromium(III) FA 10 is Pb2+ / lead The correct ions must be identified for each solution. Gives appropriate supporting evidence for two of the three ions Allow evidence for reversed Al

3+/Pb 2+ if barium chloride has been used Minimum evidence for each of the ions. Al

3+ white ppt with NaOH and NH3(aq); ppt soluble in excess NaOH, ppt insoluble in excess NH3(aq) and no reaction with HCl, H2SO4, KI Cr3+ (i) grey-green ppt with both NaOH and NH3(aq), or (ii) grey-green ppt with NaOH, soluble in excess of the

reagent, or (iii) grey-green ppt with NH3(aq), insoluble in excess of the

reagent, or (iii) Dark green solution with excess NaOH Pb2+ white ppt with NaOH and NH3(aq); ppt soluble in excess NaOH, ppt insoluble in excess NH3(aq) and white ppt with HCl or H2SO4 or yellow ppt with KI

1

1

[2]

Total [15]

If an appropriate reagent has been chosen but no observations recorded in (b)(iii): Allow deduction of the identity of the ions if the observations have been correctly recorded in the evidence section Do not however award the mark for appropriate evidence – from observations

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9701 CHEMISTRY

9701/41 Paper 4 (A2 Structured Questions), maximum raw mark 100




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1 (a) P: burns with white / yellow flame or copious white smoke / fumes produced (1)

4P (or P4) + 5O2 → P4O10 (1) S: burns with blue flame / choking / pungent gas produced (1)

S + O2 → SO2

(1) [4]

(b) (i) 2 Ca3(PO4)2 + 6 SiO2 + 10 C → 1 P4 + 6 CaSiO3 + 10 CO (2) (ii)

allotrope type of structure type of bonding

white simple / molecular covalent

red giant / polymeric covalent

(4) (iii)

P

PP

P

P

PP

P

P

PP

P

white P4 (1) red Pn (1) (in each case P has to be trivalent. Many alternatives allowable for the polymeric red P) (2) (8 max 7) [7] [Total: 11]

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2 (a) coloured ions / compounds (1) variable oxidation states (1) formation of complexes (1) catalytic activity (4 max 3) [3]

(b) (green is [Ni(H2O)6]

2+) ppt is Ni(OH)2 (1) blue solution is [Ni(NH3)6]

2+ or [Ni(NH3)4]2+ or [Ni(NH3)4(H2O)2]

2+ (1) formed by ligand exchange (1)

Ni2+ + 2OH → Ni(OH)2 (1)

Ni(OH)2 + 6NH3 → [Ni(NH3)6]2+ + 2OH (1) [4]

(5 max 4) (c) Mr = 58.7 + 48 + 6 + 28 + 32 = 172.7 (173) (1) n(Ni) = 4.00 / 172.7 = 0.0232 mol (1) mass(Ni) = 0.0232 × 58.7 = 1.36g percentage = 100 × 1.36 / 3.4 = 40.0% (1) [3]

[Total: 10]

3 (a) PbO2 decomposed into PbO (and O2). (SnO2 is stable) [1] (b) (i) PbCl4 dissociates into Cl2 and PbCl2 (white solid)

or PbCl4 → PbCl2 + Cl2 or in words (1) (1)

Cl2 + 2KI → 2KCl + I2 (1)

Eo(Cl2/Cl ) is more positive than Eo(I2/I ) (1) (ii) SnCl4 is more stable than PbCl4 / answers using Eo accepted (1) (5 max 4) [4] (c) (i) .. .. Cl:C:Cl or Cl=C–Cl (1) +

bent or non-linear or angle = 100–140o (1)

(ii) CCl2 + H2O → CO + 2HCl (1) [3]

[Total: 8]

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4 (a) hydrogen bonding (1) diag: NH2CH2CH2OH---OHCH2CH2NH2 or NH2CH2CH2OH---NH2CH2CH2OH (i.e. H-bond from OH group to either OH or NH2) (1) [2] (b) propylamine is more basic than phenylamine (1) because lone pair on N is delocalised over ring in phenylamine (so less available for

protonation) or the propyl group is electron-donating, so the lone pair is more available (1) [2]

(c) HOCH2CH2NH2 + H+ → HOCH2CH2NH3

+

or HOCH2CH2NH2 + HCl → HOCH2CH2NH3

+Cl

or HOCH2CH2NH2 + H2O → HOCH2CH2NH3

+OH (reaction with any acceptable Bronsted acid accepted) [1] (d) (i) X is CH3CH2CN (1) (ii) step 1 is KCN in ethanol, heat [HCN negates] (1) step 2 is H2+Ni / Pt or LiAlH4 or Na in ethanol [NOT NaBH4 or Sn/HCl] (1) [3]

(e) ethanolamine: Na effervescence / bubbles produced

or Cr2O7

2 / H+ colour turns from orange to green or MnO4

/ H+

purple colour disappears or PCl3 / PCl5 / SOCl2 (1) steamy fumes (1) phenylamine: Br2(aq) decolourises / white ppt formed or HNO2 / H

+ at T<10oC, then phenol in NaOH (1) coloured dye formed (1) [4] [Total: 12]

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5 (a) (i) Eo = 0.40 – (–0.83) = 1.23V (1)

(ii) 2H2 + O2 → 2H2O (1) (iii) LH electrode will become more negative (1) RH electrode will also become more negative / less positive (1) (iv) no change ecf from (iii) (1) (v) increased conductance or lower cell resistance or increased rate of reaction (1) [6] (b) (i) Eo = 1.47 – (–0.13) = 1.60V (1)

(ii) PbO2 + Pb + 4H+ → 2Pb2+ + 2H2O (1)

(iii) PbO2 + Pb + 4H+ + 2SO4

2 → 2PbSO4(s) + 2H2O (1) (iv) Eo

cell will increase (1) as [Pb2+] decreases, Eelectrode(PbO2) will become more positive, but Eelectrode(Pb)

will become more negative (1) [5] [Total: 11] 6 (a) (i) SOCl2 or PCl5 or PCl3 (1)

(ii) CH3CO2H + SOCl2 → CH3COCl + SO2 + HCl

or CH3CO2H + PCl5 → CH3COCl + POCl3 + HCl

or 3CH3CO2H + PCl3 → 3CH3COCl + H3PO3 (1) [2]

(b) (i) A is C6H5CO2C2H5 (1) B is C6H5CONH2 (1) (ii) ester (1) amide (1) (iii) nucleophilic substitution / condensation (1) [5] (c) (i) C is ClCOCOCl (1) D is ClCOCOCOCl (1) (ii) hydrogen bonding (1) (iii) because it’s an amide or not an amine or its lone pair is delocalised (over C=O)

or less available due to electronegative oxygen [NOT: E is neutral, but the diamine is

basic] (1) (iv) condensation (polymer) or polyester (1) [5] [Total: 12]

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7

C 2

Cl

NO2

HO2C H3C OCO

CO2H COOCH3

(ignore orientation) (ignore orientation of both)

H Cl C 2H OH

Cl

[6]

[Total: 6] 8 (a)

Block letter Identity of compound

J Deoxyribose (NOT “sugar” or “pentose”)

K Guanine

L Phosphate

M Thymine

All 4 correct score 3 marks, 3 score 2, 2 score 1 [3] (b) hydrogen bonds (1) between the bases (1) [2] (c) 1 RNA is a single strand; DNA is double strand (1) 2 RNA contains ribose; DNA contains deoxyribose (1) 3 RNA contains uracil; DNA contains thymine (1) 4 RNA is shorter than DNA (1) (4 max 3) [3] (d) mRNA – copies the DNA gene sequence or forms a template for a particular polypeptide / in protein synthesis (1) tRNA – carries amino acids to the ribosome (1) [2] [Total: 10]

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9 (a) spinning proton produces two spin states / magnetic moments (1) these can align with or against an applied magnetic field (1) [2] (b) field experienced by protons is influenced by adjacent atoms / protons are in two

different chemical environments (1) peaks are in the area ratio 3 : 1 (methyl to –OH protons) or are at 0.5 – 6.0δ and 3.3 – 4.0δ (1) [2] (c) (i) propanoic acid methyl ethanoate ethyl methanoate all for (2) two for (1) (ii) compound is CH3CO2CH3 or methyl ethanoate (1) the other two compounds each have 3 different proton environments, but the

spectrum shows only 2 peaks. (1) A is OCH3, B is CH3CO (1) (iii) compound – propanoic acid or ethyl methanoate the –OH proton or the H–CO proton (1) [6] (d) (i) distance between atoms / bond lengths / bond angles (1) (ii) hydrogen atoms (1) [2] [Total: 12 max 10] [Total: 10]

CH3CH2CO2H CH3CO2CH3 HCO2CH2CH3

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10 (a) ester or amide (allow nitrile) [1]

(b)

amide (1) + any one ester (1) [2] allow whole groups circled (c) (i) hydrophilic drug at C (1) hydrophobic drug at B both needed (1) (ii) (at A) the drug would be exposed to attack / breakdown / digestion (1) [3] (d) (i) at one of the –OH groups (1) (ii) volume of sphere can be large or one PEG molecule can only carry 1 or 2 drug

molecules (1) or can carry different types of drug [2] (e) more economic (1) less chance of side-effects / side effects reduced / less chance of allergic reaction (1) less risk of harming healthy tissue / organs / less chance of an overdose (1) (3 max 2) [2] [Total: 10]

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9701 CHEMISTRY





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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)

RCl

C

(H) (H)

(+)

(-)

HO

R

C

(H) (H)

ClHO

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]

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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]

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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]

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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

(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)

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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]

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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

OH

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]

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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]

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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]

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