Moles WS 1

1 B

alt
2 B

3 D

4 C

5 B

6 D

7 B

8 B

9 A

10 B

11 A

12 C

13 B

14 D

15 C

16 C

17 A

18 C

19 C

20 A

21 B

22 C

23 C

24 D

25 B

26 A

27 B

28 D

29 C

30 C

Bilal Hameed 1 Moles WS 1

 31 A

32 D

alt
Moles WS 1 2 Bilal Hameed
 (iii) sodium has mobile / free electrons / electrons free (to move throughout the structure) [1]
[2]
phosphorus is simple / covalent / molecular [1]

(iv) magnesium has two free / delocalised / outer / valence electrons per atom [1]
OR [1]
more free / delocalised / outer electrons than sodium

Moles WS 2
(b) (i) A = Mg(NO3)2 [1]
B = H2 [1]
[4]
C = NO2 OR O2 [1]
D = O2 OR NO2 [1]

(ii) any Group I carbonate OR ammonium carbonate [1] [1]

1 2016 MAR P22 Q2 [12]

alt
27.30
2 (a) (i) 0.0203 =5.46 × 10–4!(mol)
×Page Mark Scheme Syllabus Paper [1] [1]
1000 Cambridge International AS/A Level – March 2016 9701 22
(ii) (i) × 6 =3.28 × 10–3!(mol) [1] [1]
Question Answer Mark Total
© Cambridge International Examinations 2016
250
(iii) (ii) × =3.28 × 10–2!(mol) [1] [1]
25.00

(iv) Mr of FeCO3 =55.8 + 12.0 + 3(16.0) = 115.8 [1]

[2]
(iii) × Mr(FeCO3) =3.79 g [1]

(iv)
(v) ×100% = 75.9% [1] [1]
5.00

(b) (i) 2Fe3+ + Sn2+ → 2Fe2+ + Sn4+

species [1] [2]
balancing [1]

(ii) SnCl 2(aq) + 2HgCl!2(aq) !" SnCl 4(aq) + Hg2Cl 2(s)

[2]
SnCl 2 AND 2 [1]
state symbols [1]

[10]

3 (a) (i)

[3]
three bonding pairs [1]
lone pair AND octet [1]
shape = (trigonal) pyramidal [1]

© Cambridge International Examinations 2016

Bilal Hameed 3 Moles WS 2

 Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – May/June 2016 9701 23

2 2016
QuestionJUN P23 Q1 Mark Scheme Mark Total

1 (a) (i) Fe + H2SO4 → FeSO4 + H2 [1] [1]

(ii) Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O [1] [1]

–4
(iii) (0.025 × 32.0/1000=) 8 × 10 [1] [1]

alt
–4 –3
(iv) (8 × 10 × 6 =) 4.8 × 10 [1] [1]
–3 –2
(v) (4.8 × 10 × 250/25.0=) 4.8 × 10 [1] [1]
–2
(vi) (4.8 × 10 × 55.8=) 2.68 / 2.678 [1] [1]

(vii) (2.68/3.35=) 80% [1] [1]

(b) (i) covalent [1] [2]

small(er) difference in electronegativity between Fe and Cl (than between Al and Cl) [1]

(ii) FeCl 3 + 6H2O → [Fe(H2O)6]3+ 3Cl – OR [1] [1]

FeCl 3 + 6H2O → [Fe(H2O)6OH]2+ + H+ + 3Cl –

[10]

2 (a) NH3 + HNO3 → NH4NO3 [1] [1]

(b) (i) line from origin AND below left-hand end of original with peak to right of and lower than original [1] [2]
crosses original once AND above right-hand end of original AND above energy axis [1]

(ii) (curves show) more molecules with E > Ea (at higher T) [1] [2]
so greater frequency of successful (owtte) collisions / more successful (owtte) collisions per unit time [1]

(iii) catalysed Ea shown to left of original on horizontal axis [1] [2]

so more molecules with E > Ea (in presence of catalyst) [1]

© Cambridge International Examinations 2016

Moles WS 2 4 Bilal Hameed

 Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9701 21

3 2016 NOV P21 Q1

Question Answer Marks

3
1(a) 6 10 (mol) 1 1

1(b) NaOH + HCl NaCl + H2O 1 1

3
1(c) 6 10 (mol) 1 1
3
1(d) 1 1

alt
4 10 (mol)
3
1(e) 4 10 (mol) 1 1
3
1(f) 1 10 (mol) 1 1

1(g) 170 1 1

1(h) 28(.0) 1 2
Si/silicon 1

Total: 9

© UCLES 2016

Bilal Hameed 5 Moles WS 2

 Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9701 22

4 2016 NOV P22 Q1

Question Answer Mark

1(a) 0.04 OR 4 10–2 1

1(b)(i) Na2CO3 + 2HCl 2NaCl + CO2 + H2O 1

–4
1(b)(ii) 0.00075 OR 7.5 10 1

alt
1(b)(iii) 0.0015 OR 1.5 10–3 1
–2
1(b)(iv) 0.015 OR 1.5 10 1
–2
1(b)(v) 0.025 OR 2.5 10 1
–2 –2
1(b)(vi) 0.0125 OR 1.25 10 OR 0.013 OR 1.3 10 1

1(b)(vii) 40 1

Ca / calcium 1

Total: 9

© UCLES 2016

Moles WS 2 6 Bilal Hameed

 Bronsted-Lowry acid
is a proton donor / H+ (ion) donor / hydrogen ion donor

2c(ii) NH4+(aq) + H2O(l) NH3(aq or g) + H3O+(aq)

all correct species and balancing 1

correct state symbols 1

5 2018 JUN P22 Q2
2(d)(i) MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O 1
2+
2(d)(ii) (Fe is a) reducing agent / reductant 2
2+ 3+
provides/donates electron(s) / loses electron(s) / increases its oxidation number / (Fe ) becomes Fe

alt
2(d)(iii) 4 × 10−4 / 0.0004 1

2(d)(iv) 2 × 10-3 / 0.002 1

2(d)(v) 392 1

2(d)(vi) 6 1

© UCLES 2018 Page 6 of 10

Bilal Hameed 7 Moles WS 2

 Moles WS 3
1 C

alt
2 D

3 B

4 D

5 C

6 C

7 B

8 D

9 B

10 C

11 C

12 C

13 A

14 C

15 D

16 D

17 A

18 D

19 A

20 A

21 B

22 D

23 C

24 A

25 B

26 C

27 C

28 C

29 C

30 B

Moles WS 3 8 Bilal Hameed

 31 A

alt
Bilal Hameed 9 Moles WS 3
 Page 3
Moles WS 4
Mark Scheme Syllabus Paper
GCE AS LEVEL – May/June 2014 9701 23
1 [2014 JUN P23 Q2]

alt
2 (a) NH4+ + OH– ! NH3 + H2O (1) [1]

(b) (i) Initial acid = 40 × 0.4 / 1000 = 0.016 (mol) (1) [1]

(ii) 25 x 0.12 = 3.0 × 10–3 (mol) (of OH– used) (1)

1000 [1]

(iii) excess acid = OH– = 0.003 (1)

acid reacted = 0.016 – 0.003 = 0.013 (mol) [1]

(iv) NH4+:H+ = 1:1 so = 0.013 (mol NH4+) (1) [1]

(v) amount of Cu = mass / Mr = 0.413 / 63.5 = 6.5 × 10–3 (mol) (1)

so Cu:NH4 = 0.0065:0.013 = 1:2 so x = 2 (1) [2]

(vi) Mr = 399.7 (1) [1]

Total 8

© Cambridge International Examinations 2014

Moles WS 4 10 Bilal Hameed

 GCE AS/A LEVEL – May/June 2014 9701 22

Question Answers Mark Total

(e) shape of PCl 5 =(trigonal) bipyramid(al) 1

bond angles in PCl 5 = 120° and 90° 1 2

2 [2014 JUN P22 Q2] 17

2–
2 (a) (i) (The C2O4 ions) lose electrons owtte / ora 1 1
– 2– + 2+
(ii) 2MnO4 (aq) + 5C2O4 (aq) + 16H (aq) → 2Mn (aq) + 10CO2(aq) + 8H2O(l) 1+1+1 3

20.0 × 0.100 = 2(.00) × 10–3 (mol)

(b) (i) 1 1

alt
1000

(ii) MnO4! : C2O42! = 2 : 5

so amount of C2O42! = (5 / 2) × 2.00 × 10!3 = 5(.00) × 10!3 (mol) 1 1
ecf from (b)(i)

(iii) 5.00 × 10!3 × 250 / 25 = 0.05(0) (mol) 1 1

ecf from (b)(ii)

(iv) amount = mass / Mr so Mr = mass / amount = 6.30 / 0.05 = 126 1 1

ecf from (b)(iii)

(v) 126 – 90 = 36
36 / 18 = 2.00
x=2 1 1
Ecf from (b)(iv) if suitable

© Cambridge International Examinations 2014

Bilal Hameed 11 Moles WS 4

 (ii) NaCl is ionic AND giant / lattice 1
NaCl dissolves / does not react 1
SiCl 4 is covalent AND molecular / simple 1
SiCl 4 is hydrolysed / reacts 1 4

(e) shape of SF6 = Octahedral 1

bond angle = 90° 1 2
3 [2014 JUN P21 Q2] 18

2 (a) (i) !
(The MnO4 ions cause the Fe 2+
ions to) lose electrons owtte / ora 1 1

(ii) MnO4–(aq) + 5Fe2+(aq) + 8H+(aq) ! Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) 1+1+1 3

alt
20.0 × 0.020 = 4(.00) ×10 − 4 (mol)
(b) (i) 1000 1 1

(ii) MnO4– : Fe2+ = 1 : 5 1 1

so amount of Fe2+ = 5 × 4.00 × 10!4 = 2(.00) × 10!3 (mol)
ecf from (b)(i)
Page 4 Mark Scheme Syllabus Paper
(iii) 2.00 × 10!3 × 250 / 25 = 0.02(00)
GCE AS/A LEVEL – May/June 2014
(mol) 9701 21 1 1
ecf from (b)(ii)
Question Mark Scheme Mark Total
© Cambridge International Examinations 2014
(iv) 3.40 / 0.02 = 170 1 1
ecf from (b)(iii)

(v) 170 – 151.8 = 18.2 1 1

18.2 / 18 = 1.01
x=1
ecf from (b)(iv) if appropriate

9
– –
3 (a) (i) K = Cl / chloride / F / fluoride 1

H2SO4 + 2NaCl ! Na2SO4 + 2HCl (or equation with F or K for Cl) OR 1

H2SO4 + NaCl ! NaHSO4 + HCl (or equation with F or K for Cl )

ecf from identity of K so long as halide

HK is acidic / HK is a gas / an acidic gas is produced 1 3
–
(ii) L = I / iodide 1

colour = yellow 1
ecf from identity of L i.e. Cl – (white) or Br– (cream)
Ag+ + I! ! AgI (or equation with L) 3
1
AgNO3 + NaI → AgI + NaNO3 (or equation with L)
ecf from identity of L so long as halide

(iii) Br2 / bromine has fewer electrons than iodine / more electrons than chlorine 1
intermolecular / van der Waals’ forces (in Br2 / M2) weaker than in iodine / stronger than
in chlorine 1 2

(b) (i) B = chlorine / Cl 2 1

C = hydrogen / H2 1
D = sodium hydroxide / NaOH 1 3

© Cambridge International Examinations 2014

Moles WS 4 12 Bilal Hameed

 Page 3 Mark Scheme Syllabus Paper
4 [2013 JUN P22 Q2]
GCE AS/A LEVEL – May/June 2013 9701 22

25.0 × 1.00
2 (a) (i) n(H2SO4) = = 0.025 mol (1)
1000
16.2 × 2.00
(ii) n(NaOH) = = 0.0324 mol (1)

alt
1000
0.0324
(iii) n(H2SO4) reacting with NaOH = = 0.0162 mol (1)
2
(iv) n(H2SO4) reacting with NH3 = 0.025 - 0.0162 = 0.0088 mol (1)
(v) n(NH3) reacting with H2SO4 = 2 x 0.0088 = 0.0176 mol (1)
(vi) n(NaNO3) reacting = n(NH3) produced = 0.0176 mol (1)
(vii) mass of NaNO3 that reacted = 0.0176 x 85 = 1.496 g (1)
1.496 × 100
(viii) % of NaNO3 = = 91.2195122 = 91.2
1.64
give one mark for the correct expression (1)
give one mark for answer given as 91.2 – i.e to 3 sig. fig. (1)
allow ecf where appropriate
[9]

(b) NaNO3 +5 and NH3 -3 both required (1) [1]

[Total: 10]

© Cambridge International Examinations 2013

Bilal Hameed 13 Moles WS 4

 Page 2 Mark Scheme Syllabus Paper
5 [2013 JUN P21 Q1]
GCE AS/A LEVEL – May/June 2013 9701 21

1 (a) (i) NaOH + HCl → NaCl + H2O (1) [1]

(NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O (1) [1]

alt
allow ionic equations in each case

39.2 × 2.00
(ii) n(NaOH) = n(HCl) = = 0.0784 (1) [1]
1000

29.5 × 2.00
(iii) n(NaOH) = n(HCl) = = 0.059 (1) [1]
1000

(iv) n(NaOH) = 0.0784 – 0.059 = 0.0194 (1) [1]

0.0194
(v) n[(NH4)2SO4] = = 9.7 × 10-3 (1) [1]
2

(vi) mass of (NH4)2SO4 = 9.7 × 10-3 × 132.1 = 1.2814 g (1)

(vii) % of (NH4)2SO4 = 1.2814 × 100 = 43.30405405 = 43.3

2.96
give one mark for the correct expression (1) [1]
give one mark for answer given as 43.3 – i.e. to 3 sig. fig. (1) [1]
allow ecf where appropriate [9]

(b) fertiliser in the river causes

excessive growth of aquatic plants/algae or algal bloom (1) [1]
when plants and algae die O2 is used up or fish or aquatic life die (1) [2]

(c) manufacture of HNO3 or explosives or nylon or

as a cleaning agent or as a refrigerant
not detergent (1) [1]

[Total:12]

© Cambridge International Examinations 2013

Moles WS 4 14 Bilal Hameed

 Page 2 Mark Scheme Syllabus Paper
6 [2012 NOV P21 Q1]
GCE AS/A LEVEL – October/November 2012 9701 21

1 (a) ZnCO3 Zn(OH)2 ZnO

not Zn or other compounds of Zn (any 2) [2]

alt
(b) (i) to ensure all of the water of crystallisation had been driven off or
to be at constant mass (1)

(ii) mass of ZnSO4 = 76.34 – 74.25 = 2.09 g (1)

Mr ZnSO4 = 65.4 + 32.1 + (4 × 16.0) = 161.5

allow use of Zn = 65 and/or S = 32 to give values between 161 and 161.5 (1)

n(ZnSO4) = 2.09 = 0.01294 = 1.29 × 10–2

161.5

ZnSO4 = 161 gives 1.30 × 10–2 (1)

(iii) mass of H2O driven off = 77.97 – 76.34 = 1.63 g (1)

n(H2O) = 1.63 = 0.0905 = 9.1 × 10–2

18 (1)

(iv) 1.29 × 10–2 mol ZnSO4 are combined with 9.1 × 10–2 mol H2O

1 mol ZnSO4 is combined with 9.1 × 10–2

1.29 × 10–2

= 7.054 ≡ 7 mol H2O

answer must be expressed as a whole number

allow ecf on candidate’s answers to (b)(ii) and (b)(iii) (1) [7]

(c) (i) n(Zn) = n (CH3CO2)2Zn.2H2O (1)

n(Zn) = 0.015 = 2.290 × 10–4

65.4

= 2.29 × 10–4 (1)

mass of crystals = 2.29 × 10–4 × 219.4 = 0.0502655 g

= 0.05 g = 50 mg (1)

(ii) concentration of (CH3CO2)2Zn.2H2O = 2.29 × 10–4 = 0.0458

0.005
= 4.58 × 10–2 mol dm–3 (1)

allow correct answers if Zn = 65 is used [4]

[Total: 13]

© Cambridge International Examinations 2012

Bilal Hameed 15 Moles WS 4

 Page 3 Mark Scheme: Teachers’ version Syllabus Paper
7 [2012 JUN P23 Q02]
GCE AS/A LEVEL – May/June 2012 9701 23

2 (a) (i) Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (1)

35.8
(ii) n(HCl ) = × 0.100 = 3.58 × 10–3 (1)
1000

alt
35.8
(iii) n(Na2CO3) = × 10–3 = 1.79 × 10–3 mol in 25.0 cm3 (1)
2

(iv) n(Na2CO3) = 1.79 × 10–3 × 10 = 1.79 × 10–2 mol in 250 cm3 (1)

(v) mass of Na2CO3 = 1.79 × 10–2 × 106 = 1.90g

Mr of Na2CO3 = 106 (1)
mass of Na2CO3 = 1.90 g (1) [6]

5.13 − 1.90
(b) n(H2O) in 5.13 g of washing soda = = 1.79 × 10–1 mol (1)
18
n(Na2CO3) in 5.13 g of washing soda = 1.79 × 10–2 mol
n(H2O) : n(Na2CO3) = 10 : 1 (1)

or
1.90 g Na2CO3 are combined with 3.23.g H2O
3.23 × 106
106 g Na2CO3 are combined with = 180.2 g H2 (1)
1.90
this is 10 mol of H2O (1)

or
1.79 × 10–2 mol Na2CO3.xH2O ≡ 5.13 g of washing soda
5.13
1 mol Na2CO3.xH2O ≡ = 286.6 g (1)
1.79 × 10 −2
Na2CO3 = 106 and H2O = 18 hence x = 10 (1) [2]

[Total: 8]

© University of Cambridge International Examinations 2012

Moles WS 4 16 Bilal Hameed

 (1)

(ii) 180° (1) [2]

[Total: 15]
8 [2012 JUN P21 Q2]
2 (a) (NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O
correct products (1)
correctly balanced equation (1) [2]

alt
(b) (i) NaOH + HCl → NaCl + H2O (1)

31.2
(ii) n(HCl) = × 1.00 = 0.0312 = 0.03 (1)
1000

50.0
(iii) n(NaOH) = × 2.00 = 0.10 (1)
1000

(iv) n(NaOH) used up = 0.10 – 0.0312 = 0.0688 = 0.07 (1)

0.0688
(v) n[(NH4)2SO4] = = 0.0344 = 0.03 (1)
2

(vi) mass of (NH4)2SO4 = 0.0344 × 132 = 4.5408 = 4.54 (1)

4.5408 × 100
(vii) percentage purity = = 90.816 = 90.8 (1) [7]
5.00

[Total: 9]

© University of Cambridge International Examinations 2012

Bilal Hameed 17 Moles WS 4

 Page 2 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2011 9701 23
9 [2011 JUN P23 Q1]
1 Throughout this question, deduct one mark only for sig. fig. error.

(a) (i) the volume of solution A present in one ‘typical ant’ is

7.5 x 10-6 x 1000 = 7.5 x 10-3 cm3 (1)

alt
(ii) the volume of pure methanoic acid in one ‘typical ant’ is
7.5 x 10-3 x 50 = 3.75 x 10-3 gives 3.8 x 10-3 cm3
100

allow ecf on (i) (1)

(iii) no. of ants = 1000 = 263157.8947 gives 2.6 x 105

3.8 x 10-3

use of 3.75 x 10-3 gives 266666.6667 = 2.7 x 105 (1) [3]

(b) (i) the volume of solution A, in one ant bite is

80 x 7.5 x 10-3 = 6.0 x 10-3 cm3
100

allow ecf on (a)(i) (1)

the volume of pure methanoic acid in one bite is

50 x 6.0 x 10-3 = 3.0 x 10-3 cm3
100

allow ecf on first part of (b)(i) (1)

(ii) the mass of methanoic acid in one bite is

3.0 x 10-3 x 1.2 = 3.6 x 10-3 g

allow ecf on (b)(i) (1) [3]

(c) (i) HCO2H + NaHCO3 → HCO2Na + H2O + CO2 (1)

(ii) 46 g HCO2H ≡ 84 g NaHCO3 (1)

5.4 x 10-3 g HCO2H ≡ 84 x 5.4 x 10-3 g NaHCO3

46
= 9.860869565 x 10-3
= 9.9 x 10-3 g NaHCO3 (1) [3]

[Total: 9]

© University of Cambridge International Examinations 2011

Moles WS 4 18 Bilal Hameed

 100

gives 24.33 to 4 sig fig (same as data in question)

do not credit wrong number of sig figs or incorrect rounding up/down (1) [2]
10 [2010 NOV P23 Q1]
(d) Mg + Cl2 → MgCl2 (1) [1]

2.45

alt
(e) (i) n(Sb) = = 0.020 (1)
122

(ii) mass of Cl in A = 4.57 – 2.45 = 2.12 g (1)

4.57 − 2.45 2.12

n(Cl) = = = 0.06
35.5 35.5

allow ecf as appropriate (1)

(iii) Sb : Cl = 0.02 : 0.06 = 1:3

empirical formula of A is SbCl3 (1)

(iv) 2Sb + 3Cl2 → 2SbCl3 (1) [5]

(f) (i) ionic (1)

(ii) covalent (1)

not van der Waals’ forces [2]

[Total: 14]

© UCLES 2010

Bilal Hameed 19 Moles WS 4

 Page 3 Mark Scheme: Teachers’ version Syllabus Paper
GCE A/AS LEVEL – May/June 2009 9701 22
11 [2009 JUN P22 Q1]
0.72
(d) (i) n(Ti) = = 0.015 (1)
47.9

(2.85 − 0.72)

alt
(ii) n(Cl) = = 0.06 (1)
35.5

(iii) 0.015 : 0.06 = 1:4

empirical formula of A is TiCl4
Allow ecf on answers to (i) and/or (ii). (1)

(iv) Ti + 2Cl2 → TiCl4 (1)

Allow ecf on answers to (iii). [4]

(e) covalent/not ionic (1)

simple molecular or
mention of weak intermolecular forces or
weak van der Waals’s forces between molecules (1) [2]

[Total: 14 max]

2 (a) (i) Ca+(g) → Ca2+(g) + e– equation (1)

state symbols (1)

(ii) 590 + 1150 = +1740 kJ mol–1 (1) [3]

(b) (i) dissolves/vigorous reaction/

white or steamy fumes of HCl (1)
0–4 (1)

(ii) dissolves/vigorous reaction (1)

0–4 (1) [4]

(c) (i) P4S10 + 16H2O → 4H3PO4 + 10H2S (1)

(ii) P4S10 P is +5 (1)

H3PO4 P is +5 (1)

No because
there is no change in the oxidation no. of P (1)
ecf on answer to (c)(i)
and on calculated oxidation numbers [4]

[Total: 11]

© UCLES 2009

Moles WS 4 20 Bilal Hameed

 Page 5 Mark Scheme Syllabus Paper
12 [2007 NOV P2 Q3]
GCE A/AS LEVEL – October/November 2007 9701 02

(c) (i) S2Cl2 = (2 x 32.1) + (2 x 35.5) = 135.2

2.7
n(S2Cl2) = = 0.0199 = 0.02 (1)
135.2

alt
0.96
0.02 mol S2Cl2 → = 0.03 mol S
32.1

0.03 × 1.0
1.0 mol S2Cl2 → = 1.5 mol S (1)
0.02

(iii) 2S2Cl2 + 3H2O → 3S + H2SO3 + 4HCl

correct products (1)

balanced equation (1) [4]

(d) oxidation product is H2SO3 (1)

reduction product is S (1) [2]

[Total: 12]

4 (a)

H atoms must be shown.

Structure must not contain any CH3 groups (1) [1]

(b)

cis trans (1) [2]

(c) CH3CH(OH)CH2CH2CH3 (1)

CH3CH2CH(OH)CH2CH3 (1) [2]

© UCLES 2007

Bilal Hameed 21 Moles WS 4

 H2S has van der Waals’ forces only (1)

hydrogen bonds are stronger

than van der Waals’ forces or
H2S has weaker intermolecular bonds
than H2O (1) [4]
13 [2005 JUN P2 Q2]
(d) (i) 2H2S + 3O2 → 2H2O + 2SO2 (1)
from -2 (1) to +4 (1)
allow e.c.f. on equation

alt
(ii) 68.2g H2S react with 3 x 24 dm3 O2 (1)
8.65g H2S react with 3 x 24 x 8.65 = 9.13 dm3 (1)
68.2
allow 9.16 dm3 if H2S = 68 is used
allow e.c.f on (d)(i) [5]

(e) (i) an acid that is partially dissociated into ions (1)

(ii) H2S(g) + H2O(l) → H3O+(aq) + HS-(aq)

or

H2S(g) + aq → H+(aq) + HS-(aq)

or

H2S(aq) → H+(aq) + HS-(aq)

equation (1) state symbols (1) [3]

[Total: 17]

3 (a) A MgSO4
B MgCl2
C MgCO3
D MgO
E Mg(OH)2
F Mg(NO3)2

Accept name or formula

but penalise when name and formula do not agree (6 x 1) [6]

© University of Cambridge International Examinations 2005

Moles WS 4 22 Bilal Hameed

 (Any two) [2]

Total = [13]

14 [2004 JUN P2 Q2]

2 (a) 1s2 2s2 2p6 3s2 3p3 [1]

(b) 5 or V [1]

alt
(c) (i) 3NaOH + H3PO4 Na3PO4 + 3H2O [1]

(ii) (50 x 0.5) / 1000 = 0.025 (moles) [1]

(iii) conseq. on (i) 3 x .025 = 0.075 (moles) [1]

(d) (i) P4S3 + 8O2 P4O10 + 3SO2 balanced = 2 marks

(or 2P2O5)

OR + 6O2 P4O6 + 3SO2 unbalanced = 1 mark

(or 2P2O3)
[2]

(ii) P4O10 + 6H2O 4H3PO4 [1]

OR P4O6 + 6H2O 4H3PO3

SO2 + H2O H2SO3 [1]

(if SO3 then e.c.f.) Total = [9]

© University of Cambridge International Examinations 2004

Bilal Hameed 23 Moles WS 4