Chemistry, Third edition

ANSWERS TO EXERCISES AND REVISION QUESTIONS

Revision questions

7.1 (i)C  4

(ii)Mn  4

(iii)S  6

(iv)Sn  4

(v)N  3

7.2

(ii)yes

Cu^2 +(aq)2e– Cu(s) 0.34

Fe^3 +(aq)e– Fe^2 +(aq) 0.77

7.3 (i)Cu^2 +(aq)Fe(s)Fe^2 +(aq)Cu(s)

(ii)no reaction

7.4 (i)Au (oxidation number 0 to 1)

(ii)manganese in MnO 4 – (7 to 2)

(iii)MnO 4 – it is reduced to Mn^2 +.

7.5SO 2 (g)2H 2 O(l)2Fe^3 +(aq)SO 42 – (aq)

4H+(aq)2Fe^2 +(aq)

7.6Find out the reaction:

Al^3 +(aq)3e– Al(s) 1.67

Fe^2 +(aq)2e– Fe(s) 0.44

(i)From the Al electrode to the Fe electrode.

(ii)The reactions occurring are:

Al(s)Al^3 +(aq)3e–

Fe^2 +(aq)2e–Fe(s)

(iii)Al^3 +(aq)3e–Al(s) E^ —1.67

Fe^2 +(aq)2e–Fe(s) E^ —0.44

EcellE^ —R E^ —L

0.44(1.67)

1.23V

(iv)Al(s) | Al^3 +(aq) || Fe^2 +(aq) | Fe(s).

7.7 (i)Ag+, S^2 –

(ii)Yes, silver ions are reduced to silver by aluminium,

which is oxidized to aluminium ions.

(iii)Polish rubs off the sulfide layer, this method

‘regenerates’ the silver metal and it is not lost.

7.8MnO 2 (s)4H+(aq)2Cl–(aq)Mn^2 +(aq)

2H 2 O(l)Cl 2 (g)

7.9Zn4OH– [Zn(OH) 4 ]^2 – 2e–(oxidation)

NO 3 – 6H 2 O8e–NH 3 9OH–(reduction).

Multiply the oxidation reaction by 4 and add both half

equations then simplify.

4Zn(s)NO 3 – (aq)7OH–(aq)6H 2 O(l)

4[Zn(OH) 4 ]^2 – (aq)NH 3 (g)

7.10 (i)Ni^2 +(gains electrons and is reduced)

(ii)Zn (loses electrons and is oxidized).

7.11Chromium plate provides physical protection for

the iron against water and oxygen. Once it is scratched,

the chromium will lose electrons in preference to the

iron (check using anticlockwise arrows) forming a

protective layer of chromium oxide that covers the

chromium metal.

Unit 8

Exercises

8A

(i)63 u (ii)120 u (iii)26 u (iv)256 u

(v)88 u (vi)404 u (vii)20 u.

8B

(i)20 g (1 mol) (ii) 12  1023 (iii)2.4 g (0.1 mol)

(iv)159.5 g.

8C

1. (i)14 g (ii)224 g (iii)53.3 g (iv)2520 g.


2. (i)1 mol (ii)0.80 mol (iii)5.1 10 –^3 mol
   (iv)0.50 mol.
   3.Na 2 CO 3 (0.01 mol).
   4. (i)3.8 1022 (0.063 mol)
   (ii)6.6 1020 (5.5 10 –^4 mol Al 2 O 3 ).
   8D
   Volume of oil in drop   (d/2)^2  thickness of layer
   (1 molecule):
   2.54 10 –^5   102  h
   h8.1 10 –^8 cm
   Volume of 1 molecule h^3 (8.1 10 –^8 )^3
   5.3 10 –^22 cm^3.
   NA282/(0.891 5.3 10 –^22 ) 6  1023 mol–^1.
   8E
   (i)23% (ii)16% (iii)68% (iv)80%
   (v)73%.
   8F
   (i)CH 4 (ii)Al 2 O 3 (iii)Fe 3 O 4 (iv)Na 2 SO 4.
   8G
   (i)C 2 H 6 (ii)C 5 H 7 N (empirical formula)
   C 10 H 14 N 2 (molecular formula)
   8H
   CuSO 4 ·5H 2 O CuSO 4 ·3H 2 O 2H 2 O
   CuSO 4 ·3H 2 O CuSO 4 ·H 2 O 2H 2 O
   CuSO 4 ·H 2 O CuSO 4  H 2 O
   CuSO 4  CuOSO 3.
   8I
   (i)2g (ii)0.32 g (iii)38 g
   (iv)C(s)O 2 (g)CO 2 (g), 0.60 g
   (v)CuCO 3 (s)CuO(s)CO 2 (g), 0.82 g.
   8J
   (i)6dm^3 (ii)1.6 g (iii)4.8 dm^3 (iv)10 dm^3
   (v)80 cm^3 (vi)7.5 g.
   8K
   56%.
   8L
   (i)oxygen, 0.1 mol
   (ii)magnesium, 2.1 g
   (iii)copper(II) oxide, 2.4 g.

Revision questions

8.1 (i)28 u (ii)134.5 u (iii)98 u (iv)164 u

(v)322 u.

8.2 (i)35.5 g (ii)71 g (iii)124 g (iv)127 g

(v)720 g.

8.3 (i)0.94 mol (ii)0.25 mol (iii)0.25 mol

(iv)0.051 mol (v)10 mol.

8.4 (i)29.3 g (ii)11 g (iii)4g (iv)624 g

(v)6300 g.

8.57.5 g

There are 2.00/17 mol NH 3. 1 mol NH 3 has the same

number of molecules as there are in 1 mol SO 2.

2.00/17 mol NH 3 has the same number of molecules as

there are in 2.00/17 mol SO 2 or 2.00/17 64 g SO 2 

7.5 g SO 2.

8.6 (i)6g

1 mol Mg reacts with 1 mol S; 24 g Mg reacts with 32 g S;

6 g Mg reacts with 8 g S.

8.7 (i)0.8 g:

2 mol NaNO 3 produces 1 mol O 2 ; 170 g NaNO 3 produces

32 g O 2 ; 4.25 g NaNO 3 produces 32/170 4.25 g O 2.

(ii)0.6 dm^3 :

2 mol NaNO 3 produces 24 dm^3 O 2 ; 170 g NaNO 3

produces 24 dm^3 O 2 ; 4.25 g NaNO 3 produces 24/170 

4.25 dm^3 O 2.

8.8 (i)2NaHCO 3 (s)Na 2 CO 3 (s)H 2 O(l)CO 2 (g)

(ii)168 g (iii)44 g, 24 dm^3

(iv)0.60 dm^3 (600 cm^3 ).

8.951%:

7 molecular mass of water

molecular mass of MgSO 4 ·7H 2 O^100

^7246 ^18  100

8.10C 2 H 4 (the simplest, or empirical, formula is CH 2 ).

8.11Amount of Al in moles 13.50/270.50 mol.

Mass of Cl combined with Al 66.7513.5053.25 g.

Amount of Cl in moles53.25/35.51.50 mol.

Ratio of the number of moles Al atoms to number of moles

Cl atoms is 0.50:1.50 or 1:3. Formula is AlCl 3.

Unit 9

Exercises

9A

(i)10.0 mol dm–^3 (ii)0.008 mol dm–^3

(iii)50 dm^3 (iv)28.6 g.

9B

(i)K+and Cl–both 0.1 mol dm–^3

(ii)Na+and CO 3 2–are 0.250 mol dm–3and

0.125 mol dm–3, respectively

(iii)NH 4 +and SO 4 2–are 0.50 mol dm–3and

0.25 mol dm–3, respectively

(iv)Cu2+and NO 3 – are 0.15 mol dm–3and

0.30 mol dm–3, respectively.

9C

(i)0.05 mol dm–^3 (ii)0.10 mol dm–^3

(iii)0.20 mol dm–^3 (iv)0.40 mol dm–^3

(v)1.00 mol dm–^3.

9D

(i)0.53 g

Amount in mol volmolar conc 500/10000.01

5.0 10 –^3 mol

5.0 10 –^3 M(Na 2 CO 3 )g 0.53 g.

(ii)it would be more concentrated than the value

calculated

(iii)0.02 mol dm–^3

amount in mol present volmolar conc 50/1000

0.1 5  10 –^3 mol

if this amount is made up to 250 cm^3 , then

molar conc amount in mol/vol

 5  10 –^3 /0.250.02 mol dm–^3

(iv)125 cm^3

Number of moles needed volconc 1  2 2 mol

volnumber mol/conc 2/160.125 dm^3

(v)0.05 dm^3.

9E

(i)50 cm^3

1 mol H 2 SO 4 2 mol KOH

volmolar concentration H 2 SO (^4)  1
volmolar concentration KOH 2
V1.0 1
25/10004.0 2
V0.050 dm^3 (50 cm^3 )
(ii)0.071 mol dm–^3 (iii)0.263 mol dm–^3
(iv)20 cm^3.
9F
(i)6Fe^2 +(aq)Cr 2 O 72 – (aq)14H+(aq)6Fe^3 +(aq)
2Cr^3 +(aq)7H 2 O(l)
6 mol Fe^2 +1 mol Cr 2 O 72 – , so that
volmolar concentration Fe^2 + 6
volmolar concentration Cr 2 O 72 – 1
25/1000C  6
13.5/10000.02 1
C = 0.065 mol dm–^3
(ii)50 cm^3
(iii)210 cm^3
The equation is:
3MnO 4 – 5Fe^2 +5C 2 O 42 – 24H+
3Mn^2 +5Fe^3 +10CO 2  12H 2 O
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