ANSWERS TO EXERCISES AND REVISION QUESTIONScm–^1 which is recognizable as characteristic of a
hydrogen-bonded–OH group. There is no carbonyl
absorption present, suggesting that the compound is an
alcohol or phenol. The parent molecular ion possesses a
molecular mass of 32 u (too low for a phenol). This
suggests that the alcohol is CH 3 OH. This contains only
two types of protons (as indicated by the NMR spectrum)
and the H:C ratio is 4:1 as required.
20.12 (i)Chloroform contains C–H bonds which would
absorb in the same region as the hydrocarbons. CF 3 CCl 3
contains no C–H bonds.
(ii)b20 mm 2.0 10 –^2 m.cA/ b0.050/(5.0
2.0 10 –^2 )0.50 mol m–^3 or 5.0 10 –^4 mol dm–^3.
M(C 8 H 18 )114 g mol–^1 .c 114 5.0 10 –^4 103
57 mg dm–^3.
20.13 (i)
CrO 42 – (aq)Cr^3 +(aq)
6 3 (reduced)
(ii)5.0 10 –^4 mol dm–^3 5.0 10 –^1 mol m–^3. Initial
absorbance cb 200 5.0 10 –^1 1.0 10 –^2
1.0. After reaction, the concentration of Cr^3 +(aq) is also
5.0 10 –^1 mol m–^3 ; no CrO 42 – (aq) remains. Final
absorbance c b 1.0 5.0 10 –^1 1.0 10 –^2
5.0 10 –^3.
20.14The flame atomizes the sample so that atoms of the
elements are produced. The flame does not produce the
excited states of atoms, and this function is carried out by
the light sources which are known as hollow cathode
lamps. A different lamp is required for each element.
20.151,2-dibromoethane, CH 2 BrCH 2 Br (all protons in the
same chemical environment, so only one resonance in the
NMR spectrum).
20.16Butanone, CH 3 COCH 2 CH 3 :
The mass spectrum shows the molecular ion at 72,
indicating a molecular formula of C 4 H 8 O.
The infrared spectrum shows a sharp peak at about
1700 cm–1, indicating a carbonyl group.
The^1 H-NMR has a quartet at 2.5 (CH 2 protons split into
a quartet by CH 3 ), a singlet at 2.0 (CH 3 attached to CO
and too far away from the other protons for them to split
the resonance) and a triplet at 1.0 (CH 3 split by CH 2
next to it).Unit 21
Exercises
21A(^21884) Po (^21482) X 2 – (^42) He 2 +
Examination of the periodic table shows that XPb.
21B
(^21583) Bi (^21584) X+ 10 e–
Inspection of a periodic table shows that XPo
(polonium).
21C
P and S are isotopes of the same element.
21D
From Chapter 20,
c/ 3.00 108 /10–^10 3.0 1018 Hz
EhNA3.99 10 –^13 3.0 1018
1.2 106 kJ mol–^1
106 kJ mol–^1.
21E
(^238) U has an extremely long half-life (Table 21.2). Its
decay would be represented by a horizontal straight line
as for^137 Cs in Fig. 21.3.
21F
(i)^235 U decays faster than^238 U, so that more^235 U
would have been present millions of years ago.
(ii)(a)
(^20187) Fr (^19785) At 2 – (^42) He 2 +
astatine
(^23190) Th (^23191) Pa+ 10 e–
protactinium
(b)
For francium, it is easy to show that after 10 half-lives
there is only a tiny fraction of radionuclide left. The
reduction in mass after 10 half-lives is actually 1/(2 2
2 2 2 2 2 2 2 2)1/2^10 1/1024.
If 0.0100 g is reduced by 1024, only about 1 10 –^5 g of
Fr would be left. In fact, 100 h represents 7.5 107 half-
lives of Fr! We can safely assume that after 100 h, all the
Fr would be gone.
For thorium, 100 h is four half-lives so that the original
mass will fall by a factor of 2 2 2 2 16. One-
sixteenth of 0.0100 g is 0.000 625 g.
These calculations may also be made using the
formulamtmoe–kt, with kbeing calculated from the
equation
t (^1) ⁄ 2 0.693k
For example, for thorium,
k0.693t 1 0.693/(25 60 60)7.7 10 –^6 s–^1.
⁄ 2
mtm 0 e–kt0.0100 e–(7.7^10 –^6 3.6^105 )
(where 100 h 3.6 105 s)0.0100 0.0625
0.000 625 g (as above).
Revision questions
21.1^147 N^42 He^2 +^178 O^2 +^11 H
21.2 (i)^31 H^32 He+ 10 e–
(ii)^63 Li^10 n^31 H^42 He
21.3From the half-life of^21084 Po (140 days) we calculate
thatk5.7 10 –^8 s–^1.
(a)t0.994.605/5.7 10 –^8 8.0 107 s or
926
days
(b)t0.9996.908/5.7 10 –^8 1.21 108 s or 1403
days
21.4Lots to think about here. Alternative fuels and their
availability, the pollution resulting from conventional
fuels against the hazards of nuclear power. The safety
aspects of nuclear fusion are discussed in
Environmental Physics, by E. Boeker and R. van
Grondelle, John Wiley, 1995.
21.5 (i)Radon gas.
(ii)The harm caused depends upon the activity of the
source (i.e the half-life and mass of the radionuclide), the
type of nuclear radiation, the distance of the individual
from the source, the health and age of the individual, the
area or part of body exposed, the duration of exposure,
the degree of protection provided etc.
Unit 22
Exercises
22A
(i)
0.012 g in 250 cm^3 is 0.048 g in 1 dm–^3 (or 48 ppm (mg
dm–^3 )).
0.048 g in 1 dm–^3 is
0.0487.7 10 – (^4) mol dm– 3
62
(ii)No.
(iii)
%N^2 ^1480 ^100 35%
22B
CO(g) + O 2 (g)CO 2 (g) – oxidation of CO
2NO(g)N 2 (g) + O 2 (g) – reduction of NO to N 2
22C
Nuclear power, wind energy, solar energy, using
hydrogen as a fuel (it burns to form H 2 O).
22D
(i)C 6 H 12 O 6 (aq)6O 2 (g)6CO 2 (g)6H 2 O(l)
(ii)waste contains 600 105 mg glucose or 600
102 /180 mol glucose. This needs
600 102 (^6) mol O
180 2
or
600 (^102) 6 32 g O
180 2 in 10^5 dm^3
which is 0.64 g in 1 dm^3 , or 640 mg dm–^3 (or ppm).
Revision questions
22.1
[Pb^2 +]10 mg dm–^3
^1010 ^1
1000 g dm
– (^3)
1000 207 mol dm
- 3
Rate4.8 10 –^5 mol dm–^3 min–^1
22.2 (i)C 8 H 18 (l) 121 ⁄ 2 O 2 (g)8CO 2 (g)9H 2 O(l)
(ii)200 g octane is 200/114 mol.
This would produce 200/114 8 mol CO 2 at 100%
efficiency. At 95% efficiency, 200/114 8 95/100
13.3 mol CO 2 (about 585 g).
22.3Limestone is calcium carbonate, which reacts with
the acid and neutralizes it.
22.4Amount in mol of CO PV/RT
1.5 105 10 –^3 /8.3145 298
0.0606
The flask is released into a sealed room of volume 100
m^3. The concentration of gas (in mol m–^3 ) is
n/v0.0606/1000.000 606 mol m–^3
6.06 10 –^4 mol m–^3
17 mg m–^3
This is well below the dangerous level. However, higher
concentration will be experienced near the bottle.22.5 100 (^10) 1g
1000
22.6 (i) 15 1.25g1.88 10 – (^2) g
1000
(ii) (^85) 8.9 10 – (^4) mol dm– 3
1000 96
22.7pH of lake before pollution is 10–8.40= 4.0 10 –^9
mol dm–^3 , pH of lake after pollution is 10–8.10= 7.9
10 –^9 mol dm–^3. So the hydrogen ion concentration has
doubled.
22.8We need to add ozone such that the concentration
of the stratosphere increases by 1 10 –^11 mol dm^3.
Number of moles of O 3 concentrationvolume
1 10 –^11 1 1022 1011 mol. Given that M(O 3 )
48 g mol–^1 ,
mass of O 3 48 1011 g or 4.8 109 kg.
cost of O 3 4.8 109 200
£960 000 000 000
Comment
The mass of ozone required is fantastically large.
Ozone is made by passing electricity through oxygen. It
would require about 2 1013 kJ of energy to make 4.8
million tonnes of ozone. To put this into perspective, we
would require a 1000 MW nuclear reactor to operate for
about 8 months in order to provide all the required
energy! The rapid decomposition of ozone on surfaces
makes the whole operation even more technically
difficult.
451
●
0230_000118_2 9 _Ans.qxd 3/2/06 2:22 pm Page 451