Chemistry
Chemistry Past Paper Solutions for CAPE ® Unit 2
2007-2015
Granville Delahaye
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Contents
Past Paper Solution
Page
 CAPE Chemistry Unit 2 Paper 2 May/June 2007……………………………….….…… 1
 CSEC Chemistry Unit 2 Paper 2 May/June 2008…………………………………..…… 10
 CAPE Chemistry Unit 2 Paper 2 May/June 2009……………………………….……… 16
 CAPE Chemistry Unit 2 Paper2 May/June 2010……………………………….……… 23
 CAPE Chemistry Unit 2 Paper 2 May/June 2011……………………………….……… 30
 CAPE Chemistry Unit 2 Paper 2 May/June 2012……………………………….……… 36
 CAPE Chemistry Unit 2 Paper 2 May/June 2013……………………………….……… 42
 CAPE ChemistryUnit 2 Paper2 May/June 2014……………………………….……… 48
 CAPE ChemistryUnit 2 Paper2 May/June 2015……………………………….……… 54
C APE Chemistry
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1|Page CAPE Chemistry Unit 2
Solution to Question 1:
A)
(i)
Br2 + Cl2
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(iii)
2 BrCl
should
be
conducted in fume cha-
(state symbols not
mber or wear gas mask or
required)
any other reasonable sug-
(ii)
gestion.
2
BrCl 
Kc = 
Br2  Cl 2 
B)
Experiment
Solution to Question 2:
A)
(i)
 0.0546 
=
 0.0389  0.0111
2
K eq
=6.90
Equilibrium attained
Halide
NaCl
Accurate analysis of
the reactants and product.

Constant environmental conditions (pres-
(i)
Misty fumes
HCl
(ii)
the concentration of
Observation
Products
NaBr

Reaction
(ii)
Brown fumes
Br2
(iii)
SO2
(white) choking fumes
of gas
NaI
(iv)
Purple fumes and black
I2
precipitate
H2S
(v)
Smell of rotten eggs or
foul smell
sure and temperature).
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B)

Increasing polarizability
(ii)
Green algal bloom
(iii)
Eutrophication
(i)
Use of graph
down the group.

Increase in Van der
waals/
weak
london
B)
SAMPLE [PO43-] mg/dm3
force down the groups
increased
induced
dipole.

Induced dipole interacttions.
C)
1
0.70
2
0.59
3
0.62
(ii)
Cl2(g)+2NaOH(aq)  NaCl(aq) +
Average  PO 43 
NaClO(aq) + H2O(l)

Oxidation state change : 0 
-1 and +1
1.91
3
 0.64(mg / dm3 )

O  -1 and +1
NaCl
NaClO
To 2 significant figures
Solution to Question 3:
A)
(iii)
(i)


fertilizer

detergents
[PO43-] within
acceptable limits
(organic phosphate)
pesticides
0.70  0.59  0.62
3
Solution to Question 4:
A)
Buffer solution - regulates pH by
responding to small additions of
acid or alkali.
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3|Page CAPE Chemistry Unit 2
B)
(i)
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Strong Acid Response

H2CO3  CO2 + H2
releasing CO2 from
H+ ions react with
the blood
anions from the salt

Concentration of H+

decreased
(ii)
equilibrium shifts to
the left
Strong Alkali Response

OH– ions react with
weak acid molecules

H+ ions reabsorbed as
–
OH ions neutralized
D)
(i)
H+  CH3COO- 
Ka =
CH3COOH
to H2O
C)
(ii)
(i) a.

Equilibrium shifts to
produce H2CO3

H2CO3 dissociates to
+
increase H concentration in the blood or
lower pH of blood .
K CH COOH 
H +  = a  3
  CH COO- 
 3

-5
1.75×10 ×0.025
=
0.010
=4.4×10-5 mol dm -3

pH=-log H + 

=-log 4.4×10-5
b.
=4.4


Deep rapid breathing, clean lungs of
CO2 or decrease CO2
concentration.

Equilibrium will shift
to the left to release
CO2 from blood or
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
4|Page CAPE Chemistry Unit 2
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

Solution to Question 5:
A)
-3
6×10-3 k× 2×10
=
3×10-3 k× 1×10-3
(i) a. Using expts (4) and (5)
doubling of rate
n
n
2=2n
or any combination
Doubling of [R - X]


n  1  1st order
B)
Overall order m + n =2
Rate equation
1:1 proportionality
Rate = K [R - X] [NaOH]
1st order
1×10 
-3
OR
=
 0.5×10 
-3
k 2 10 3 
k 1 10 3 
2=2
C)
3 x 10-3 = K(6 103 )(1103 )
m
3 10   5 10
K=
m
3
6 10
m
6
2
Units  dm3 mol-1 s-1
m=1
(ii)
Using expt. (1) or any other expt
Using excperiments (1)
and (2) or any comb-
D)
Rate
determining
step
bimolecular, or SN2.
ination
Doubling of [NaOH] 
doubling of rate
1:1 proportionality  1st
order
OR
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is
5|Page CAPE Chemistry Unit 2
E)
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Higher temperature cause increase in:

Average kinetic energy particles.
Solution to Question 6:
A)
OR

Rate increases

Compound
Type of Bonding
Na2O
Ionic
MgO
Ionic
Al2 O3
Ionic – covalent
AlCl3
Covalent
SiCl4
Covalent
PCl5
Covalent
Average speed of particles
OR

Number of collisions

Number of collisions with E
act.

Rate
Higher concentrations

Number of collisions per unit
time increases

B)
Probability of favorable coll-
C)
(i)
Basic
(ii)
Atmospheric
(i)
Na2O(s) + H2O(l) 
2NaOH(aq)
isions increases

Rate increases

Number of molecules with
(ii)
SiCl4(l) + 4H2O(l)  Si
(OH)4(s) +4HCl(aq)
(iii)
K.E  Ea
PCl5(s) + 4 H2O (l) 
H3PO4(aq) + 5HCl(aq)
Increased surface results in

Probability of favourable collisions increase or number of
D)
NaOH : pH 13 to 14
HCl
: pH 1 to 2
collisions increase.
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6|Page CAPE Chemistry Unit 2
E)
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SiCl is a simple covalent mol-
state down the group
ecule with weak intermolecular
OR Decrease in stability of +4
forces.
state down the group
SiO2 is a giant covalent structure.
A large number of covalent
bonds need to be broken to break
E)
down the giant lattice structure.
E  = 0.15V
Pb4+ + 2e-  Pb2+
E  = +1.8 V
Cr2O72- + 14H- +6e- 
Solution to Question 7:
A)
Sn4- + 2e-  Sn2-
7H2O
As you go down the group
= 2Cr3- +
E  = +1.33
electrical conductivity
With Sn2+ E  cell = +1.33 - (+0.15) =
Increases with increasing met-
+1.18
allic character
With Pb2+ E  cell = + 1.33 - (+1.18)=
-0.47
B)
(i)
CO2 — covalent
(ii)
SiO2 — covalent
(iii) & (iv)
-ve E  value , cell for Sn2+ supports
reduction of Cr2O72-
GeO2 + PbO2 intermediate
(F)
carbon does not and so CCl4 does not
(covalent and
react with water
ionic)
C)
SiCl4 + 4H2O  Si(OH4) + 4HCl
CO2 + SiO2 — acidic (covalent
White ppt
structure)
GeO2 , SnO2, PbO2 - amph-
Solution to Question 8:
oteric
A)
(intermediate
ionic/
covalent structures)
D)
Si has available 3d orbitals while
Increase in E
Ge 
(i)
Fractional distillation
(ii)
Pb in-
dicates increase in stability of +2
6|Page CAPE Chemistry Unit 2

Fractions are separated according to
boiling point.
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
Higher
mass
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molecular
(or
vegetation.
boiling
point) mass fractions
at the bottom/ lower
C)
Pb is a neurotoxin or
(i)
nerve poison OR causes
molecular mass (or
damage to the brain and
B.P) fractions at the
nervous system.
top of the fraction-
CO will compete with O2
ating tower.
B)
(i)
for haemoglobin OR will
cause oxygen starvation
NO2 is formed from nit-
which can be fatal.
rogen and oxygen be-
cause of high tempera-
Pb comes from the anti-
(ii)
tures.
knock agent added to g
N2(g)+ 2O2(g)  2NO2(g)
gasoline. This forms PbO
2NO(g) + O2(g) 
burned.
2NO2(g) or
CO is formed to the
N2(G) + 2O2(g) 
incomeplete combustion
of petrol.
2NO(g)
S comes from fuel itself
(iii)
S(s) + O2(g)  SO2(g)
Unleaded gasoline is now
available on the market.
OR
2H 2S(g) + 3O2(g) 
2H2O(l)+ 2SO2(g)
(1 for equation, 1 for
balancing)
(ii)
Both NO2 and SO2 will
form acid rain. Acid rain
destroys buildings and
7|Page CAPE Chemistry Unit 2
D)

Chose proximity for raw
materials.


Port/ deep harbour facilities.
Good infrastructure e.g.
roads.

Isolation from residential
areas.

Power supply
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8|Page CAPE Chemistry Unit 2

Stable geological area.

Skilled labour force.
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B)

Causes breathing problem
Solution to Question 9:
A)
(i)

mation of photoch-
Formation of Ozone
emical smog.
hv
O2(g) +  O.(g) + O. (g)

free radicals


O3(g) + M(g) (M optional)
Along with the formation
of ozone, degradation
C)
O3(g) +  O2(g) +O (g)
hv
low flammability

relatively unreactive

low boiling point

easy liquefaction by compressing
NO g) +O3  O2 + NO2
(ii)
(ii)
ozone
protects

Destroys organic ma-

OR
rays.
Destroys vegetation
(i)
levels constant naturally.
from
materials
tter or animals
also takes place to keep
The
Destroys
such as rubber
O.(g) + O2(g) + M(g) 

Takes part in the for-
the
harmful
Prevents
layer
mutations

foaming agents ;

UV
skin cancers, gene
refridgerants
foams
earth
cataracts,


dry cleaners
fire extinguishes
D)

They have long residence times,
persist in atmosphere.

The regeneration of Cl as it
destroys O3 makes it potent.
 Ozone causes breathing
8|Page CAPE Chemistry Unit 2
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9|Page CAPE Chemistry Unit 2
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problems.
 Takes
part
formation
of
in
the
photo-
chemical smog.
 Destroys materials such
as rubber.
 Destroys vegetation
 Destroys organic matter
9|Page CAPE Chemistry Unit 2
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10 | P a g e C A P E C h e m i s t r y U n i t 2
Solution to Question 1:
A)
Reagent Condition
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B)
A → B bromine/ Br2 UV
light/heat
C → A hydrogen/ H2 Pd/ Ni
catalyst
C → B hydrogen bromide/ HBr
(or condition)
Mechanism
A → B free radical
substitution
C)
OBS
1o
2o
3o
Given
Colour
No
change
C→A
( from
C → B electrophilic addition
to
orange
green)
Product
Carboxylic
acid or
ketone
aldehyde
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change
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b.
Solution to Question 2:
A)
Points
Theoretical principle

Separation of components between mobile and

Solvent front (a)

Distance travelled by
component (b)
stationary phases.
Use

Pesticide analysis / forensic tests / purification
of
natural
C)
Rt =

Mobile phase = gas.
(i)
products/

separation of compon-
Stationary phase =
liquid
ents of mixtures. Any
other correct answer.
(ii)
B)
b
a

Non- volatile oil (Paraffin) –
Stationary
(i) a. The solvent system that
Inert gas – Mobile
carries components to be
separated.
(iii)
Retention times Y > Z >
X.
b. The solid support on which
the sample is adsorbed.
(iv)
(ii)
Polarity
The relative concentrations of the components.
(iii) a.
Solution to Question 3:
A)
Make sure the container is clean/
rinse container thoroughly/ securely stoppered container.
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12 | P a g e C A P E C h e m i s t r y U n i t 2
B)
(i)
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Pb2+ - test with KI or S
Disadvantage – format-
SO4
ion of harmful organic
2-
(aq) or
CrO
2-
4(aq)
halides. Any other rea-
NO3- - diphenylamine or
sonable suggestion.
brown ring test
(ii)
Pb2+ + 2I-  PbI2 (yellow
(iii)
Desalination/ ozonolysis
(i)
Thermal pollution
ppt); PbSO4 (white ppt);
PbCrO4 (reddish brown
D)
ppt)
Equation not necessary.
Diphenylamine forms a
blue colouration in the
presence of NO3-. OR
Brown ring at the surface
(ii)

algal growth.

(FeSO4
and
conc. H2SO4) are added.
Decrease the solubility of oxygen in
of water (interface) when
reagents
Increase in plant/
the river.

Both can result in
oxygen starvation of
aerobic organisms or
C)
(i)
Sedimentation; filtration;
aeration; addition of
charcoal; purification on
(iii) Heat exchange system;
ization; precipitation of
water in tank before
addition of Cl2 or sterilsuspended solids
(Al2(SO4)3) .
(ii)
death of organisms.
Advantage – continuous
holding
discharge;
or
any
cooling
other
answer suggesting dec-
rease in temperature of
water before disposal.
sterilization of water or
destruction of harmful
bacteria.
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Solution to Question 4:
A)
(i)
Functional groups

Alkene

Ester

Ketone

Phenyl
(ii)
Reagent
Functional
Group
Change
HCN
Solution to Question 5:
A)
Principles

Used to separate components with relatively close
boiling points.
Br2

Vapour of component of
higher boiling point condenses on fractionating column.
B)

Vapour of component of
lower boiling point rises to
the top of the fractionating
column.

Distillate of lower boiling
component collected first.
C)
Appropriate example – NO2+,
Br+
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14 | P a g e C A P E C h e m i s t r y U n i t 2
B)
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Solution to Question 6:
A)
N2(g) + 3H2(g)
2NH3(g)
Ammonia is used as fertilizer OR
in the production of fertilizer.
B)
(i)
A decrease in Keq means
the equilibrium has
shifted to the left and
there is a decrease in
C)
the amount of NH3 pro-
(i)
Ethanol

duced.
molecules
have a polar OH
(ii)
group.
Liquefying the ammonia
Allows for bonding

causes the equilibrium to
between H2O and
shift to the right which
C2H5OH molecules.
Mutual

means more NH3 is
attraction
produced.
between ethanol and
This is based on Le
water.
(ii)
(
D)
Chatelier’s principle.
% C2H5OH
20
) x 100 = 80%
25
Pure water is obtained and the
azeotrope.
B)
(iii)

(200 atm).


Rum

Perfume

Petroleum
Moderate
temper-
ature (400 -500℃) .

E)
Increase in pressure
Use of catalyst (Fe +
Fe2O3 + other metals)
High pressure - gives
high yield as equilibrium
shifts to the right.
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Temperature - low tem-
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
Safety drills
perature gives high yield
but the process is slow
and uneconomical so a
moderately high temperature provides a compromise between yield
and rate.
Catalyst - lowers the
activeation energy which
speeds up reaction
rate. Low E2 results in
lower operating temperature.
C)
(i)

Abundant labour
force

Transportation

Available source of
energy

Availability of raw
materials
(ii)

Wear helmets, masks,
protective clothing

Observe all clearly
marked safety signs

Mark hazardous
materials clearly

Exit signs should be
clearly marked
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Solution to Question 1:
A)
(i)
b.

Lone pair on phen-
ylamine is delocalized into 𝜋 system of
benzene ring.

(ii)
of lone pair to accept
Condition for Stage I: 55
proton.
-60°C
(iii)

Greater
hydrogen
bonding of ammon-
Reagent for Stage II: Sn/HCl
ium ion with water
or Fe/HCl
B)
decreases availability
provides great stabi-
(i)
lity.
C)
(i)
(Cl- is optional)
(ii) a. Phenylamine is less basic
than ammonia.
(ii)
Reagent: HCl + NaNO2
(or KNO2) OR HNO2/
HCl
Condition: < 5℃
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17 | P a g e C A P E C h e m i s t r y U n i t 2
(iii)
(iv)
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Sodium hydroxide
is not written)
solution.
No. of moles of Na2CO3
remaining = (
Yellow
= 0.002
0.004
)
2
Solution to Question 2:
A)
(i)
(iii)
Equivalence Point
reacted = (0.005 -
The stage at which the
0.002) = 0.003
reaction is complete or
the two solutions have
(iv)
reacted exactly (stoic-
Conc. Of barium ions =
0.003 x 40 = 0.120 mol
hiometric amounts of
dm-3
reagents have reacted)
(ii)
No. of moles of BaCl2
0.003
 25
1000
= 0.120 mol dm-3
OR
End Point
The point at which there
is a change in the colour
D)
(i)
of the indicator.

of
oxalic acid placed in
B)

flask.
A known excess of sodium car-

bonate is reacted with BaCl2.

The
remaining
solution


Ba2+(aq) + CO32-(aq) →
of
flask
Oxalic acid titrated
hot with manganate
BaCO3(s)
(ii)
Contents
heated (to - 80℃ ).
ration
(i)
(Equal) volume of dil.
H2SO4 added.
of
Na2CO3 is determined by tit-
C)
Stated volume
(VII) sol-ution.
Na2CO3: HCl = 1:2
(assume mole ratio if this
17 | P a g e C A P E C h e m i s t r y U n i t 2
(ii)
Potassium manganate
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(VII) solution

Energy/Power generation
(fossil fuels)

Solution to Question 3:
A)
(i)
Combustion engine
Nitrogen fixation:

conversion of atmos-
Deforestation

Biological – use of
Costly policies

bac-teria in plants
Result in unemployment

Price of essential goods and
rogen compounds.
D)
(ii)
(legumes)

Industrial Production

pheric nitrogen to nit-

Transportation/ Internal
Athospheric
electrical
services increases
–
discharge
in the atmos-phere /
E)
(i)
brown
lightning

Industrial
process
B)
–
pink/red/ reddish brown/
haber
(ii)
Reagent – (Conc.)
NaOH(aq)/ caustic soda
Acid Rain
(iii)
Colour – white
Appearance – gelatinous
N2(g) + O2(g)  2NO(g)
(precipitate).
2NO2(g) + O2(g)  2NO2(g)
O2(g) + NO2(g) + 2H2O(l) 
(iv)
4HNO3(aq)
Process


2 NO2(g) + H2O(l)  HNO2(l) +
HNO3(l)
C)
Filtration
Heating
Solution to Question 4:
A)
(i)a.
Stereoisomerism
Compounds that differ

CFC’s
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only
in the spatial arr-
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angement of atoms/
groups in molecule.
b.
Chiral centre
Carbon atom which is
bonded to four different a
atoms/ groups.
b. pH=13
(ii)
Optical isomerism
c. pH=7
(iii)
B)
(i)
Condensation
(iv) a. Ph=2
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(ii)
(ii)
Categories


(iii)
Stretching
Bending
Steps

Grind
solid
with
excess KBr to a fine
consistency

Press mixture into a
tablet/ pellet
OR

(iii)
with drop of liquid
Amide (Amido)
hydrocarbon (paraffin oil)
Solution to Question 5:
A)
(i)

Two Principles

Molecules

Absorption
undergo
occurs
when vibration cause
a net change in dipole
moment of molecules.

The energy of vibrational transitions cor-
The mixture (mull) is
pressed between plates of NaCl
changes in modes of
vibration.
Solid finely ground
B)
(i)
A: O – H (alkane)
B: C – H (carboxylic
acid)
C: C = O (carboxylic
acid)
responds to IR spectrum.
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(ii) Structure of Y
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Solution to Question 6:
A)
(i)

C)
Advantages

Environmentally
solvent

(ii)
friendly
Cheap/ cost effective (less
fuel)

Water insoluble product

Does not destroy/ decompose extract.
D)
(i)
Disadvantages

Partially hydrolysed
product

Difficulties in
removing all the
water

Yield will be lower
Chlorine

Hydrogen

Sodium Hydroxide
At the Anode
Chlorine ions are converted to
chlorine gas by oxidation
(electron loss)
2Cl-  Cl2(g) + 2e-
At the Cathode
Water is decomposed to
produce hydrogen by reduction
(electron gain)
2H2O(l) + 2e-  H2(g) +
2OH-(aq)
Sodium Hydroxide
Hydroxide and sodium in the
(ii)
Length of Extractor
Increased surface area
due to the length of the
anode compartment are removed as sodium hydroxide.
Na+(aq) + OH-(aq) → NaOH(s)
extractor, increases efficiency of the process.
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B)

Insecticides

Bleaches/ disinfectants

Antiseptics

Weedkillers

Chlorinated organic compounds/ solvents
C)

SO2 is a reducing agent
which prevents oxidation of
the product, thus retarding
food spoilage.

Disadvantage – distorts the
taste of the food.
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Solution to Question 1:
A)
(i)
I: Hot KMnO4, OH- / H+
(ii)
Addition polymerization
(iii)
An relevant example : PVC,
(ii)
II: Concentrated H2SO4/
polyethene (do not accept
H2 O
B)
polyalkene)
(i)
(IV)
(ii)
(V)
No. There is no
present.
C)
(ii)
Polymer – a large
molecule made up of
D)
Amino acid
many identical repeating subunits called
Monomers.
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Wavelength: the dista-
(ii)
nce between successive
peaks (or troughs) of
waves of radiation.
(iii)
E)
Frequency: The number
of waves passing a given
Similarity
point per second or the
Both contain the amide linkage
number of cycles per
second.
B)
Difference
Proteins are formed from amino
(i)
1 x 10-5
(ii)
9.5 x 107 Hz
(iii)
Infra-red
Radio Waves
(iv)
acids while Nylon 6.6 is formed from diacid.
(butane – dicarboxylic acid) and
the diamine (1,6 – diamino-
C)
(i)

is weighed using an
hexane).
analytical balance (usually
Solution to Question 2:
A)
(i)
Electromagnetic radiation consist of oscillating
electric and magnetic
fields of energy which
can be transmitted through space.
24 | P a g e C A P E C h e m i s t r y U n i t 2
Accurate mass of sample
1.0
mg/ml
for
organic).

Sample dissolved in appropriate
solvent
and
made up to the mark in a
volumetric flask. (2 or 5
ml) for organic compounds)
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
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If sample has any par-

sent within or without
clear, transparent, homo-
the sample).
genous solution).
D)
(i)
Energy absorbed by one
is blanked, sample sol-
molecule:
ution absorbance is chec-
E = hv = 6.63 x 10-34 Js  1.5
ked at a particular wave-
 1015 s-1
length. (190 - 900 nm)

inter-
constituents (those pre-
suitably filtered to give
After spectrophotometer
from
ference by other solution
ticulate matter, it must be

Freedom
= 9.9 x 10-19 J
If absorbance value is
(ii)
greater than 1.0, serial
Energy absorbed by one
mole of substance:
dilutions are necessary.
E mole =E molecule×L
=9.9×10-19 J×6.02×1023 mol -1
(ii)

=6×105 Jmol -1
Stability of complexing
reagent

Ability to develop colour
with sample

Stoichiometric reactivity
with
desired
reagent
sample

Solution to Question 3:
A)
(i)
SO2, NO2, CO2
(ii)
CO2: respiration
Transparency in the UV/
NO2: oxidation of N2 due
ion
N2 + 2O2 → 2NO2
to lightning
visible wavelength reg
Specificity to the sample
SO2: Volcanic activity
reagent

Ability to function in a
common solvent
25 | P a g e C A P E C h e m i s t r y U n i t 2
(iii)
CO2(g) + H2O(i)
H2CO3(aq)
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SO2(g) + H2O(i)

H2SO3(aq)
toxic metal ions

Or
buildings
2SO3(aq)

destroys vegetation

ensure the container is
SO3(g) + H2O(i)
C)
H2SO4(aq)
(i)
2NO2(g) + H2O(i)
clean.
HNO3(aq) + HNO2(aq)

(i)
In industrialized and
populated
areas there
(ii)
KI or any other
halide

combustion of fuels

emission
of

gases
Chromate ; SO42-
NO3-
from industrial plants

and factories
(ii)
Pb2+

the:

rinse container
thoroughly.
would be an increase in

increases rate of corrosion of metals and
SO2(g) + O2 (g)
B)
increases solubility of
Acidified FeSO4 and
deforestation via bur-
conc. H2SO4 (brown
ning
ring test)

vehicular emission
(boiled)
Effects of acid raid:

destroys aquatic life

leaches soil nutrients
Cu or Zn with NaOH
(iii)
Pb2+

KI – yellow ppt.

Other halides or SO42- white ppt.
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
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Chromate – yellow ppt
(ii)
NO3
Acidified FeSO4 + conc.
H2 SO4 to yield brown
ring
at
junction
(or
interface)

Cu or Zn with NaOH releases NH3 with boiling.
B has no chiral carbon
Solution to Question 4:
A)
atom.
Molecular formula = C9H18O
the same, but different structural formula.
D)
Cracking-process involving
(i)
the breaking down of large
alkanes to smaller hydro-c
B)
alkene
C)
carbons.
A and B: (primary) alcohol and
(ii) a. C8H18  C4H8 + C4H10
(i)
b.

Use for Br2(aq) or Br2 in
organic solvent KMnO4

Bromine decolourized
or KMnO4 goes colourless

Only alkene (butane)
gives a positive test
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C)
Solution to Question 5:
A)

(75.77×34.97)+(24.23×36.96)
=35.45
100
Determination of relative isotopic masses and relative
isotopic abundance.

Distinguish between molecules

of
similar
D)
relative
Two uses of chromatography:

Separation of dyes and inks
molecular mass.

Pesticide analysis
Prediction of possible ide-

Forensic testing

ntities of simple organic
Purification of natural products
molecules based on fragmentation pattern.
Solution to Question 6:
B)
(i)
M: 78
A)
Fractional distillation
M + 1 = 79
(ii)
43 - (CH3CHCH3)+
B)
63 - (CH3CH35Cl)+
M - CH3CH35ClCH3
M + 1 - CH3CH35Cl13CH3
2- chloropropane
(i)
Reforming
(ii)
CH4(aq) + H2O(g)  CO(g)
+3H2(g) OR CH4(aq) +
65 - (CH3CH37Cl)+
(iii)
Air (liquid)
2H2O(g)  CO2(g) + 4H2(g)
C)
High Pressure favours reduction in
volume
Low Temperature favours exothermic reaction (forward)
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D)
Moderate
temperature
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and
catalyst increases the rate of
equilibrium concentration.
E)
Pollutant
Possible
Polluting Effect
Source
-
(NO3)
Fertilisers
- Eutrophication
Synthetic
/ algal bloom
detergents
- Reduction in
Untreated
O2 for marine
sewer waste
life
or BOD
3-
(PO4)
Fertilisers
Eutrophication /
Synthetic
algal bloom
detergents
Untreated
sewer waste
2+
Pb
(aq)
Water
- Cumulative
pipes, paint,
poison affects
lead
gut
batteries
and nervous
(any
system.
relevant
- Can be
source)
carcinogenic
and
teratogenic
- Brain damage
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Solution to Question 1:
A)
(i)
Test: KMnO4/ H+ or
(ii)
philic substitution or u
+
K2Cr2O3/ H
unimolecular nucleoph-
Obs: purple to colourless or
orange
to
green
iclic substition
when
reacted with compound A.
(ii)
Snl or 2-step nucleo-
C)
(i)
Step 1: conc. H2SO4
Step 2: conc. H2SO4 and H
Obs: Silver mirror when
H2O (H2SO4(aq) )
reacted with compound B.
Step 3: KMnO4 l H+ or K
(iii)
Test: 2,4 – dinitrophenyl-
K2Cr2 O2 l H+
hydrazine
(ii)
Obs: Yellow ppt with
Compound
B)
(i)
Oxidation
Solution to Question 2:
A)
Both refer to the closeness of
agreement between two or more
measurements. Whereas precision refers to measurements of the
same quantity. Accuracy refers
to the closeness of a measurement to the true value of the
quantity being measured.
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B)
Burette,
pipette,
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measuring
cylinder, volumetric flask.

Condition a 10cm3 pipette.

Fill pipette to the mark with
water at room temperature.
C)

(i)
Student
S.D.
1
0.065
2
1.15
3
3.46
4
0.082
Transfer contents of pipette
to the beaker (without spilling).

Weigh accurately beaker
and water.

Record weight in a suitable
table.

Repeat steps 3-6 until consistent water masses are
(ii)
obtained.
Student
1 – precise but not
accurate
Solution to Question 3:
A)

2 – accurate but not
Crude oil is a mixture of
hydrocarbons.
precise

3 – neither precise nor
Heating separate components according to volatility
accurate
(b.pts).
4 – both accurate and

precise
Components are collected by
use of a fractionating tower
D)
(more volatile collected first).

Weigh a (clean, dry, empty)
beaker on an analytical balance.
31 | P a g e C A P E C h e m i s t r y U n i t 2
B)
Cracking: the breaking up of
large molecules into smaller
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ones.
(ii)
Types of structural
Reforming: the rearrangement of
isomers:
atoms in molecules to form new

Chain

Functional Group

Positional
structures (molecules).
C)
C8 H18  C4H10 + 2C2H4
D)
The production of insulators
(clothing)/ textiles/ plastics/ solvents/ pharmaceuticals
E)
(i)
A : Yeast
B: starch or sugar
(sucrose)
(ii)
Conical flask – effervescence
Beaker – white ppt formed
(iii)
CxHy + (x +
High temperatures will
kill the yeast.
10cm3 35cm3 20cm3
Structural isomerism – refers to
Mole ratio:
x=2
compound with the same
(i)
y
) O2 → x CO2 +
4
y
H 2O
2
Solution to Question 4:
A)
B)
32 | P a g e C A P E C h e m i s t r y U n i t 2
3.5
2
x + = 3.5
= 3.5 -2
Molecular formula but different structural formula .
1
= 1.5
y=6
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MF = C2H6
C)
bond (inductive effect).
Order of increasing acidity.

Alcohols
<
Solution to Question 5:
phenols
<
A)
(i)
between molecular energy
carboxylic acids alcohols are
levels.
weakest because R group is
electron
releasing
Electronic absorption
which
(ii)
makes the negative charge
localized on the oxygen,
therefore, it is less likely to
release a proton

Phenols
are
because
of
weak
the
acids
electron
withdrawing ability of the
Under
phenyl ring making the
phenolate
ion
energy
level
diagram it is important to
resonance
point out only three possible
stabilized.

the
transitions can occur in the
Carboxylate ions are reson-
ultraviolet region of the
spectrum
ance stabilized.
which
are:
n   ,    and n  

B)
(i)


Chromophore: a group of
atoms in a molecule res-
They are also less likely to hold on to
the proton because the electronegative oxygen pulls electrons
ponsible for (producing)
the absorption of electromagnetic radiation.
towards itself away from the O – H
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Chromophore
(ii)

in blood
– benzene structure or

-conjugated system of
Using A = E C 1
1.2 = 288  1 C
C
Determining the extent of
cyanide pollution of water
double bonds.
(iii)
Determining amount of urea

Analysis of iron in iron
tablets
Solution to Question 6:
A)
1.2
288 1
A: nitrogen-fixing bacteria
(nitrifying bacteria)
B: proteins
 C = 4.2 x 10-3 mol
C: soil nitrates
dm-3
D: nitrogen oxides; (NOx)
C)

Prepare number of standard
solution.

Take absorbance reading of

each solution.
Plot curve of absorbance vs

Concentration. of unknown
B)


34 | P a g e C A P E C h e m i s t r y U n i t 2
Using of internal combus-
cars, etc).

C)
cose in blood
fuels
(burning of fossil fuels in
using calibration curve.
Determining amount of glu-
fossil
tion engine for transport
can then be determined

of
(industry)
concentration
D)
Burning
Deforestation
Ozone formation:
Dissociation of oxygen into
atoms by UV radiation
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O2(g)  O(g) + O+(g)
Ozone is formed by reaction of
oxygen atoms and molecular
oxygen
O2(g) + O+(g) + M(g)  O3(g) +
M+(g)
Ozone breakdown:
Ozone decomposes to molecular
oxygen and oxygen atoms by
lower energy UV radiation
hv
O3(g) 
 O2(g) + O(g)
Ozone molecules and oxygen
atoms produce two molecules
of oxygen
O3(g) → O(g) + 2O2(g)
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to colourless
Solution to Question 1:
A)
Isomeric Alcohols MF – C4H10O
2 – methyl-2-propanol no
reaction.
E)
Positive test with K2Cr2O7 l H+
Orange/ yellow blue/ green
Solution to Question 2:
A)
Forensic testing or any other
reasonable suggestion.
B)
Structural Isomerism
C)
Stereoisomerism (optical
B)
Experimental Procedure

bottom
isomerism)
D)
Observation on treatment with
KMnO4 l H+
1 – butanol
with
a
solvent
(mobile phase).

Apply the mixture to be separated as a spot at a short dis-
purple to col-
tance from one end of the
ourless
2 – butanol
Obtain a beaker and cover its
TLC plate.
purple to col-
ourless
2 – methyl-1-proponal purple
36 | P a g e C A P E C h e m i s t r y U n i t 2

Dip the end of the plate
below the spot into the
solvent
(make
sure
the
solvent level does not cover
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the spot).

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D)
Allow the solvent to climb the
phases
plate until it nears the top of

Silica (SiO2)

Alumina (Al2O3)
the TLC plate.

Commonly used stationary
Mark the level the solvent
has reached on the plate
(alternatively
make
E)
(i)
the
mark first and remove the
plate
when
the
solvent
reaches the mark)
A
2.4
 0.35
6.8
B
5.1
 0.75
6.8
A is more attracted to the

Allow the plate to dry.

Stain the TLC plate to make
stationary phase/ has a
the components visible or
greater retention time.
mark the spots using UV
B is more attracted to the
(ii)
light to highlight.
C)
Rr Values
Stationary phase is a solid TLC
and is the material which holds
the solute on the plate.
Mobile phase is a liquid and
carries the solute along the
stationary phase on the TLC
plate.
37 | P a g e C A P E C h e m i s t r y U n i t 2
mobile phase/ has a
smaller retention time.
(iii)
Factors influencing Rr
value

Nature of solute

Nature of stationary
phase

Nature of mobile
phase
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material/wastes – pollution
Solution to Question 3:
A)
and genetic mutations.
Paper

Reuse: Newspaper as absorbing
material,
off-cuts
C)
Carbon dioxide, hydrogen
used for note-pads, etc. any
sulphide, methane.
other reasonable application.

Recycle: Making a papier
maché,
reconstitution
of
CO2, H2S, CH4
D)
(i)
glucose and then to
cellulose to reproduce recycled
paper.
Any
ethanol by enzymes pro-
other
duced by yeast.
reasonable application.

Sucrose is converted to
C12H22O11 + H2O  2C6H12O6
Reduce: In offices reduce
frequency of written (typed)
Sucrose
memoranda. Use email for
C6H12O6  2C2H5OH(aq) + 2CO2
Ethanol
communications: any other
reasonable application.
Glucose
(ii)
Acetic acid/ Ethanoic acid,
B)

Waste streams contaminate
ethanol Oxidation (Redox).
and degrade land.

Radioactive gases – free
radicals – released into the
air – carcinogenic.

Large land use areas are
associated with reactor systems, cooling systems, artificial lakes and buffer areas
with potential radioactive
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Solution to Question 4:

A)
attached to the carbon atoms of
With aqueous Bromine
(i)
There are no common groups
the double bond.
Liquid bromine
(ii)
Electrophilic Addition
(iii)
Cold KMnO4 l H+
Solution to Question 5:
A)
Fig 1: Sintered glass crucible –
Filtration by suction and drying
(iv)
Hot KMnO4 l H
of ppt in oven/ furnaces.
+
Fig 2: Suction (Buckner) Funnel –
filtration by suction.
B)
B)
Compound A does not exhibit
transisomerism.
(i)
[ester(toluene)]eqm
=r
[ester(water)]eqm
(partition coefficient)
C)

Both groups to each carbon
of the double bond are the
same.
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(ii)
(iv)

Temperature

Immiscible solvents

1.6
 0.016
100
(do not react with

each other)

Conc, of Y in ether =
8.4
 0.042
200
Solute does not react
with solvents

(iii)

Conc, of Y in water =
Partition coefficient =
0.042
 2.625
0.016
Organic compounds

generally are more
With respect to K,
ether and water.
soluble in nonpolar
solvents than in water

(polar). Solvents are
Solution to Question 6:
immiscible.
A)
Location of an industrial plant:
An organic comp-

ound is partitioned
Proximity to workforce

between two solvents
Good transportation

Readily accessible energy
by shaking.

Organic compound is
needs

recovered/ obtained
Readily accessible raw materials.
by distilling of the

solvent after separa-
Political situation

tion of two layers.
Site development

Expansion possibilities

Any other reasonable factor.
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B)
Safety concentration:

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C)
High pressure process –
2UK: reactants + products
possibility of explosions
2CO + O2  2CO2
2 UK: reactants + products
Measures: Any one

2CO + 2NO  2CO2 + N2
Materials used in construction should be able to
D)
(i)
SO2 + H2O  H2SO3
withstand high pressure.

Produces acid rain.
Workers should be fully
OR 2SO2 + 2H2O + O2
aware of evacuation pro-
 2H2SO4
cedures.

Systematic safety drills
should be practiced.
Safety concern:
(ii)
Production of smog ;
adverse effects on respiratory system.
Product spill in transportation
is likely.
Measures: Any one

Clean up measures should be
developed.

Should partner with public
services in evacuation exercises.
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Solution to Question 1:
A)
(i)
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B)
(i) a. Yellow/orange ppt is
formed.
b.
KMnO4 decolourized.
c.
Z is an aldehyde.
d.
Z is an aromatic
aldehyde. Or Z is not a
(ii)
reducing sugar..
B)
(ii)
(iii) a. P – saponification
b. Q – transesterification
C)
(iv) a. P – soap
b. Q – biodiesel fuel
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Solution to Question 2:
(iii) a. C: Butanoic acid
A)
b.

Vibrations in bonds must match
frequency in the IR region of
spectrum.

Molecules must have a dipole
at 1710 cm-1
moment.
OH at 3400 cm-1
B)

Monitoring air pollutants.

Laboratory identification of

D)
(i)

organic compounds.
solid in mortar to
Identification of functional
form a mull.
groups.

Add mineral oil to

Analysis of petroleum hyd-
Smear samples between NaCl plates.
rocarbons.
C)
OR
(i) a. A: Acetone

Grind solid with KBr.

Form a pellet under
pressure.
b.

Insert sample in
machine.

at 1715 cm-1
OH at 3400cm-3
(ii) a. B: 2 – propanol
(ii)
Record readings.
They are ideal because
NaCl is transparent to
infrared radiation.
b. OH at 3350 cm-1
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Solution to Question 3:
(ii)
A)
(i)
Leaching
(ii)
Pesticides;
detergents,
herbicides,
fungicides
Excessive
Nutrients  algal
growth  Bacterial
decomposition on death
(any other appropriate
(NO3- ; PO43-)
source).
B)
(i)
(removal of oxygen)
Nitrates: FeSO4(aq); conc
H2SO4 (Brown Ring Test)
(iii)
A brown ring forms at
due to dissolve oxygen.
junction of liquids
(ii)
Phosphate: Ammonium
molybdate; yellow ppt
OR
Ammonium
molybdate
Test; Blue solution
C)
(i)

Diffusing from atmosphere (air)

Dissolving
the
by-
product of photosyn-
Corrosion in boiler pipes
D)

Screening

Sedimentation

Filtration

Coagulation

Flocculation

Chlorination
Solution to Question 4:
A)
(i)
NaOH
(ii)
thesis (aquatic plants)

Aeration from rapid
move-ment.
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B)
(i)
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Step I : Conc HNO3 and
Solution to Question 5:
con.
A)
H2SO4
of the solute concentrations
Step II: Br2 and AlBr3 l
immersed in solvents at a given
FeBr3
temperature.
Or
Step III: Sn and conc HCl
(ii)
Partition coefficient is the ratio
C1 = k (k = partition coefficient)
H2SO4 + HNO3 → NO2+
C1 = conc. of solute in solvent 1
C2 = conc of solute in solvent 2
B)
(iii)
(iv)
Electrophilic Substitution
NO2 is a meta director so
nts to the meta position.
(i)
Concentration either
k
x
10  5.36
1 x
10
Nitrobenvene< benzene<
methylbenzene

(ii)
7.36x = 5.36
x=
C)
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Concentration water
0.854
∴ 10  5.36 = k
0.159
10
Let mass found in water = x g
it directs the substitute-
C)
Using
5.36
= 0.728 g
7.36
(i)
Steam distillation
(ii)
Solvent extraction
(iii)
Fractional distillation
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(iv)
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Simple distillation

3H2O + Al2O3 + 2NaOH →
2NaAl(OH)4 (aq)
D)
NaAl(OH)4 (aq) →

NaOH(aq) +
Liquid of composition x
heated produce vapour.

Composition a1.

Vapour on condensing pro-
Al(OH)3 (s)
2Al (OH)3 (g) → Al2O3 (s) +
3H2O(g)
duce liquid composition x1.

Liquid has greater conce-
B)
(i)
ntration of A than B.

Red mud (sodium hydroxide residue)
Repeated vapourization and
(ii)
condensation produce liquid

A (ditillate).

A)
Filtrate seeded to precipitate
Al(OH)3

Al(OH)3 heated producing
aluminium oxide.
(Allow use of CO2 instead of
Removes
valuable
space
from
agricultural produc-
with NaOH

Disfigures the envir-
land
Ore is crushed and treated
Mixture is filtered
and
onment
Solution to Question 6:

flora
fauna


Harms
tion or housing.
C)
(i)
Anode: 2O2-(l)  O2 (g) +
4e
Cathode: Al3+(l) + 3e 
Al (l)
seeding)
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(ii) a. Reasons for high cost:

Large quantity of
energy needed

Need to replace
carbon anodes

Degree of purity of
metal
b. Recycling reduces:

Exploitation of
resources

Environmental
scarring

Energy demands

Pollution effects
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Solution to Question 1:
A)
(i)
To produce
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phenylamine is
bromine
calized into the benzene
radicals by breaking Br –
ring thus reducing ele-c
Br bond or cleavage or
homolytic
fission
ctron availability.
to
bromine radicals.
delo-
C)
(i)
Kb > 9.38
(ii)
Ethanamide reduces
(ii)
further the electron availability of lone pair on
nitrogen.
B)
(i)
 RNH 3+   OH - 


Kb  
 RNH 2 
(ii)
Ethylamine
(iii)
The presence of ethyl
group enhances the availability of lone pair on
the nitrogen. The lone
pair on nitrogen in
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TABLE 1: LABORATORY
D)
(ii)
RECORD
TEST
Functional Groups
X – rays – A
Observation
Add SOCl2
- Vigorous
cautiously.
- Dense white
very
Infrared – B
B)
C = frequency x λ
3.0 x 108 = 4.5 x 1015 x λ
reaction
fumes
- Pungent,
λ=
choking or
irritating smell
Add
Yellow
AgNO3(aq)
precipitate
and boil
= 6.7 x 10-8 m
C)
gently.
Add Br2(aq)
White
precipitate
formed
Antiseptic smell
Add Br2(aq)
3.0 108
4.5 1015
Bromine

Detect pollutants in the air.

Iron in iron tablets

Moisture content of soil

Metals in alloys
decolourised
D)
(i)
Step 1 – Produce S from
mixture to be analysed
Solution to Question 2:
and weighed.
A)
Step 4 – To ensure removal
(i)
of all traces of water.
(ii)
Suction funnel, crucible,
water aspirator, vacuum
pump. Any other reasonable
apparatus.
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(iii)
E)
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Oven, desiccator,
(ii)
Temperature 4500C –
crucible,
5000C
Bunsen burner.
Pressure 200 atm
Mass of washing soda = 6.44g
(iii)
Mass of anhydrous Na2SO4 =
Fractional distillation of
liquid air.
37.69 – 34.25 = 2.84g
(iv)
Mass of water = 6.44 – 2.84 =

materials.
3.60g Moles of water =

3.60g/18g mol-1 = 0.2 moles
Availability of cheap
energy sources.

Moles of Na2SO4 = 2.84/142 g
Impact on environment.
mol-1 =0.02
Na2SO4
Proximity of raw

H 2O
0.02
0.02
0.2
0.02
1
10
Impact on residents
in nearby communities.
B)
(i)
Increase in temperature
will result in the equi-
Na2SO4 .10H2O
librium shifting to the
left (exothermic in forward reaction) and dec-
Solution to Question 3:
A)
(i)
N2(g) + 3H2(g)
2
NH3(g) -92
KJmol-1
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(ii)
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The forward reaction favours
that is attached directly to three
an increase in pressure due
other carbon atoms.
to lowering of the total number of moles. There should be
B)
(i)
a higher yield of ammona.
C)
(i)
Compound A is 2bromo-butane.
(ii)
A – Steam reformin/
Optical isomerism,
presence of a chiral
cracking
carbon.
B – Dissolving of CO2 in
H2O /Caustic soda
D – Condensation /
cooling
(ii)
Finely divided iron
(iii)
Liquid
Positional Isomerism
(iii)
Solution to Question 4:
A)
Primary alkyl halide: halogen
atom attached to a carbon that is
attached directly to one other
carbon.
Secondary alkyl halide: halogen
atom is attached to a carbon
atom that is attached directly to
C)
(i)
2-bromo-2-methyl
propane
two other carbon atom.
Tertiary alkyl halide: halogen
atom attached to a carbon atom
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(ii)
(iii)
Recorder – Shows fragments as peaks according
to relative abundance.
B)
Ratio of relative abundance
indicating number of carbon
atoms.
C)
D)
(i)
Creamy yellow precipitate formed.
(ii)
NaBr(aq) + AgNO3(aq) 
(i)
88
(ii)
43
(iii)
CH3+, C2H5+
C3H7+
NaNO3(aq) + AgBr(s)
(iv)
OR
Ag+(aq) + Br-(aq)  AgBr(s)
Solution to Question 5:
A)
(i)
Electron beam – bom-
D)
bards molecules, producing
molecules,
ions
and fragments.
(ii)
Magnetic field – Separates fragments based on
M/z ratio
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B)
Solution to Question 6:
A)
(i)
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(i)
Yeast is added to glucose
/carbohydrates. The zy-
Destruction involves the
mase in yeast acts on the
ultra violet radiation
glucose to produce eth-
breaking down the oz-
anol and carbon di-
one to oxygen molecules
and atom.
oxide.
O3(g)  O2(g) + O(g)
C6H12O6(aq)  2
C2H5OH(aq) + 2 CO2(g)
Ozone and oxygen atoms
react to provide oxygen m
(ii)
Fractional distillation.
molecules.
O3(g) + O(g)  2O2(g)
(ii)
Use of CFC in aerosol
can
C)
Ethanol has a wide range of
application as solvents for fragrances, lotions, fuels for gasoline, extraction of essences
from fruits and spices. Ethanol is
(iii)

Excessive radiationskin
cancer;
free
radicals production
an active ingredient in alcoholic beverages which can
sold locally and internationally.
in the body impairs
the immune system.

Cataract in the eye.

Respiratory ailments;
asthma, bronchitis.
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Solution to Question 1:
A)
(i)
Cracking
(ii)
(ii)
Bromine with propane in
sunlight.
Vapour passed over catalyst/ Thermal crack
ing Al2O3/SiO2 at 450oC.
Propagation:
high temperature (greater than 600oC).
(iii)
B: methane, CH4
C: propene, CH3CH=CH2/
CH3CHCH2
B)
(i)

There is the unpairing of 2s2 electrons
and one electron enters the empty P orbital.

The four orbitals are
hybridised to give 4
sp3 orbitals.

Two steps are required
in the propagation stage,
if two steps are present
and no fish hook arrow.
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
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If one of the two steps is
missing and fish hook is
missing no marks allotted.
Termination: Br  Br  Br : Br
OR
OR
Bromine with propene
(without sunlight)
The two first steps can be
merged with the appropriate
All arrows should be full
arrows not fish hook.
curved arrow.
(iii)
Free radical substitution
OR
Electrophilic addition
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C)
(i)
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The reddish brown

colour of the bromine is
by drop near the end
decolourised in sunlight.
point

(ii)
Adding titrant drop
Taking
successive
The purple colour of the
readings until cons-
permanganate solution
tant
is immediately decolo-
obtained.
urised/ pale pink.

volumes
Taking
are
successive
readings until the
volume
Solution to Question 2:
A)
(i)
Accuracy:

Reading burette at



Removal
of
filter
0.1cm3.
B)
funnel after adding

High level of purity.
titrant to burette.

Atmosphere stability
Reading burette to 2

Large molar mass
decimal places.

Reasonable solubility in titration
Eliminate any bubbles in burette tap.

Absence of hydrate water.
Retort stand must be

Cheap and readily available.
on a flat surface

Do not blow out
remaining liquid at
the tip of the pipette.
(ii)
not
differ by more than
eye level

does
Precision:
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C)
Draw line of best fit
(i)
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
use six points
This is repeated until a total
of 45 cm3 of acid are added
or until the temperature
stabilized.
End point volume: 24 
(ii)
0.5
(iii)
25 cm3 NaOH contain =
2.0  25
=0.05 moles
1000
Moles NaOH: HCl= 1:1
Moles HCl=0.05 moles
24.5cm3 HCl contains
0.05 moles.
1000 cm3 contain
0.05 1000
24.5
=2.04 M HCl
D)

25 cm3 NaOH added to a
Solution to Question 3:
A)

polystyrene cup using a

Increased pressure/ conc. Of
reactants.
pipette.

(Solution allowed standing
Decreased temperature

Use of catalyst
for few minutes) and temperature noted.

5 cm3 portions of acid added
from burette, mixture stirred
and temperature noted.
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B)
(i)
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Le Chatelier’s Principle:
(v)
if one or more factors
Safety considerations

Reaction
between
that affect an equilibr-
SO3 and water is
ium are changed, the
highly exothermic.
position of equilibrium

shifts in the direction
Clouds of sulphuric
acid are produced.
which opposes the cha-

nge.
Sulphuric acid can
cause burns to the
skin and flesh.
(ii)


Low temperature

High pressure

The sulphur trioxide
cause blindness if it
gets into the eyes.
(iii)
is
dissolved
into
conc. Sulphuric acid
to form oleum.

Sulphuric acid can
The oleum is diluted
with water to form
conc. Sulphuric acid.
Solution to Question 4:
A)
Additional polymerisation involves the linking of one type of
monomer with double or triple
bonds to give the polymer as the
only product.
Condensation
polymerisation
involves the linking of two types
of monomers OR monomers
(iv)

H2SO4(l) + SO3(g) 
H2S2O7 (l)

H2S2O7(l) +H2O(l) 
2H2SO4(l)
58 | P a g e C A P E C h e m i s t r y U n i t 2
with bifunctional group (two
functional groups) to give the
polymer and a small molecule
such as water or hydrogen
chloride.
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B)
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Carboydrates/ Polysaccharides
Proteins
E,g., glycogen, cellulose, starch,
pectin,
albumin,
keratin,
giving
monomers:
collagen.
C)
Solution to Question 5:
A)

D)
Chromatography
involves
the separation of compo-
The link between the two
nents of a mixture between
two phases.
monomers
is

Involves the partitioning of
components
because one end of unit is
stationary
between
phase
a
and
a
mobile phase.
and the other end is
. The

Partitioning occurs because
the
remaining part of repeating unit
components
mixture
will
of
the
experience
different absorption forces
has
and
with the stationary phase.

Have
different
solubility
with the mobile phase.
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B)
(i)
The components can be
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

of
will react with the
required.


(ii)
2.5
 0.2
13.50
12.5
R f of R 
 0.9
13.5
R f of Q 
R has a greater solubility
in the mobile phase than
Q which has a stronger
tendency to absorb onto
the stationary phase. OR
Q is more polar so it is
large
amount
Cannot separate compounds
Samples can be removed
Coloum chromatography:
Large amounts of material can
be separated and collected.

Used for pesticide analysis

Large columns can be used
for purification.

Can be used to prepare
compounds.

Fractions can be collected
for analysis.
stronger absorbed onto
the stationary phase than
R which is less polar.
C)
Thin layer chromategraphy:
Separates small amounts of
compounds.
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are
from analysis.
Dis tan ce moved by solute
Dis tan ce moved by solvent
(dyes,
of similar Rf values.
coloured compound.
Rf 
separate
Less useful if quantification
visualizing agent. This
component and form a
(iii)
to
amino acids, plant pigments)
detected by a reagent
called a locating agent or
Used
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layer in the atmo-
Solution to Question 6:
A)
sphere.

Destruction of the
ozone layer allows
harmful UV radiation to reach the
surface of the earth.

The
increase
UV
radiation could resB)
ult in any of the
Health concern
following:
Use of asbestos diaphragm

skin cancer etc.
Asbestos deteriorates with

use overtime.

with ozone to cause
its
carcinogenic.
C)
(i)
declined:

It was discovered in
the 1970’s that CFCs
accelerates the depletion of the ozone
61 | P a g e C A P E C h e m i s t r y U n i t 2
conversion
to
oxygen.
Possible leakage of chlorine
Production of CFCs has
broken
radicals which react
Asbestos is considered to be
gas into the environment.
are
to give chlorine free
respiratory problems.

CFC’s
down in stratosphere
Dry asbestos fibres can be
inhaled/ ingested causing

cateract,
(ii)
Products of combustion
of fossil fuels- CO2:

CO2 is a “greenhouse
gas”.

Increased
concent-
ration of CO2 contributes to the green-
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house effect i.e. global warming.

Negative effect on
the atmosphere through
melting
glaciers
and
of
ice-
caps  increase in
sea levels.

Decreases productivity of the land.
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