Possible structure is:
14 a empirical formula
%
÷ Ar
÷ smaller
value
Carbon
Hydrogen
85.7
14.3
85.7
= 7.136
12.01
14.3 = 14.187
1.008
7.136 = 1
7.136
14.187 = 1.988...
7.136
Empirical formula is CH2. [1]
Molecular ion peak of mass spectrum at
m/z = 84.
mass of molecular formula 84
=
= 6 then
As
mass of empirical formula 14
molecular formula is 6 × empirical formula
Molecular formula is C6H12. [1]
b IHD = 1 [1]
c The IR spectrum has a strong peak at 2950 cm−1
corresponding to C–H bonds [1] and a weaker
peak at 1620 cm−1 corresponding to C=C in
alkene. [1]
The mass spectrum has fragments at the following
m/z values:
41, due to C3H5+
55, due to C4H7+
69, due to C5H9+
[max 2 marks for any two sensible suggestions
including charges]
NMR has signals at:
1.0 corresponding to 6 protons, possibly two CH3
groups; triplet signal indicates 2 adjacent protons,
possibly two CH3 both adjacent to a CH2 group.
[1]
2.0 corresponding to 4 protons, possibly two CH2
groups; quartet signal indicates 2 adjacent protons,
possibly two CH2 each adjacent to a CH3 group,
i.e. two ethyl groups. [1]
[Alternatively, 1 mark for identifying both
environments, 1 mark for identifying the number
of adjacent protons in both signals]
4.6 corresponding to 2 protons, possibly CH2=C
group [1]; singlet signal indicates no adjacent
protons. [1]
H
CH2CH3
C
H
C
CH2CH3
Figure A11.15 Possible structure of compound A.
[2 marks for correct molecule, 1 mark for alkene
with formula C6H12 such as hex-2-ene (splitting
pattern is incorrect for this molecule) that
matches their interpretation of the spectra]
Chapter 12: Materials (Option A)
Exercise 12.1
1 a i
ceramic
ii metal
iii polymer
iv composite
b A measure of:
i
the degree to which a substance conducts
electricity.
ii how easily a liquid or gas can pass through a
substance.
iii the ability of a substance to return to its
original shape once a stretching force has been
removed.
iv how easily a material breaks when it is
stretched.
v how easily a material can be hammered into
sheets or can be drawn into wires.
c i
high melting points
ii good electrical conductors
iii malleable
iv permeable
v good electrical conductors
vi brittle
vii brittle
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2
Question
Electronegativity
difference
Average
electronegativity
a TiO2
3.4 − 1.5 = 1.9
2.45
b Cu
0
Class of
substance
ionic
metallic
3.4 − 1.6 = 1.8
1.9
borderline
ionic/polar
covalent
d Ge
0
2.0
metallic
e H2O
3.4 − 2.2 = 1.2
2.8
polar covalent
f AlCl3
3.2 − 1.6 = 1.6
2.4
borderline
ionic/polar
covalent
g MgBr2
3.0 − 1.3 = 1.7
2.25
ionic
c Al2O3
112
112 moles of electrons would produce
3
moles of aluminium = 37.3 mol
v m(Al) = n × Ar = 37.3 × 26.98 = 1007 g or 1 kg.
3 a A homogenous mixture of a metal with other
metals or non-metals.
b The properties of a pure metal rarely exactly suit a
given use. The properties of an alloy can be
adjusted by adjusting the composition to match a
particular use.
c The presence of atoms of a different size prevents
the layers of atoms sliding over each other easily.
iv
This larger atom the top layer sliding
across when a force is applied.
force
Table 12.1
Exercise 12.2
1 a redox reactions
b The higher the metal is in the activity series, the
more difficult it is to extract from its ore.
c SnO2 + C → Sn + CO2
d 2Al + Cr2O3 → Al2O3 + 2Cr
2 a i
Al(OH)3 + NaOH → NaAl(OH)4
ii amphoteric
iii 2Al(OH)3 → Al2O3 + 3H2O
iv dehydration
b The melting point of aluminium oxide is very high
and so the energy costs would be prohibitive if
molten aluminium oxide was used. The melting
point of cryolite is lower than that of aluminium
oxide and so the aluminium oxide is dissolved into
the molten cryolite, lowering the running costs.
c i
Al3+ + 3e− → Al
ii 2O2− → O2 + 4 e−
iii 2Al2O3 → 4Al + 3O2
iv The oxygen reacts with the carbon anodes
turning them to carbon dioxide:
C + O2 → CO2
d i
Q = It = 3000 × (60 × 60) = 1.08 × 107 C
ii moles of electrons
amount of charge
=
amount of charge in 1 mol of electrons
1
08 × 1107 = 112 mol
= .08
96500
3+
iii Al + 3e− → Al
Figure A12.1 Structure of an alloy.
4 a Paramagnetism is a weak attraction to a magnetic field.
It occurs in substances that have unpaired electrons.
b Diamagnetism is a very weak repulsion from a
magnetic field. It occurs in substances with paired
electrons.
c i
O O
Figure A12.2 Electron arrangement in O2
Diamagnetic as all the electrons are paired.
Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 and
Cl− 1s2 2s2 2p6 3s2 3p6
The 3d5 electrons in Fe3+ are unpaired
), and so FeCl3 is paramagnetic.
(
iii Zn2+ 1s2 2s2 2p6 3s2 3p6 3d10 and O2− 1s2 2s2 2p6
All the electrons are paired and so ZnO is
diamagnetic.
1 unpaired electron
d Na 1s2 2s2 2p6 3s1
ii
As 1s2 2s2 2p6 3s2 3p6
4s2 3d10 4p3
Cr 1s2 2s2 2p6 3s2 3p6
4s1 3d5
3 unpaired electrons
(4p electrons are )
6 unpaired electrons
)
(3d electrons are
Sc 1s2 2s2 2p6 3s2 3p6
4s2 3d1
Ti 1s2 2s2 2p6 3s2 3p6
4s2 3d2
1 unpaired electron
2 unpaired electrons
(3d electrons are )
ANSWERS
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least paramagnetic → most paramagnetic
Na = Sc < Ti < As < Cr
5 a OES is optical emission spectroscopy.
MS is mass spectrometry.
b A plasma is a state of matter that occurs above the
temperature of a gas in which most of the (in this
case) argon particles are ionised to form positive
ions and electrons.
c The argon is subjected to a spark form an electric
discharge in the presence of oscillating
radio-frequency radiation.The radio frequency
induces a strong electromagnetic field and this causes
the electrons to move in a circular path due to its
coiled shape and so the electrons collide with other
argon atoms causing them to also lose electrons,
become ionised and generate very high temperatures.
d It causes ionisation and excitation.
e i Technique 1 is ICP-OES.
Technique 2 is ICP-MS.
ii A calibration graph using substances of known
concentration.
f monitoring of pollutant levels in the environment,
e.g. mercury levels in shellfish, lead levels in petrol,
or in paint, or in tissue samples
Exercise 12.3
1 a Homogeneous catalysts are in the same phase as
the reactants. Heterogeneous catalysts are in a
different phase from the reactants.
b Heterogeneous catalysts work by providing a
surface with active sites onto which the reactant
molecules are adsorbed. The reactants are held
close to each other and in the correct orientation.
Weak bonds formed between the reactants and the
catalyst surface weaken the bonds within the
reactants and lower the activation energy for the
reaction; the product molecules are formed. The
products are then desorbed leaving the surface of
the catalyst available for new reactant molecules.
c large surface areas
d Homogeneous catalysts work by providing an
alternative reaction mechanism, with a lower
activation energy (E2 and E3 < E1 in Figure
A12.3). This often involves forming a more stable
activated complex or intermediate as can be seen
in the diagram.
activated complex
uncatalysed reaction
E1
E3
E2
reactants
intermediate
catalysed reaction
E1 = activation energy of
uncatalysed reaction
∆H
products
E2/E3 = activation of
catalysed reaction
Figure A12.3 Energy profile diagram showing the action of a
homogeneous catalyst.
Exercise 12.4
1 a The molecules are rod-shaped and distributed
randomly but aligned in the same direction as each
other.
b Thermotropic liquid crystals are pure substances
whereas lyotropic liquid crystals are solutions.
c i
Figure 12.2: thermotropic liquid crystal.
Figure 12.3: lyotropic liquid crystal.
ii Thermotropic, temperature,
lyotropic, concentration.
d
Thermotropic
Lyotropic
Examples of this type of liquid
crystal include biphenyl
nitriles.
An example of this type of
liquid crystal is soap and
water.
These liquid crystals are pure
substances.
These liquid crystals are
solutions.
This liquid-crystal phase is
dependent on temperature.
This liquid-crystal phase
depends on concentration.
Table 12.2
e Elasticity, optical properties (whether planepolarised light is transmitted or not), electrical
conductivity.
2 a electric field
b in order to be able to change orientation when an
electric field is applied
c chemically stable
respond quickly to an electric field (rapid
switching speed)
temperature range for the liquid-crystal state must
be suitable for the application
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Exercise 12.5
1 a A polymer is a long-chain molecule made by
joining lots of smaller monomer units together.
b i
Thermoplastics soften and melt when heated
and harden again when cooled.
ii Weak intermolecular forces such as London
forces. These are easily broken and reformed
which allows the chains of the polymer to slide
across one another making the plastic flexible.
c i
A thermosetting polymer is a pre-polymer in a
soft solid or viscous state that changes
irreversibly into a hardened thermoset by curing.
ii Thermosetting polymers have covalent cross
links between their chains.
d i
Elastomers return to their original shape once
a stretching force has been removed.
ii Elastomers tend to have coiled chains with a
few covalent cross links between them. When
stretched the chains uncoil. When the force is
removed the cross links pull the chains back
into their coiled shape.
2 a There are more, and therefore stronger,
intermolecular forces in molecules with longer
chains. These hold the chains more lightly together
making the polymer stronger and more energy is
required to overcome these forces, making the
melting points higher as well.
b The more branched structure of LDPE means that
the chains cannot pack as tightly together as the
chains in HDPE and so LDPE has fewer, and
therefore weaker, London forces between its chains.
This means that LDPE has a lower melting point
and is weaker than HDPE.
c i
H CH3 H CH3 H CH3
H H H H H H
isotactic
CH3 H H CH3 H CH3 CH3 H
H H H H H H H H
atactic
Figure A12.4 Isotactic and atactic polypropene.
Isotactic polypropene is more crystalline due
to more regular and closer packing.
iii 2- methylpropene has two CH3 groups on the
same carbon atom so different orientations are
not possible.
ii
Plasticiser molecules fit between the polymer
chains and lubricate their movement allowing
the chains to side over one another more easily.
ii Plasticised PVC can be used for insulating
electrical cables, toys, waterproof materials, and
food wrap.
Unplasticised PVC is used for window frames,
drainpipes and guttering.
i
When heated, the hydrocarbon turns to a gas
and expands.
ii The polymer expands becoming a foam and
has a much lower density.
atom economy =
molar mass of desired products
× 100
total molar mass of all reactants
01) + (10
10 × 1.01)
(6 × 1122.01
=
× 100
01) + (12
12 × 1.01) + (1 × 1166.0000)
(6 × 1122.01
= 82.01%
Using the same method as in part a, atom
economy = 45.83%
Using the same method as in part a, atom
economy = 26.00%
d i
e
3 a
b
c
Exercise 12.6
1 a Bottom-up describes manipulating atoms or
molecules using chemical or physical means to
build up a larger structure.
Molecular self-assembly describes the situation
when atoms or molecules spontaneously arrange
themselves into nanoparticles either on a surface or
in solution.
b The formation of soap molecules into spheres/
folding of a protein chain/spiralling of the DNA
double helix.
c Physical techniques involve the manipulation of
atoms or molecules using a probe. Chemical
techniques involve using specific chemical
reactions to position the atoms in a molecule.
2 a In graphene the carbon atoms are in a planar
hexagonal arrangement in a single layer with each
carbon atom attached to three others. The fourth
electron is held in a p orbital perpendicular to the
plane of the hexagons.
Carbon nanotubes have a similar arrangement but
the layer is curled into a hollow tube. The ends of
the tube can be closed or open. Closed ends
include pentagons as well as hexagons.
ANSWERS
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The electrons in the p orbitals of the carbon
atoms overlap each other and form a
delocalised π system. The electrons are free to
move through this and so the nanotube can
conduct electricity.
ii Carbon nanotubes are strong as there are
strong covalent bonds between the carbon
atoms in the walls of the tube.
i
CVD is chemical vapour deposition
ii HIPCO is high-pressure carbon monoxide
process
The oxidation state of carbon in C6H5CH3 is − 8
whereas in carbon nanotubes it is zero. As the 7
oxidation state increases, this is oxidation.
The electrodes can be made from graphite and an
arc passed between them in an inert atmosphere of
low-pressure helium or nitrogen.
If oxygen were present then carbon dioxide would
be formed.
2CO(g) → C(s) + CO2(g)
b i
c
d
e
f
g
Exercise 12.7
1 a The composition of the plastic, the availability of
oxygen, the temperature of incineration.
b i
carbon dioxide and water
ii carbon monoxide and water
iii carbon dioxide, water and hydrogen chloride
9n
O2 →
c i
[CH2CH(CH3)]n +
2
3nCO2 + 3nH2O
ii [CH2CH(CH3)]n + 3nO2 → 3nCO + 3nH2O
5n
iii [CH2CHCl]n +
O2 →
2
2nCO2 + nH2O + nHCl
d i
polychlorinated dibenzodioxins
ii dioxin
iii These molecules are carcinogenic.
2 a They have high electrical resistance.
b
Cl m
Cl n
c fatty tissue
d The molecules have strong covalent bonds between
the atoms and microbes do not possess enzymes
capable of breaking down these molecules.
3 a plasticiser
b The bonding is weak and intermolecular and so
the plasticisers are easily lost.
c The plasticiser molecules are fat-soluble and so can
dissolve into the fatty food which they are
wrapped around.
4 a Advantages:
Reduces the use of raw materials such as crude oil
which would be needed to make replacement
plastics.
Reduces the amount of carbon dioxide released
into the atmosphere compared to incineration;
CO2 is a greenhouse gas linked to global warming.
Reduces the release of potentially harmful
chemicals into the atmosphere such as dioxins and
PCDD compared to incineration.
Reduces the demand for land, the spoiling of the
environment and loss of wildlife habitat that is
required by landfill.
Reduces the potential for harmful chemicals such
as phthalate esters from plastic waste contaminating
soil and water which could be caused by disposing
of in landfill.
Disadvantages:
Plastic waste is difficult to recycle as it must be
sorted into its different types.
This is difficult to do mechanically. It is mostly
done manually and labour costs are therefore high.
The recycling of plastics is an energy-intensive
process. Once sorted by type, the plastic waste
needs to be processed by heating to melt and
remould it or by using the plastic as a chemical
feedstock; both processes require a lot of energy.
b They enable the type of plastic to be more easily
identified and recycled.
c Figure 12.4a is di-n-octylphthalate. There is a
strong absorbance between 1750 cm−1
corresponding to the C=O bonds in the ester
groups, at 1275 cm−1 corresponding to the C–O in
the ester as well between 2850 and 3090 cm−1
corresponding to the C–H bonds in the polymer
chain.
Figure A12.5 A typical PCB.
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Figure 12.4b is polystyrene. There are strong
absorbances between 2850 and 3090 cm−1
corresponding to the C–H bonds in the polymer
chain and in the phenyl group.
( 81 ) + 1 = 2 atoms
1
Figure 12.7b: (8 × ) + (6 × ½) = 4 atoms
8
1
Figure 12.7c: (8 × ) = 1 atom
8
iii Figure 12.7a: 8 ×
Exercise 12.8
1 a Resistance is caused by the collision between the
electrons and the lattice of positive ions which
impedes the movement of electrons through the
structure.
b As the temperature increases, the vibrations of the
positive ions increase and so they block the path of
the electrons through the material more than at
lower temperatures.
c Superconductors have zero resistance.
d As an electron approaches a positive ion, the
neighbouring positive ions move towards it,
becoming more closely packed and creating larger
gaps in the lattice elsewhere. A second electron is
attracted to this positive region; it and the first
electron form a Cooper pair which can move
freely through the structure.
2 a i
Meissner effect
ii Moving electrons on the surface of a
superconductor create a mirror image of the
external magnetic field. The external and
mirror-image magnetic fields oppose
each other.
b
Resistance
type 2 superconductor
type 1 superconductor
metal
Temperature/k
c X-ray crystallography
d Using nλ
λ = 2dsinθ
n = 1 (as this is a first-order reflection)
λ = 5.53 × 10−10 m (the wavelength of the X-rays)
d = path length
θ = 14.3° (the angle at which the X-rays hit the
plane of atoms)
1 × 5.53 × 10 −10
= 2.80 × 10−10 m
d = nλ =
2 sinθ
2 × sin 14
14.3
For a simple cubic structure, the length of the unit
cell equals twice the atomic radius as can be seen in
Figure 12.8 in the tip box adjacent to this question.
length of unit cell
r
r
Figure 12.8 (from Workbook page 178) A simple cubic
structure.
Therefore, the atomic radius = 1.40 × 10−10 m
e i
ii
iii
Figure A12.6 Variation of resistance and temperature of a type 2
superconductor.
iv
3 a The simplest repeating unit from which a crystal
structure can be built.
b i
a is body-centred cubic, b is face-centred
cubic, c is simple cubic
ii The atoms in a have a coordination number
of 8, in b the coordination number is 12, in c
the coordination number is 6.
iv
mass of a Cu atom
mass of one mole of atoms
63.55
=
=
number of atoms in one mole 6.02 × 1023
= 1.056 × 10−22 g
A face-centred cubic unit cell contains
4 atoms (see Question 3 part b iii).
mass of one unit cell = 4 × 1.056 × 10−22
= 4.222 × 10−22 g
volume of a unit cell = (3.61 × 10−10 )3
= 4.70 × 10−29 m3
mass of unit cell
4.222 × 10 −22
=
density =
volume of unit cell 4.70 × 10 −29
= 8 975 000 g m−3 or 8.98 g cm−3
ANSWERS
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Exercise 12.9
2 a
1 a A long-chain molecule that is formed when
monomers with two functional groups are joined
by a condensation reaction. That is a reaction in
which a covalent bond is formed and a small
molecule such as water is released.
b
O
O
O
C
ester linkage
N
C
n HO
C
O
O
OH + n H2N
C
C
O
O
H2N
H
H
N
N
+ 2n –1 H2O
n
Figure A12.8 Equation for the formation of Kevlar®.
b It is able to form hydrogen bonds between the
amide links in different chains (see Figure A12.9).
These are numerous and relatively close together
giving a strong structure.
The benzene rings in adjacent chains are also held
in stacks by London forces.
H
amide linkage
Figure A12.7 Ester and amide linkages.
c One monomer needs to have two carboxyl groups
and the other two hydroxyl groups.
Alternatively, a polyester can be made from a single
monomer with both groups.
d One monomer needs to have two carboxyl groups
and the other two amine groups.
Alternatively, a single monomer with both groups
could be used.
e
• Addition polymers are normally made from
only one monomer whereas condensation
polymers commonly have two.
• Monomers for addition polymerisation are
alkenes and contain the C=C functional group.
Those for condensation polymers have different
functional groups; dicarboxylic acids and
diamines or dicarboxylic acids and diols.
• Addition polymerisation does not produce any
other products whereas in condensation
polymerisation, a small molecule such as water
is eliminated from each ‘join’.
• Addition polymers tend to be non-polar whereas
condensation polymers always contain polar
groups; the ester and amide linkages are polar.
• Condensation polymers can be hydrolysed in
acid or alkaline conditions as they contain ester
or amide groups. Addition polymers tend to be
chemically inert due to the unreactive C–C
bonds.
• Some condensation polymers can be
biodegradable due to the ester and amide
linkages. Addition polymers are not biodegradable.
C
C
C
O
O
H
H
N
N
H
H
N
N
Hydrogen
bonds
C
C
O
O
H
H
N
N
C
C
O
O
H
H
N
N
C
C
O
O
H
H
N
N
C
C
O
O
H
H
N
N
Figure A12.9 Structure of Kevlar®.
c The sulfuric acid can hydrogen bond to the amide
groups instead of the amide groups hydrogen
bonding to each other and this breaks the
connections between the chains.
H
H
O
O
S
O
O
O
O
O
C
C
O
O
H
H
S
H
O
O
O
H
H
N
N
O
O
H
O
O
H
H
O
S
S
O
Figure A12.10 Kevlar® in concentrated sulfuric acid.
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3 a Condensation polymers contain ester or amide
linkages and these are susceptible to hydrolysis.
These are the same function groups as are found in
some foods (fats and proteins) and so some
organisms can digest these.
b
H
O
H
O
C
CH3
C
3 a a lone pair of electrons which they can donate/
must be Lewis bases
b They are joined by a coordinate bond, the
electrons are donated from the ligand into a vacant
orbital of the central atom/ion.
c a ligand that can form two dative coordinate bonds
to the central metal atom/ion
d
O
O
H
Figure A12.11 Monomer of PLA.
c They do not use crude oil as a raw material which
saves a valuable and finite natural resource.
Bioplastics have a lower carbon footprint and the
carbon dioxide released when these materials
decompose/are digested/are incinerated matches
that which was taken in by photosynthesis when
the plants were grown.
d Large amounts of land are needed to grow sufficient
crops to supply the bioplastics industry. This reduces
the land available for growing food crops, which
could lead to food shortages and price rises.
The demand for land can lead to deforestation and
loss of habitat.
Exercise 12.10
1 a Transition metals have variable oxidation states and
can take part in redox reactions. As many biological
processes are redox reactions, the heavy metal ions
can interfere with the normal pathways of these
biological processes.
b i
Fe3+ + •O2− → Fe2+ + O2
Fe2+ + H2O2 → Fe3+ + •OH + OH−
ii Fenton reaction
2 a i
sodium hydroxide solution (or any soluble
hydroxide solution).
ii Cr3+(aq) + 3OH− → Cr(OH)3(s)
b i
hydrogen sulfide
ii Pb2+(aq) + H2S(g) → PbS(s) + 2H+(aq)
c The heavy metal ions can be removed from the
water by being adsorbed on the surface of a
material such as a clay or zeolite. The heavy metal
ions stick to the surface.
d Chelation uses a chelating agent which is a
substance that can hold the metal ion in a ring and
binds tightly to it.
C
C
CH2
–O
O
N
C
C
CH2
O–
H
H
C
C
H
H
CH2
C
CH2
C
N
O
O–
O
O–
Figure A12.12 EDTA.
[Ni(H2O)6]2+ + EDTA4− →
[Ni(EDTA)]2− + 6H2O(l)
ii There is an increase in entropy during the
reaction as there are more product molecules
than reactant molecules.
solubility product constant
Ksp = [Mn+][Xn−]
i
Ksp = [Ag+][Br−]
ii Ksp = [Mg2+][OH−]2
iii Ksp = [Fe3+]2[CO32−]3
i
PbCO3(s) Pb2+(aq) + CO32−(aq)
ii Ksp = [Pb2+][CO32−]
iii Ksp = (2.72 × 10−7) × (2.72 × 10−7)
= 7.40 × 10−14
i
Zn(OH)2(s) Zn2+(aq) + 2OH−(aq)
ii Ksp = [Zn2+][OH−]2
iii Ksp = (1.96 × 10−6 mol dm−3) ×
(2 × 1.96 × 10−6 mol dm−3)2
= 3.01 × 10−17
i
Ag2SO4(s) 2Ag+(aq) + SO42−(aq)
ii Ksp = [Ag]2[SO42−]
Let the [Ag2SO4] = S, then [Ag+] = 2S and
[SO42−] = S
Ksp = (2S)2 × S = 4S3
e i
4 a
b
c
5 a
b
c
−5
S = 3 1.2 × 10 = 0.014 mol dm–3
4
d i
Hg2CO3(s) Hg22+(aq) + CO32−(aq)
Ksp = [Hg22+][CO32−]
As [Hg22+] = [CO32−] = solubility of Hg2CO3
ANSWERS
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ii
then solubility = √ 3.
3.6 × 10−17
= 6.0 × 10−9 mol dm−3
Ksp = 3.6 × 10−17 = [Hg22+] × 0.10
−17
Solubility of Hg2CO3 = [Hg22+] = 3.6 × 10
0.110
= 3.6 × 10−16 mol dm−3
This is significantly lower than its solubility in
water.
iii Increasing the concentration of carbonate ions
shifts the equilibrium to the left and the
mercury(I) chloride will precipitate out. This
aids its removal from the solution.
Exam-style questions
1
Metals
Ceramics
Polymers
Melting points:
high or low?
high
high
low, although
thermosets burn
rather than melt
Electrical
conductivity:
good or poor?
good
poor
poor
Permeability:
high or low?
low
high
low
Elasticity:
high or low?
low
low
generally low, apart
from elastomers
Brittleness:
high or low
low
high
low
Malleability/
ductility:
high or low?
high
low
variable
Table 12.3
b
c
d
2 a
b
[1 mark for each row]
Most ceramics are ionic or have giant covalent
structures. [1]
They do not contain charged particles that are free
to move. [1]
Metals and ceramics have giant structures with
strong forces of attraction between their particles
and so have high melting points. [1]
Polymer chains are held by only weak
intermolecular forces. [1]
These require relatively little energy to overcome
them and the melting points are low. [1]
areas B and C [1]
Fe2O3 + 2Al → 2Fe + Al2O3 [1]
Fe2O3 is reduced and Al is oxidised. [1]
c Al has the electron configuration
1s2 2s2 2p6 3s2 3p1.
It has one unpaired electron and is paramagnetic.
Al3+ has the electron configuration 1s2 2s2 2p6.
It has no unpaired electrons and is diamagnetic.
[1 mark for correct electron configurations, 1 mark
for correctly assigning as para- and diamagnetic]
d Alloys contain atoms of different sizes. The larger
size of some atoms prevents the layers of atoms
from sliding over one another making the metals
stronger. [1]
e i
As the electrons in the excited plasma fall back
down to lower energy levels, light is emitted/
an emission spectrum is produced. [1]
The wavelengths of the lines in the emission
spectrum are characteristic of a particular
element. [1]
The light is passed through a prism into its
separate wavelengths and the wavelength and
intensity of these are measured. [1]
The concentration of a substance can be
measured by comparing the intensity against
that of solutions of known concentration
using a calibration graph. [1]
ii From the graph, the concentration of the
sample is 0.92 ppm.
1 ppm = 1 mg dm−3 = 1 × 10−3 g dm−3
therefore 0.92 ppm = 0.92 × 10−3 g dm−3
So 100 cm3 of solution contains
100
0.92 × 10−3 ×
= 0.92 × 10−4 g [1]
1000
Percentage manganese in the 0.10 g sample
0.9922 × 110 −4
=
× 100 = 0.092% [1]
0.110
3 a A zeolite is a ceramic crystalline material. [1]
b They have relatively high surface areas. [1]
c The toxicity/effect on human health is
unknown. [1]
d Zeolites have a cage-like porous structure [1]
through which only particles of a specific size or
shape can fit. [1]
e The specificity of the catalyst.
The efficiency of the catalyst/by how much it
increases the rate of reaction.
The conditions required for the reaction
(temperature/pressure and so on)
If the catalyst is affected by any of the likely
impurities/is inhibited or poisoned.
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f
4 a
b
c
b
5 a
b
c
d
6 a
The ease of separating the products from the
catalyst.
The environmental impact of its use and/or
disposal.
[3 max]
No effect on atom economy.
Branched polymer chains cannot pack as tightly
together [1] so have weaker intermolecular forces
between the chains [1] giving them lower melting
points and making them softer. [1]
Suitable examples for both branched and
unbranched. HDPE, buckets; LDPE,
plastic bags. [1]
Phthalate esters are used as plasticisers [1]. They
allow the polymer chains to slide over one another
more easily [1] so the plastic is more flexible [1].
Suitable examples for both, e.g. PVC with a
plasticiser, waterproof material/food wrap; PVC
without a plasticiser, drain pipes. [1]
If the side chains of a polymer are all on the same
side (isotactic) then the chains can pack more
tightly together [1] compared to when the side
chains are randomly arranged (atactic) so have
stronger intermolecular forces between the chains
[1] giving them higher melting points and making
them stronger. [1]
Suitable examples for both, e.g. isotactic
polypropene, crates; atactic polypropene, fibres. [1]
The volatile hydrocarbon turns to a gas on heating
and expands causing the polymer to expand as
well. [1] This lowers the density of the polymer. [1]
Suitable example, e.g. expanded polystyrene for
packaging. [1]
It has a rod-like shape and has a polar group at one
end. [1]
It causes their orientation to change to line up
with the applied current. [1]
i
lyotropic [1]
ii concentration [1]
They are all aligned in the same direction but are
randomly distributed and are free to move. [1]
HCl is also formed. [1]
O
N
H
(CH2)6
N
H
C
O
(CH2)8
C
b
c
d
e
f
7 a
b
b
[1 mark for amide link, 1 mark for rest of structure,
must show at least one repeating unit]
hydrogen bonds [1]
i
a pre-polymer that changes irreversibly into a
polymer network by curing [1]
ii covalent bonds [1]
iii hard/brittle/doesn’t soften on heating/burns/
rigid [1]
i
Landfill – leaking of contaminants into ground
water/loss of habitat/spoil the environment/
waste of natural resources. [1]
Incineration – toxic gases produced/CO2
produced/greenhouse gases produced/waste
of natural resources (credit this only once). [1]
ii (C16H30N2O2)n + 20nO2 →
14nCO2 + 2nHCN + 14nH2O
[1 for correct products, 1 for balancing]
iii Energy is required to sort, wash and clean the plastic
waste/energy is need to melt the plastic in order to
reform it into new objects/energy is needed to
break down the polymers into their monomers in
order to use them as a chemical feedstock.
[Any 2]
i
starch/corn/maize/sugar [1]
ii Microorganisms do not have the correct
enzymes to break them down. [1]
i
The plasticisers used in PVC are fat-soluble
and can enter the food chain. [1]
They are thought to affect hormone action/to
be carcinogenic. [1]
ii PVC produces dioxins/PCDDs/toxic fumes
when burnt. [1]
X-ray crystallography [1]
i
body-centred cubic
ii face-centred cubic
iii Corners = 8 × 1 = 1
1 8
Face = 6 ×
=3
2
Centre = 1
Total = 5 ions [1]
i
r
5.56 x 10 –10 m
r
r
r
N.B.
The unit cell is to the centre
of each ion, not the edge.
Figure A12.14 Face of a fcc crystal.
Figure A12.13 Nylon 6,10.
ANSWERS
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length of diagonal = 4 × ionic radius [1]
diagonal = 2 × (5.56 × 10 −10 )2
= 7.86 × 10−10 m [1]
Therefore atomic radius
7.86
86 × 110 −10
= 1.97 × 10−10 m [1]
4
mass = mass of atoms in unit cell
ii Density =
volume
volume of unit cell
Mass of one atom of calcium = 40.08 23 g [1]
6.02
02 × 110
Number of atoms in a fcc unit cell = 4 [1]
=
40.08 × 4
Mass of atoms in unit cell =
6.02
02 × 11023
= 2.66 × 10−22 g [1]
Volume of unit cell = (5.56 × 10−10 )3
= 1.72 × 10−28 m3 [1]
mass of atoms in unit cell
Therefore density =
volume of unit cell
−22
2
.
66
6
6
×
10
1
=
1.7722 × 110 −28
= 1.55 × 106 g m3 or 1.55 g cm−3 [1]
8 a Pb2+(aq) + 2OH−(aq) → Pb(OH)2(s) [1]
Pb2+(aq) + H2S(g) → PbS(s) + 2H+(aq) [1]
b Ksp = [Pb2+][OH−]2 [1]
[Pb2+] = Ksp
K sp
[OH − ]2
=
1.4433 × 110 −20
(0.001)2
= 1.43 × 10−14 mol dm−3 [1]
c i
ii
Ksp = [Pb2+][I−]2 [1]
let solubility be S, then
[Pb2+] = S and [I−] = 2S
so Ksp = 4S3 = 8.5 × 10−9
8.5 × 10 −9
Solubility = S = 3
= 1.29 × 10−3
4
−3
mol dm [1]
PbI2 is more soluble at higher temperatures [1]
Chapter 13: Biochemistry
(Option B)
Exercise 13.1
1 a
b
c
d
metabolism
catabolism
anabolism
i 6CO2 + 6H2O → C6H12O6 + 6O2
ii respiration
e They maintain the amount of oxygen and carbon
dioxide in the atmosphere at an approximately
constant level.
2 a A reaction in which two or more molecules join
together with the formation of a covalent bond
and the elimination of a small molecule such as
water.
b A reaction in which a covalent bond in a molecule
is broken by the reaction with water.
c Protein, disaccharides, polysaccharides, specific
examples such as starch, triglycerides.
d Digestion of proteins, fats, disaccharides,
polysaccharides.
Exercise 13.2
1 a H2NCH(R)COOH
b Amphoteric means that they act as both acids and
bases.
They are amphoteric because they contain both an
acidic and a basic group.
The carboxyl group can donate a proton so is
acidic:
H2NCH(R)COOH  H2NCH(R)COO– + H+
The amine group can accept a proton so is basic:
H2NCH(R)COOH + H+  H3N+CH(R)COOH
c H3N+CH(R)COOH  H3N+CH(R)COO– + H+
 H2NCH(R)COO– + 2H+
At low pH (high H+ concentration), the above
equilibrium lies to the far left and the amino acid
will be positively charged.
At high pH (low H+ concentration), the above
equilibrium will lie to the far right and the amino
acid will be negatively charged.
d zwitterion
e isoelectric point
f Amino acids are soluble in water as they form ions
in solution.
They have relatively high melting points compared
to organic molecules of a similar molecular mass as
they exist as zwitterions in the solid state with
relatively strong electrostatic/ionic forces of
attraction between the ions.
2 a the linear sequence of amino acids
b the regular folding of the amino acid chain into an
ordered array
c the folding of sections of a protein which gives the
protein its overall 3-dimensional shape
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