Worksheet 3 Answers
1.
Reactants
Products
-
ethanol
Ethene and oxygen
Ethylene oxide and water and
catalyst
Ethene and chlorine and oxygen
-
Uses
Solvent for cosmetics, medicines,
cleaning products, flavourings etc.
fumigant
-
Antifreeze, production of polymers
Vinyl chloride
-
2. a) monomers, polymerisation, polymer
b) addition, double
c) polyethylene, thousand
d) temperature, branching, low density
e) pressure, catalyst, unbranched
3. a) The difference in density is caused by the branching in low density polyethylene. The polymer chains
cannot pack together closely when they are highly branched, causing spaces between them. This decreases the
overall density.
b) HDPE would have a greater mechanical strength because its high density makes it rigid.
ci) HDPE – needs to withstand high temps and being full of water, which is quite heavy.
ii) LDPE – Doesn’t need to withstand high temps, fairly small volumes <2L.
d) Polyethylene bags are not biodegradable and cause the deaths of many animals that get caught in them or
try to eat them.
4.
Polymer
Name of Monomer
Structure of
Monomer
LDPE
Ethylene
CH2=CH2
HDPE
Ethylene
CH2=CH2
PVC
Vinyl chloride
H
Cl
C=C
H
polystyrene
5.
Item
Drain pipe
Carrier bag
Packaging
Carpet
Electrical wiring
insulation
Soft drink bottles
styrene
Common Uses
Wrapping material,
drink bottles
Carry bags,
utensils, toys
Electrical
insulation, pipes
Important
Properties Related
to Use
Flexible, easily
moulded
Rigid, easily
moulded, stable to
heat and light
Strong, rigid,
waterproof
H
H
C=C
H
H
Insulated
containers, tool
handles
Good insulator,
very strong
Disadvantage
Not as strong or
tough
Not biodegradable
Not biodegradable
Not biodegradable
Decomposes in UV
light
Not biodegradable
Old Material
Ferrous material
New Material
PVC
Advantage
Does not corrode
Paper
Straw
Wool
Rubber
HDPE
Polystyrene
polyacrylonitrile
PVC
Stronger
Doesn’t go mouldy
Lasts longer
Lasts longer
glass
LDPE
Doesn’t break
1
Worksheet 4
1) Alternative sources of the compounds presently found in crude oil need to be found because oil will
probably run out soon, and also because oil-based polymers are not biodegradable.
2) The major use of crude oil is petrol for transport.
3) Alternative sources of chemical energy could include ‘solid petroleum’ from tar sands, producing liquid
fuel from coal, or by using microbes to produce oil.
4) Builder materials means the basic materials that contain C atoms which can be used for making polymers.
5) Other possible sources of builder materials are cellulose from biomass and coal.
6a) A condensation polymer is one where a small molecule, usually water, is eliminated from between two
monomers when the polymer is formed.
b) n C6H12O6  H-(O-C6H10O4)n-OH + (n-1)H2O
glucose  cellulose + water
7) Biopolymers are those produced from substances made by living organisms. Two biopolymers currently
produced from cellulose include rayon and cellulose acetate.
8) A current biopolymer development is the genetic engineering of cotton and oil seed crops to produce
biopolymers
Worksheet 5
1a) 1-hexanol
b) 2-butanol
c) ethanol
d) 3-pentanol
e) 5-bromo 3-hexanol
2a)
d)
b)
e)
c)
3)
4) Alkanols have a much higher boiling point than the corresponding alkanes because of the –OH group. The
O and the C and H have very different electronegativites, resulting in polar molecules. This group undergoes
strong hydrogen bonding between alkanol molecules which holds them together and increases the amount of
energy (temperature) required to separate them. Alkanes only have relatively weak dispersion forces holding
molecules together.
2
5a) The larger the molecules, the less soluble it is in water.
b) Hydrogen bonding
c) Longer chain alkanols are less soluble in water because there are many non-polar bonds with dispersion
forces but few polar bonds to hydrogen bond with.
6a) Ethanol has both polar and non-polar bonds in the molecule. Water has only polar bonds and hexane only
has non-polar bonds. So ethanol’s polar bonds can hydrogen bond with water and hence dissolve. Ethanol’s
non-polar bonds can have dispersion forces with hexane and hence dissolve. But polar water and non-polar
hexane cannot form any intermolecular bonds and hence cannot dissolve in each other.
b)Ethanol is widely used as a solvent in industry because it can dissolve both polar and non-polar substances
easily and allow them to mix.
7a) CH3CH2OH  CH2=CH2 + H2O
b) CH2=CH2 + H2O  CH3CH2OH
dehydration
hydration
8a) CH3CH-OHCH2CH3  CH3CH=CHCH3 (2-butene) + H2O
b) CH2=CHCH2CH2CH3 + H2O  CH2-OHCH2CH2CH2CH3 (1-pentanol) or CH3CH-OHCH2CH2CH3 (2pentanol)
c) CH3CH2CH-OH CH2CH2CH3  CH3CH2CH=CHCH2CH3 (3-hexene) + H2O
d) no such thing as 2-propene, but…
CH3CH=CH2 + H2O  CH3CH-OHCH3 (2-propanol) or CH3CH2CH2-OH (1-propanol)
Worksheet 6
1) 2 C2H5OH + 7 O2  4 CO2 + 6 H2O
2a) Ethanol can be obtained from plant material through the processes of fermentation and distillation. New
plants can be grown, then harvested and used in the production of ethanol.
Petrol is not renewable despite also originating from plants. It is formed over millions of years by decaying
plant matter under heat and pressure and cannot be quickly reformed.
b) Advantages: Renewable, potential to reduce greenhouse gas emissions.
Disadvantages: large areas of land needed with associated agricultural problems, disposal of large amounts of
fermentation liquors.
3a) Fermentation is the process which converts glucose into ethanol and carbon dioxide.
bi) Starting ingredients for fermentation are grain/fruit mashed with water and yeast.
bii) Conditions for successful fermentation are no O2, 37oC.
biii) C6H12O6(aq)  2 CH3CH2-OH(aq) + 2 CO2(g)
4) The molar heat of combustion of a compound is the heat released when 1 mole of a substance undergoes
complete combustion at constant temperature and pressure.
5)
Hcomb = - 1360 kJ/mol = - 1 360 000 J/mol
m (H2O) = 250g
T = ?
m (ethanol) = 2.0g
M (ethanol) = (2x12.0) + (6x1.0) + (16.0) = 46.0
30% heat loss
C = 4.2 JK-1g-1
3
n (ethanol) = m/M = 2.0/46.0 = 0.043478
But only 70% efficient, so 0.7 x 0.043478 = 0.030434 moles actually goes into heating up the water.
H of this combustion = - 1 360 000 x 0.030434 = - 41 391 J
H = - mCT
- 41 391 = - (250) x (4.2) x T
T = 39 oC
If initial temp was 20 oC and change in temp is 39 oC, then final temp is 59oC.
6a)
m (H2O) = 200g
m (ethanol) = 82.5 – 81.2 = 1.3g
T = 26oC
C = 4.2 JK-1g-1
M (ethanol) = (2x12.0) + (6x1.0) + (16.0) = 46.0
Hcomb = ?
H for this combustion
= - mCT
= - (200) x (4.2) x (26)
= - 21 840 J
n (ethanol) burnt = m/M = 1.3/46.0 = 0.02826
Number of 0.02826 in one mole = 1/0.02826 = 35.3857
So 35.3857 x - 21 840 J = - 773 kJ/mol
Therefore, from the experimental data, Hcomb = 773 kJ/mol (remember heats of combustion are written as
positive).
b) The calorimeter has heat losses and is not 100% efficient. Heat can escape between the flame and the
container, and from the top of the water and the walls of the aluminium container to the atmosphere.
c) Put a lid on the aluminium beaker made of foam or polystyrene (insulating material) and wrap the beaker in
an insulating material. Place the flame as close to the bottom of the beaker as possible.
7) 4 CH + 5 O2  4 CO2 + 2 H2O
2 CH3OH + 3 O2  2 CO2 + 4 H2O
2 C8H18 + 25 O2  16 CO2 + 18 H2O
C2H5OH + 3 O2  2 CO2 + 3 H2O
C3H7OH + 5 O2  3 CO2 + 4 H2O
CH4 + 2 O2  CO2 + 2 H2O
CH
CH3OH
CH4
C2H5OH
C3H7OH
C8H18
1:1 ratio CH to CO2 therefore 455
1:1 ratio CH3OH to CO2 therefore 726
1:8 ratio C8H18 to CO2 therefore 8 x 5460 = 43680
1:2 ratio C2H5OH to CO2 therefore 2 x 1360 = 2720
1:3 ratio C3H7OH to CO2 therefore 3 x 2021 = 6063
1:1 ratio CH4 to CO2 therefore 890
Least CO2 released per kJ
Most CO2 released per kJ
4
Worksheet 10
1a) False – A displacement reaction involves the transfer of electrons between a metal and a metal ion.
b) True
c) False – When copper loses two electrons to form Cu2+ it is oxidised.
d) True
e) True
f) False – For positive monatomic ions the oxidation state can vary.
g) False – The oxidation number of manganese in MnO2 is +4.
h) True
i) False – In the reaction CuO + H2  Cu + H2O copper goes from an oxidation state of +2 to 0.
j) True
2a) Any of Mg, Al, Zn, Cr, Fe. Not K, Na, Li, Ba, Ca as they will react with water in solution before they
react with the ions in solution.
b) As above plus Fe, Ni, Sn, Pb, H.
3i) right
bi) 2Ag+ + Ni  2Ag + Ni2+
ii) 3Fe2+ + 2Al  3Fe + 2Al3+
iii) Fe + Hg2+  Fe2+ + Hg
ii) left
iii) right
4a) The strongest reductant makes the other atom gain electrons most easily, so it is the one that loses
electrons most easily, hence it’s the most active metal.
Strongest  Al, Fe, Pb, Cu  Weakest
b) The strongest oxidant is the one that makes the other atom lost electrons most easily, so it’s the atom that
most wants to gain electrons itself.
Strongest  Ag+, Ni2+, Al3+, K+  Weakest
5a) Zn +Cu2+  Cu + Zn2+
b) Pb + 2Ag+  2Ag + Pb2+
or
+
2+
+
2+
c) Fe + Na + Pb  Na + Fe + Pb
Pb + 4Ag+  4Ag + Pb4+
or
2Fe + Na+ +3Pb2+  Na+ + 2Fe3+ + 3Pb
6a) K 1+ Br 1b) Mg 0
c) Al 3+ O 2-
d) Fe 2+ Cl 1e) I 0
f) Fe 3+ Cl 1-
7a) Species oxidised – Bromine
Oxidant – Chlorine
b) Species oxidised – Carbon
Oxidant – Iodine
c) Species oxidised – Iodine
Oxidant – Manganese
Species Reduced – Chlorine
Reductant – Bromine
Species Reduced – Iodine
Reductant – Carbon
Species Reduced – Manganese
Reductant – Iodine
5
Worksheet 12
1a) galvanic/voltaic, electricity, (ignore, ignore,) anode, cathode.
b) displacement, electrolyte, salt, ions, cations, cathode, negative, anode
c) electrodes, cathode, oxidation, anode, negatively, anode, cathode, electrode.
d) ignore
2) cathode - Ag
anode - Cu
direction of electron flow is Cu to Ag.
salt bridge connects the two beakers.
3) 2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq)
4) oxidation
reduction
Cr(s)  Cr3+(aq) + 3ePb2+(aq) + 2e-  Pb(s)
5) anode – Fe
cathode – Pb
2+
oxidation
Fe(s)  Fe (aq) + 2ereduction
Pb2+(aq) + 2e-  Pb(s)
6) oxidation half-reaction
Ag(s)  Ag+(aq) + ereduction half-reaction
Cl2(s) + e-  2Cl-(aq)
overall reaction
2Cl (aq) + Ag(s)  Cl2(g) + Ag+(aq)
shorthand representation Ag | Ag+ || Cl- | Cl2, Pt
7)
8) Magnesium will displace the mercury.
oxidation half-reaction
Mg(s)  Mg2+(aq) + 2ereduction half-reaction
Hg2+(aq) + 2e-  Hg(s)
2+
overall reaction
Hg (aq) + Mg(s)  Hg(l) + Mg2+(aq)
6
12a) Using table of standard reduction potentials: Ag+, Cu2+, Sn2+, Na+, Ba2+
Using Activity Series of Metals: Ag+, Cu2+, Sn2+, Ba2+, Na+
b) Mg, Cr, Pb, Au
13)
14)
15)
Ni(s)  Ni2+(aq) + 2eFe3+(aq) + e-  Fe2+(aq)
+0.26V
+0.77V
Ni(s) + 2 Fe3+(aq)  Ni2+(aq) + 2 Fe2+(aq)
+1.03V
2I-(aq)  I2(aq) + 2eCl2(g) + 2e-  Cl-(aq)
-0.62V
+1.36V
I-(aq) + Cl2(g)  I2(aq) + Cl-(aq)
+0.74V
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O
Fe2+  Fe3+(aq) + e-
+1.51V
-0.77V
MnO4-(aq) + 8H+(aq) + 5Fe2+  Mn2+(aq) + 5Fe3+(aq) + 4H2O
+0.74V
Because the EMF is positive the reaction will occur.
Worksheet 13
v)  ray
vi) transuranic
vii) carbon-14
viii) lead
1i) radioactive
ii) isotopes
iii) particle
iv)  particle
ix) nuclear fission
x) half-life
xi) Geiger-Muller counter
2.
Alpha Particle
Beta Particle
Gamma Particle



Helium nucleus
4
2+
Low
electron
0.00055
1Medium
EM radiation
N/A
0
High
Large deflection
Medium deflection
No effect
Symbol
Identity
Relative mass
Relative charge
Penetrating power
Effect of electric
and magnetic fields
3ai) protons – 35, neutrons – 44, electrons – 35
ii) protons – 35, neutrons – 46, electrons – 35
iii) protons – 38, neutrons – 52, electrons – 38
iv) protons – 53, neutrons – 70, electrons – 53
bi) 131
53
I
4a) 222
86
e) 131
53
Rn
I
ii) 193
77
Ir
b) 218
84
f) 24
12
Po
iii) 121
51
c) 218
84
Sb
Po
Mg
7
iv) 123
37
Rb
d) 90
39
Y
5ai) 230
90
Th 
ii)
234
92
U

iii)
221
87
Fr

6)
7)
226
88
Ra
232
90
Th 
228
88 Ra 
228
89
228
90
Ac 
Th
8i) stable

He
+
226
88
Ra
bi) 27
12
He
4
2 He
+
230
90
Th
ii)
+
217
85
At
iii) 90
38
4
2
4
2


4
2
222
86
He

+
228
88
228
89
0
-1
e
+
0
-1
e
+
He
+
4
2

Rn
218
84
Po

0
-1
I

0
-1
Sr

Mg
131
53


214
82
e
e
0
-1 e
+ 27
13
+
131
54
+
90
39
Al
Xe
Y
Pb
Ra
Ac
228
90 Th
224
88
ii) unstable
Ra
iii) unstable
iv) unstable
9) Produced in nuclear reactors by neutron bombardment.
Produced in linear accelerators or cyclotrons by firing ion beams at target nuclei.
Discovered in fallout from hydrogen bomb test blasts.
(Your answer should be more in depth)
10)
Keep dividing by 2 until you get to less than 0.1% then count the number of half-lives.
100%  50%  25%  12.5%  6.25%  3.125%  1.56%  0.78%  0.39%  0.195%  0.097%
10 half-lives x 2.4 x 104 years per half-life, so 2.4 x 105 years
Radioactive decay of iodine-131
12
Mass remaining (g)
10
8
6
4
2
0
0
10
20
30
40
Time (days)
11a)
8
50
b) Approx. 3.5g
c) Approx. 20 days
12) Cloud chamber, Geiger-Muller counter, scintillation counter, electroscope, photographic film (Your
answer should be more in depth and describe how each works – see your notes)
Worksheet 14
1a and b
Use
i) Thickness
gauge
Property
 Low energy emission
 Long half-life


ii) Detecting
leaks
iii) Destroying
cancer

Short-lived isotope




Gamma emitter
Medium-long half-life
Intense amounts of
radiation released




iv) Sterilising
medical
supplies

v) Diagnosing
blood
circulation
problems
vi) Smoke
detector

Short half-life



Medium-long half-life
Alpha emitter

vii) Internal
irradiation of
tumours in
thyroid

Gamma emitter of
medium intensity
 Long-half-life
Has to be taken up by
the thyroid gland only
 Short half-life

Explanation of why property is desirable
The thin film needs to absorb a lot of the energy so we can
get a useful reading
Doesn’t need to be changed often so minimises exposure to
workers
To minimise exposure to the environment, especially
aquatic life
Need high energy to destroy biological molecules (eg DNA)
to disrupt growth of cancerous cells
Doesn’t need to be changed often so minimises exposure to
workers
Needs to penetrate deep tissue with a lot of energy
Need high energy to destroy biological molecules (eg DNA)
to disrupt growth of bacteria and fungi
Doesn’t need to be changed often so minimises exposure to
workers
To minimise exposure of patient to radiation
Doesn’t need to be changed often because people are not
reliable at checking household equipment
 Radiation (He ions) have to be big enough to be stopped by
smoke particles in the air and so prevent the ions reaching
the detector (hence triggering the alarm)
 To minimise exposure of patient to radiation (therefore use
Iodine radioisotope)
 To minimise exposure of patient to radiation
bi) Shielded radioactive source on one side of the material to be measured, radiation detector on the other
side. Radiation passes through the material – a certain amount will pass through if the material is a specified
thickness. Too much radiation means it’s too thin. Not enough radiation means it’s too thick.
ci) Co-60 and Sr-90
ii) Na-24
iii) Co-60
iv) Co-60 and Cs-137
v) Tc-99m
vi) Am-241
vii) I-131
9
2a) C-14 and K-40 have half-lives similar to the geological time scales they are providing information on.
You can fit several half-lives of the radioisotope within the time scale which will allow you to accurately
determine the age of the specimen.
2b) C-14 couldn’t be used to estimate a time scale of billions of years because it would have all decayed
away; similarly K-40 can’t be used for time scales of thousands of years because hardly any would have
decayed – neither will provide you with useful information.
3) A radioactive tracer is a radioactive isotope placed into a fluid system such as the bloodstream or water
pipes.
Examples are Tm-99m (bloodstream) and Na-24 (water or oil pipes).
 They have to have short half-lives so they don’t expose people or the environment to too much
radiation.
 They need to be soluble in water so they are carried along with the flow.
 They need to be gamma emitters so the radiation can be detected through thick tissue or underground.
It is hazardous to use radioisotopes as they can damage biological molecules such as DNA and cause sterility
and cancer. Scientists need to wear radiation badges to make them aware of whether they have too much
exposure to radiation. They need to use protective shielding when preparing radioisotopes for use and store
them in containers with thick lead shielding.
10