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```Answers to CSEC®
Chemistry Examination
Practice
Section A
1
2a i
Final volume (cm3)
Initial volume (cm3)
Volume used (cm3)
2
12.60
0.70
11.90
23.80
12.20
11.60
3
26.90
15.30
11.60
[5]
ii 11.60 cm3 (average of readings 2 and 3) [1]
iii Amount in mol = concentration × volume
Amount in mol of KMnO4 = 0.025 mol dm−3 × 0.0116 dm3
= 2.9 × 10−4 mol [1]
iv Mole ratio of Fe2+ to MnO4− is 5 : 1
amount in mol of Fe2+
in 25 cm3 of solution = 5 × 2.9 × 10−4 mol
= 1.45 × 10−3 mol [1]
v 100 cm3 of solution is 4 × 25 cm3, so in 100 cm3 of this
solution there are 4 times as many Fe2+ ions:
amount in mol of Fe2+
in 100 cm3 of solution = 4 × 1.45 × 10−3 mol
= 5.80 × 10−3 mol [1]
Mass of Fe2+ in 100 cm3
of solution = 56 g mol−1 × 5.80 × 10−3 mol
= 3.25 × 10−1 g [1]
vi 100 cm3 of solution contains 5 tablets. [1]
Mass of Fe2+ in one tablet =
3.25 × 10 −1 g
5
= 6.5 × 10−2 g = 65 mg [1]
vii Two tablets give a dose of 2 × 65 mg = 130 mg; three
tablets give a dose of 3 × 65 mg = 195 mg
The daily dose should be 2–3 tablets. [1]
b i
Graph plotted for [5] marks. (Appropriate title [1], axis
labels [2], points plotted accurately and lines drawn [2].)
Cooling curve of naphthalene
100
Temperature (°C)
80
60
40
20
0
0
1
2
3
4
5
Time (min)
6
7
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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Answers to CSEC® Chemistry Examination Practice
ii 80 °C [1]
iii As it cools it will naturally crystallise. Stirring will
prevent this from happening. [1]
iv Test
Observation
A small amount of solid M was heated A brown gas evolves which turns moist
in a dry test tube.
blue litmus red.
Another gas evolves which rekindles a
glowing splint.
To a sample of a solution of M,
Pale green precipitate [1]
Precipitate remains in excess [1]
until in excess.
To another sample of solution M, a few Brown fumes evolved which turn moist
blue litmus red.
by concentrated sulfuric acid.
3a i
Graph plotted for [6] marks. (Appropriate title [1], axis
labels [2], key for different currents [1], points plotted
accurately and best-fit lines drawn[2].)
Mass of silver deposited on two
watch casings at different currents
6
Key
1.5 A
Mass of silver deposited (g)
5
3.0 A
4
3
2
1
0
0
5
10
20
15
Time (min)
25
30
ii 1.1 g [1]
iii Initial mass = 3.5 g
From graph, mass of silver deposited after
22 minutes = 4 g [1]
Total mass of watch casing = initial mass + mass deposited
= 4.0 g + 3.5 g = 7.5 g [1]
iv Silver metal [1]
v Ag+(aq) + e− → Ag(s) [1]
vi Q = It
Q = 3 A × 30 × 60 s = 5400 A s = 5400 C [1]
96 500 C would deposit 1 mol of silver, which has a mass
of 107.87 g [1]
Let y g be the mass deposited by 5400 C.
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CSEC® Chemistry Examination Practice © Keane Campbell 2016
Inference
NO2(g) [1]
O2(g) [1]
Fe2+ ion present
NO3− ion present [1]
Section A
y=
5400 C × 107.87 g
96 500 C
[1]
y = 6.04 g [1]
vii Yes, there is a difference. The theoretical mass >
experimental mass. [1]
Reason, one of the following (for [1] mark):
n
Incorrect reading of the mass deposited at the cathode.
n
Time was not properly recorded.
n
n
Variation of value of current during experiment.
b Use a magnet to remove all the iron filings.
Add water to the mixture of calcium carbonate and copper
sulfate.
Filter to separate the suspension of calcium carbonate and
copper sulfate mixture.
Evaporate 2/3 of the filtrate and leave the copper sulfate to
crystallise. [4]
c Table completed for [4] marks.
Test
To a sample of solution FA1, aqueous
ammonia was added until in excess.
Barium chloride followed by dilute nitric acid
was added to another sample of solution
FA1.
another sample of solution FA1.
4a i
Observation
No precipitate.
Inferences
Ca2+ ion present. [1]
White precipitate. [1]
Precipitate dissolves. [1]
SO32− ion present.
Colour change from purple to SO32− ion present.
colourless. [1]
Osmosis [1]
ii There is a higher concentration of water [1] inside the
leaves of the weeds than in the salt solution on the
outside surface. [1]
iii Water travels from a hypotonic solution (lower
concentration of solute; higher concentration of water)
to a hypertonic solution (higher concentration of solute;
lower concentration of water) [1] which dehydrates the
cells present in the weeds [1], thereby disrupting the
internal water balance of the plant cells, causing the
plant to collapse. [1]
b i
Simple distillation [1]
ii Electrolysis [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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Answers to CseC® Chemistry exAminAtion PrACtiCe
iii Diagram for [6] marks. (Bulb included [1], electrodes
correctly labelled [1], crucible included [1], molten
sodium chloride [1], ions at their correct electrode [1])
+
–
anode
cathode
Cl–
Na+
molten sodium
chloride
heat
c
Test
X is heated strongly in a boiling tube.
A blue litmus paper was placed at the
mouth of the boiling tube.
The boiling tube was connected to
another tube containing acidiﬁed
potassium manganate(VII).
To 1 ml of solution X is added silver
nitrate followed by aqueous ammonia.
Aqueous sodium hydroxide is added to
1 ml of solution X and heated.
Observations
• Choking odour gas evolves which
turns moist blue litmus red.
• Colour change from purple to
colourless. [1]
Inferences
SO2 present. [1]
Gas is acidic. [1]
Reducing agent present. [1]
No observable change.
No Br− , Cl− or I − ions. [3]
Colourless gas with pungent
odour which:
• turns moist red litmus blue
• gives dense white fumes with
HCl(g).
NH4+ ions present. [1]
OH− (aq) + NH4+(aq) →
H2O(l) + NH3(g) [1]
NH3(g) + HCl(g) → NH4Cl(s) [1]
d (NH4)2SO3 [1]
5 a Heat of neutralisation is the heat change when 1 mol of
water is produced in a reaction between an acid and an
alkali. [1]
Heat of solution is the heat change when 1 mol of solute
dissolves in such a volume of solvent [1] that further dilution
by the solvent produces no further heat change. [1]
b i
Amount in mol = molar concentration × volume
Amount in mol of NaOH used = 0.5 mol dm−3 × 0.05 dm3 [1]
= 2.5 × 10−2 mol [1]
OR
1000 cm3 NaOH solution contains 0.5 mol of NaOH [1]
50 cm3 of NaOH4 solution will contain
50 cm 3 × 0.5 mol
1000 cm 3
= 2.5 × 10−2 mol [1]
ii Amount in mol = concentration × volume
Amount of H2SO4 = 0.5 mol dm−3 × 0.025 dm3 [1]
= 1.25 × 10−2 mol [1]
OR
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CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section A
1000 cm3 H2SO4 solution contains 0.5 mol of H2SO4 [1]
25 cm3 of H2SO4 solution will contain
25 cm 3 × 0.5 mol
1000 cm 3
= 1.25 × 10−2 mol [1]
iii Mass of solution = 75 g [1]
Temperature change = 21.1 °C − 27.3 °C = −6.2 °C [1]
Heat change for the reaction = 75 g × −6.2 °C × 4.2 J g−1 °C−1
= −1953 J [1]
iv H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
H2SO4 is the limiting reagent. In the reaction, the mole
ratio of H2SO4 to H2O is 1 : 2. [1]
1.25 × 10−2 mol of H2SO4 will produce 2 × 1.25 × 10−2 mol
of H2O. [1]
Heat of neutralisation =
−1953 J
2 × 1.25 × 10 −2 mol
= −78 120 J mol−1 [1]
v The density of the solution is 1 g cm−3. [1]
Negligible heat is lost to the surroundings. [1]
c The rate of reaction will decrease. [1]
d Graph for [2] marks.
Total volume of product
0.5 mol dm−3
0
e i
0.25 mol dm−3
Time from start of reaction
Cation: Zn2+ [1]; Anion: NO3− [1]
ii 2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g) [2]
iii Pb2+, Al3+ and Zn2+ ions [1]
iv The hydroxides of Al, Pb and Zn are amphoteric and
react with excess sodium hydroxide forming soluble salts
which caused the precipitate to disappear. [2]
6a i
The white stain on the shirt is a compound of
aluminium. [2]
ii 1) Mass of deodorant used [1]
2) Precipitate [1]
iii • Weigh 2 g of the deodorant and crush with water,
using a mortar and pestle, to make a solution.
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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• Add 2–3 drops of NaOH to the mixture of deodorant. [1]
A white precipitate forms. [1]
• Add an excess of NaOH to the same mixture.
Precipitate dissolves in excess. [1]
• Add 2–3 drops of NH3 to a separate deodorant
mixture. [1] White precipitate forms. [1]
• Add an excess of NH3 to the above deodorant
mixture. White precipitate remains. [1]
• To another sample of the deodorant mixture add
2–3 drops of KI. [1] No precipitate. [1]
iv Al3+(aq) + 3OH−(aq) → Al(OH)3(s) [1]
Mass of precipitate
v 1) Sketch graph for [2] marks.
2) Yes [1]. The Al(OH)3 [1] is amphoteric and therefore
will react with the NaOH to form a soluble salt, hence
the precipitate will dissolve. [1]
vi Add conc. H2SO4 and Cu turnings [1] to the solid and
warm gently. [1]
If nitrate ions are present, a blue solution will form [1]
and a brown gas will be emitted. [1]
OR
Brown ring test: Make a solution of the solid. Add
iron(II) sulfate solution and mix. [1]
Add conc. H2SO4 down the side of the tube. [1]
H2SO4 sinks. [1] If nitrate ions are present, a brown ring
forms between the two liquid layers. [1]
b i
Test tube A [1]
ii 1) Temporary hardness: calcium hydrogencarbonate [1]
2) Permanent hardness: calcium and magnesium
sulfates [1]
7 a Avogadro’s Law states that equal volumes of all gases, under
the same conditions of temperature and pressure, contain
the same number of molecules. [1]
b i
From the equation, 1 mol of CaCO3 will produce 1 mol of
CO2 [1]
At rtp 1 mol of any gas occupies 24 dm3
1 mol of CaCO3 will produce 24 dm3 of CO2 [1]
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CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section A
So 100 g of CaCO3 will produce 24 dm3 of CO2 [1]
22.4 g of CaCO3 will produce
22.4 g × 24 dm 3
100 g
= 5.38 dm3 [2]
ii 24 dm3 (1 mol) CO2 has a mass of 44 g [1]
5.38 dm3 CO2 will have a mass of
= 9.86 g [1]
c i
5.38 dm 3 × 44 g
24 dm 3
Q: CuSO3 [1]; R: CuCl2 [1]; V: BaSO3 [1]; X: SO2 [1]
ii Reduction [1]
iii 2Cr2O72−(aq) + 14H+(aq) − 6e− → 2Cr3+(aq) + 7H2O(aq) [2]
iv S – Blue precipitate that does not dissolve in excess. [1]
T – White precipitate. [1]
U – Turns purple in sunlight. [1]
W – Precipitate dissolves. [1]
v Ag+(aq) + Cl−(aq) → AgCl(s) [2]
vi QSO3(s) → QO(s) + SO2(g) [2]
vii Copper carbonate, CuCO3 [2]
8a i
Calcium hydroxide [1]
ii Ca(OH)2(s) → CaO(s) + H2O(g) [2]
iii Lime raises the pH level of the soil so that the plant
roots are better able to absorb the necessary nutrients
from the soil. [1]
Ca(OH)2(s) + 2H+(aq) → Ca2+(aq) + 2H2O(l) [1]
iv Base reacts with the ammonium salt to produce
ammonia gas [1] thus nullifying the effect of the
fertilisers. [1]
2OH−(aq) + NH4+(aq) → NH3(g) + 2H2O(l) [1]
b i
Polystyrene is a good insulator [1], which reduces
heat loss to the surroundings or heat gain from the
surroundings. [1]
ii Completed table for [4] marks.
(cm3)
0
5
10
15
20
25
30
35
40
Temperature of solution
(°C)
26.0
27.5
28.5
31.5
34.5
36.0
39.0
36.0
35.5
iii A neutralisation reaction is a reaction between a base
and an acid to form a salt and water. [1]
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iv Graph plotted for [2] marks.
Thermometric titration of slaked lime and
an acidic soil solution
45
40
Temperature of solution (°C)
35
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
v 30 cm3 [1]
vi Amount (in mol) of acid used =
1.5 mol dm−3 × 0.03 dm3
= 4.5 × 10−2 mol
Amount (in mol) of
slaked lime used =
1 mol dm−3 × 0.025 dm3
= 2.25 × 10−2 mol [1]
Mole ratio of acid to slaked lime =
4.5 × 10−2 : 2.25 × 10−2
= 2 : 1 [1]
c ZnCO3(s) → ZnO(s) + CO2(g) [1]
The formation of a white precipitate with slaked lime
indicates the presence of carbon dioxide; hence the
compound is a carbonate. [1]
Pb2+, Al3+, Zn2+ and Ca2+ ions all produce a white precipitate
with NaOH added dropwise. The ions Pb2+, Al3+, Zn2+
dissolve in excess NaOH. Only Zn2+ will dissolve in excess
NH4OH. [1]
Zn(OH)2(s) + NH4OH(aq) → [Zn(NH3)4](OH)2(aq) [1]
M is ZnCO3 [1]
9a i
Any one (for [1] mark each) from:
• The density of the solution is 1 g cm−3.
• Negligible heat is lost to the surroundings.
• The specific heat capacity of the solution is the same
as that of water, which is 4.18 J g−1 °C−1.
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CSEC® Chemistry Examination Practice © Keane Campbell 2016
section A
ii
Completed table for [5] marks.
Temperature after reaction (°C)
Temperature before reaction (°C)
Temperature change (°C)
Total volume of solution (cm3)
Total mass of solution (g)
36.0
26.0
10.0
75.0
75.0
iii Amount in mol of NaOH = 1.0 mol dm−3 × 0.05 dm3
= 5.0 × 10−2 mol [2]
iv Amount in mol of HCl = 2.0 mol dm−3 × 0.025 dm3
= 5.0 × 10−2 mol [2]
v
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) [2]
vi 1 : 1 [1]
vii Heat energy evolved = 75 g × 10 °C × 4.18 J g−1 °C−1
= 3135 J [2]
viii Heat of neutralisation =
−3135 J
5.0 × 10 −2 mol
[1]
Heat of neutralisation = −62 700 J mol−1 [1]
bi
ii
The dryness in Robyn’s hair is due to the presence of
sulfate compounds in her shampoo. [2]
Aqueous Ba(NO3)2 or BaCl2 [1]; aqueous HCl or HNO3 [1]
iii CO32− ions [1]; SO32− ions [1]
iv Add a sample of the shampoo to H+/K2Cr2O7; if it turns
from orange to green then it contains SO32− ions. [1]
Heat a sample of the shampoo and place lime water
above it; if it forms a white precipitate, then it contains
CO32− ions. [1]
10 a i
ii
bi
Yes. [1] Salt reduces the water content in meat, making
the water unavailable for chemical reactions that cause
decay. [1] High concentrations of salt also interfere with
the replication of microorganisms. [1]
Osmosis [1]
Gas syringe [1]; correct set up [1].
gas syringe
gas
reactants
ii
In the gas syringe [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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Answers to CSEC® Chemistry Examination Practice
iii Stop watch [1]
iv Graph for [3] marks.
Reaction of 4.1976 g of sodium carbonate and
31.68 cm3 of 2.5 mol dm−3 hyrodochloric acid
Volume of CO2 produced (cm3)
120
100
80
60
40
20
0
0
50
100
150
200
250
300
350
Time (s)
1) The initial stage of the graph [1]
2) Rate =
change in volume of product
time taken for the change
[1]
This is the gradient of the curve at time = 60 s.
Drawing a tangent gives:
rate =
108 cm 3
215 s
[1]
rate = 0.50 cm3 s−1 [1]
Reaction of 4.1976 g of sodium carbonate and
31.68 cm3 of 2.5 mol dm−3 hyrodochloric acid
Volume of CO2 produced (cm3)
120
100
80
60
40
20
0
0
50
100
150
200
250
300
350
Time (s)
10
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section A
v Curve added to graph as shown [1].
Reaction of 4.1976 g of sodium carbonate and
0.5 mol dm−3 and 2.5 mol dm−3 hyrodochloric acid
Volume of CO2 produced (cm3)
120
100
80
60
40
20
0
0
50
100
150
200
250
300
350
Time (s)
vi It would affect the initial rate of the reaction. [1]
There would be no effect on the total volume of carbon
dioxide produced. [1]
A change in particle size changes the surface area of
soda ash; the greater the surface area, the faster the
reaction rate. [1]
vii No change in the initial rate. [1] There is a 2 : 1 mole
ratio between hydrochloric acid and carbon dioxide
so a decrease in volume of acid used will decrease the
amount of acid available to react and so will reduce the
volume of carbon dioxide liberated. [1]
c Table completed for [4] marks.
Test
A small amount of solid T was placed in
a test tube and heated over a Bunsen
flame. The gas was bubbled through
lime water.
Sodium hydroxide was added to a
solution of T until in excess.
A solution of potassium iodide was
added to a separate portion solution of T.
To a third portion of solution T, excess
Observation
White precipitate. [1]
Inference
CO23− ions were present.
A white precipitate was formed that
dissolved in excess.
No observable change.
Pb2+, Al3+ or Zn2+ ions are
present. [1]
Pb2+ ion not present. [1]
Precipitate did not dissolve. [1]
Al3+ ions were present.
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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Answers to CseC® Chemistry exAminAtion PrACtiCe
Section B
2 a The structural formula shows the arrangement of atoms in
the molecule of a compound. [1]
b [2]
H
CH3
OH
N
C
O
c i
Alcohol [1]
ii 1) Upward delivery [1]; because the gas is lighter than
air. [1]
2) Covalent [1]
3)
H
×
[1]
H
4) Use a lighted splint. [1] A popping sound identifies
the gas as hydrogen. [1]
d Amide or peptide linkage [1]
e A polymer is a large macromolecular structure which
consists of repeating units. [1]
f Condensation polymerisation [1]
H
h
HO
or
N
CH3
C
[1]
O
3a i
Note that a crucible is used and not a beaker because it
is molten NaCl. [2]
+
–
anode
cathode
Cl–
Na+
molten sodium
chloride
heat
ii Anode: 2Cl−(l) − 2e− → Cl2(g) or 2Cl−(l) → Cl2(g) + 2e−
Correct state symbols [1]; balanced equation with correct
numbers of electrons [1]
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CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section B
Cathode: Na+(l) + e− → Na(s)
Correct state symbols [1]; balanced equation with correct
numbers of electrons [1]
iii Q = It
Q = 5 A × 10 × 60 s [1]
Q = 3000 A s
Q = 3000 C
From the equation: 2Cl−(l) − 2e− → Cl2(g) [1]
2 mol of electrons = 2 faradays
2 faradays will produce 1 mol of chlorine gas
2 × 96 500 C will produce 24 dm3 chlorine gas (at rtp) [1]
3000 C will produce y mol
y=
b i
3000 C × 24 dm 3
2 × 96 500 C
= 0.373 dm3 [1]
Displacement reaction [1]
ii 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) [1]
c In dilute NaCl, there will be a preferential discharge of
the H+ ion over the Na+ ion, thereby producing hydrogen
gas [1], and preferential discharge of the OH− ion over the
Cl−, thereby producing oxygen gas. [1]
In molten NaCl there is only one cation, Na+ and one anion,
Cl− to be discharged. Hence the products are sodium and
chlorine gas. [1]
4a i
Oxygen/air [1] and water [1]
ii Ore I: haematite [1]; chemical formula: Fe2O3 [1]
Ore II: magnetite [1]; chemical formula: Fe3O4 [1]
iii Coke [1] and limestone (CaCO3) [1]
iv Limestone is used to remove impurities such as sand
from the mixture. [1]
CaCO3(s) → CaO(l) + CO2(g) [1]
Note that the state symbol of CaO is a liquid because of
its molten state.
CaO(l) + SiO2(s) → CaSiO3(l) (slag) [1]
b i
R2O5(s) + 5CO(g) → 2R(l) + 5CO2(g) [1]
ii For each mole of oxide, 5 mol of CO2 are produced.
Mr of the oxide = (2 × 90) + (5 × 16) = 260
1500
260 mol
1500
260 mol [1]
1500 g of oxide are used, which is
So amount of gas produced = 5 ×
At stp 1 mol of gas occupies 22.4 dm3
CSEC® Chemistry Examination Practice © Keane Campbell 2016
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Answers to CSEC® Chemistry Examination Practice
3
So volume of CO2 gas produced =
5 × 1500
260 × 22.4 dm
= 646 dm3 [1]
iii From the equation, each mole of oxide produces 2 mol
of R.
So 1500 g of oxide will produce
R = 1040 g [1]
5a i
1500
260
× (2 × 90) g of
Green chemistry [1]
The utilisation of a set of principles that reduces or
eliminates the use or generation of hazardous substances
[1] in the design, manufacture and application of
chemical products. [1]
ii Atom economy [1]
iii Any three of the following: [3]
• prevention of waste generation
• design less hazardous chemical synthesis
• design safer chemicals and products
• use safer solvents/reaction conditions
• increase energy efficiency
• use renewable feedstocks
• avoid chemical derivatives
• use catalysts
• monitor to prevent pollution
• minimise the potential for accidents.
b i
Any two (for [1] mark each) from: garbage from
neighbouring streets; major untreated sewage from
poorly functioning plants; waste effluent from industrial
plants; land reclamation and engineering such as
the causeway bridge that have interfered with the
tidal currents and waves, reducing water circulation;
accidental oil spills from ships.
ii Eutrophication [1]
iii 1) Any two (for [1] mark each) from: small and
shrivelled seeds and fruits; poor development of root
systems; weak stalks; crops show less resistance to
diseases and moisture stress.
2) Any one (for [1] mark) from: stunted plant; distorted
shapes in leaf; development of dead areas on leaves,
fruit and stems.
3) Any two (for [1] mark each) from: yellowing
(chlorosis) of leaves; stunted, spindly plants; less
14
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section B
tillering in small grains; low protein content in
seed and vegetative parts; fewer leaves; higher
susceptibility to weather stress, pests and diseases.
6a i
Two or more different crystalline or molecular forms of
the same element. [1]
ii In diamond each carbon atom is strongly covalently
bonded to four other carbon atoms. [1]
In graphite each carbon atom is weakly covalently
bonded to three other carbon atoms, forming layers.
The bonding between layers is weaker than the strong
covalent bonds within layers. [1]
iii 1) Tetrahedral lattice [1]
2) Hexagonal lattice [1]
iv
[1]
Diamond
Diamond
[1]
Graphite
Graphite
v He should use graphite. [1] Because it has free mobile
electrons, graphite is able to conduct electricity. [1]
Diamond does not conduct electricity as there are no
free mobile electrons. [1]
b i
AI3+ AI3+
AI
AI
O
O
O
O
2-
O
2-
O
2-
[2]
ii Molar mass of Al2O3 = (2 × 27) + (3 × 16) = 102 g mol−1 [1]
iii Al2O3(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2O(l)
[1] balancing; [1] correct formula
7 a I: ethyl ethanoate [1]
II: 3-ethyl-2-methylpentane [1]
III: 2-chloro-2-methylbutane [1]
IV: 2-chloro-3-methylbutane [1]
V: but-2-ene or 2-butene [1]
b An organic compound that contains hydrogen and carbon
atoms only. [1]
c II (or 3-ethyl-2-methylpentane) and V (or but-2-ene) [1]
d I: esters [1]; II: alkanes [1]; V: alkenes [1]
e i
H
H
H
H
C
C
C
H
H
H
O
C
O
H
[1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
15
Answers to CSEC® Chemistry Examination Practice
ii Colour change from orange to green [1]
f i
III: 2-chloro-2-methylbutane [1] and
IV: 2-chloro-3-methylbutane [1]
ii I: ethyl ethanoate [1]
8 a [8]
Compound and conditions
Ions
i Concentrated hydrochloric acid using From H2O:
H+, OH−
inert electrodes
From HCl:
H+, Cl−
ii Aqueous copper sulfate using active
electrode
From H2O:
H+, OH−
From
CuSO4:
Cu2+, SO42−
From H2O:
H+, OH−
From
CuSO4:
Cu2+, SO42−
iii Aqueous copper sulfate using inert
electrodes
b i
Ionic equation for the preferential
reaction at
Cathode
Anode
2H+(aq) + 2e− →
2Cl− (aq) − 2e− →
H2(g)
Cl2(g)
or
2Cl− (aq) →
Cl2(g) + 2e−
Cu2+(aq) + 2e− →
Cu(s) − 2e− →
Cu(s)
Cu2+(aq)
or
Cu(s) →
Cu2+(aq) + 2e−
Cu2+(aq) + 2e− →
4OH− (aq) − 4e− →
Cu(s)
2H2O(l) + O2(g)
or
4OH− (aq) →
2H2O(l) + O2(g) + 4e−
Number of moles of HCl used:
1000 cm3 HCl(aq) contains 0.5 mol HCl
50 cm3 HCl contains
50 cm 3 × 0.5 mol
1000 cm 3
= 0.025 mol HCl [1]
Number of moles of NaOH used:
1000 cm3 NaOH(aq) contains 0.25 mol NaOH
25 cm 3 × 0.25 mol
1000 cm 3
25 cm3 NaOH contains
= 0.00625 mol NaOH [1]
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Since NaOH and HCl react in the ratio 1 : 1, in this
reaction:
0.00625 mol NaOH reacts with 0.00625 mol HCl to form
0.00625 mol H2O
Final volume = 75 cm3, so final mass = 75 g
(assuming the density of water = 1 g cm−3)
Temperature change = initial temperature − final
temperature
Temperature change = (19 − 24.5) °C = −5.5 °C [1]
Heat energy evolved in neutralisation
of 0.0625 mol = 75 g × −5.5 °C × 4.18 J g−1 °C−1
= −1724.25 J [1]
16
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Changes in electrolyte
Electrolyte becomes
more dilute.
No change in the
electrolyte.
Electrolyte becomes
acidic.
Section B
−1724.25 J
So heat of neutralisation = 0.00625 mol
=−
275 880 J mol−1 or
−275.88 kJ mol−1 [1]
Potential energy
ii
Activation
energy
NaOH(aq) + HCl(aq)
Energy
released
NaCl(aq) + H2O(l)
[2]
Reaction progress
9 ai
ii
Electrolysis [1]
Lowers the melting temperature of pure alumina [1]
iii Molecular formula: Na3AlF6 [1]; role: catalyst [1]
iv 1) the cathode: Al3+(l) + 3e− → Al(l) [2]
2) the anode: 2O2−(l) − 4e− → O2(g) [2]
v
Q = It, Q = 100 000 A × (12 × 3600) s [1]
Q = 4.32 × 109 C
3 faradays will produce 1 mol (27 g) of Al [1]
3 × 96 500 C will produce 27 g of Al
Hence 4.32 × 109 C should produce
4.32 × 109 C × 27 g
3 × 96 500 C
= 402 901 g of Al [1]
b Any two (for [1] mark each) from:
•
low density – as light as aluminium
•
much stronger than aluminium
•
more resistant to corrosion than aluminium.
c Aluminium is more reactive than Cu [1] so it displaces it
from its salt. [1]
10 a i
Any three (for [1] mark each) from:
• emissions from volcanoes
• combustion of fossil fuel
• broken thermometers releasing mercury liquid
• broken fluorescent bulbs releasing the mercury
vapour they contain
CSEC® Chemistry Examination Practice © Keane Campbell 2016
17
Answers to CSEC® Chemistry Examination Practice
• waste water containing mercury released from the
manufacture of caustic soda and chlorine by the
flowing mercury cathode cell.
ii
Any three (for [1] mark each) from:
• loss of coordination of movement
• muscle weakness
• memory loss, reduced mental function
• kidney failure, respiratory failure and death
• irritability, mood swings
• impaired speech, hearing and vision
• feeling ‘pins and needles’ in hands and feet.
b iInhalation (lungs) [1]; dermal (skin) [1]; ingestion (gut) [1]
ii
Any three (for [1] mark each) from:
• dysfunction of muscles and kidney
• brittle bones and poor posture
• reduced IQ levels in children
• weakness in fingers, wrists or ankles
• anaemia
• a low number of blood cells
• speech and hearing impairment.
c i
ii
Global warming [1]
Any two (for [1] mark each) from: planting of trees;
using renewable energy; using hydrogen-powered
vehicles
11 a i
13 p
13 n
[2]
ii
Electron configuration of W is 2, 8, 3 [1]
iii Group: 3 [1]; Period: 3 [1]
iv Neon [1]
v
bi
Yes [1]; isotopes [1]
2W(s) + 6HCl(aq) → 2WCl3(aq) + 3H2(g)
Balancing [1]; correct formula [1]
18
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section B
ii
2 × 26 g of W will produce 3 × 22.4 dm3 of H2 at stp. [2]
5 g of W will produce
5 g × 3 × 22.4 dm 3
2 × 26 g
= 6.46 dm3 of H2(g) at stp. [2]
iii Any one (for [1] mark) from:
• manufacture of ammonia
• manufacture of methanol
• extraction of metals
• hardening of oils
• hydrogen fuel cells
• rocket fuel.
12 a i
H
H
H
H
C
C
C
H
H
H
H
H
[1]
H
H
H
C
C
C
H
H
H
B
H
H
H
C
C
H
O
H
C
O
H
H
[1]
OH
[1]
C
H
H
O
C
C
C
H
H
E
ii
H
H
H
H
C
C
C
H
OH H
O
H
H
H
C
C
C
H
H
H
H
[1]
F
H
[1]
iii Oxidation [1]
iv Propanol [1]
v
React both propene and B with H+/KMnO4 at rtp or Br2
in CCl4 [1] in the dark. [1]
Propene will decolourise the Br2(l) from a red-brown
liquid to colourless [1] or change H+/KMnO4 from
purple to colourless.
B will give no reaction with either of the two reagents [1]
vi
bi
ii
CH3 H
C
C
H
H
n
[1]
Moles of C in 1 mol of G =
24 g
12 g mol −1
= 2 mol [1]
Mass of H atoms
= mass of G − (mass of C atoms + mass of O atom)
mass of H atoms = 46 g − (24 g + 16 g) = 6 g
Moles of H in 1 mol of G =
iii
H
H
H
C
C
H
H
O
6g
1 g mol −1
= 6 mol [1]
H
[1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
19
Answers to CSEC® Chemistry Examination Practice
13 a To determine the effect of varying concentrations on the
rate of reaction between Ca and HCl [1]
b Concentration of HCl [1]
c Any two (for [1] mark each) from: mass of Ca turnings;
surface area of Ca turnings; volume of HCl
d Time [1]
e Experiment II [1]
Greater concentration [1]
Increased collisions due to more molecules in a given
volume [1]
f [2]
Total volume of product
1.5 mole dm−3
0
0.5 mole dm−3
Time from start of reaction
g Ca(s) + 2HCl(aq) → CaCl2 + H2(g) [1]
hi
1) Calcium turnings: amount in mol =
5.5 g
40 g mol −1
= 1.375 × 10−1 mol [1]
2) Hydrochloric acid:
amount in mol = conc. × volume
= 0.5 mol dm−3 × 0.01 dm3
= 5 × 10−3 mol [1]
ii
i
Hydrochloric acid [1]
Temperature or surface area [1]
14 a i
ii
Neutralise the acidity [1]; precipitate the impurities [1]
The mud is returned to the fields and used as fertilisers. [1]
The bagasse is burnt to heat water to produce steam,
which is used to generate electricity. [1]
iii 1) Any three (for [1] mark each) from: filtration;
vacuum distillation; crystallisation; centrifugation
2) Centrifugation [1]
bi
ii
20
Brine [1]
The OH− ion will be preferentially discharged [1] because
it is lower in the electrochemical series than Cl−. [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section B
iii Anode: 2Cl−(aq) − 2e− → Cl2(g) [1]
Cathode: 2H+(aq) + 2e− → H2(g) [1]
iv Solution became alkaline [1]
H+(aq) and Cl−(aq) are removed, leaving Na+(aq) and
OH−(aq). [1]
15 a i
ii
Acidic oxide [1]; reducing agent [1]
Bubble the gas through either:
H+/KMnO4 – colour change from purple to colourless
or
H+/K2Cr2O7 – colour change from orange to green. [1]
Choking odour [1]
iii SO2 + H2O → H2SO3 [1]
iv Any two (for [1] mark each) from:
• Decreases the quality of agricultural products – less
nutritional value.
• Decreases the yield of agricultural products –
reduction in the proportions and quantity.
• Decreases soil quality – increased hydrogen prevents
plants from growing quickly or large.
v
Lime – Ca(OH)2 [1]
Ca(OH)2(s) + 2H+(aq) → Ca2+(aq) + 2H2O(l) [2]
b Any of the following methods:
Boiling – this converts the soluble Ca(HCO3)2 to insoluble
CaCO3. [1]
Ca(HCO3)2(aq) →
→ CaCO3(s) + H2O(l)+ CO2(g) [1]
heat
OR
Use of sodium carbonate – this removes both temporary
and permanent hardness. Dissolved carbonate magnesium
or calcium ions are precipitated as insoluble carbonate. [1]
Na2CO3(aq) + Ca2+(aq) → CaCO3(s) + 2Na+(aq) [1]
OR
Use of ion exchange resins (zeolite) – the magnesium or
calcium ions in the water displace sodium ions as they pass
through the column. Calcium and magnesium ions are now
absorbed onto the resin. [1]
Ca2+(aq) + 2Na–zeolite → Ca–zeolite + 2Na+(aq) [1]
c Phosphate (PO43−) ions [1] OR nitrate (NO3−) ions [1]
Any two (for [1] mark each) from:
•
low oxygen levels for aquatic life
•
low species diversity
CSEC® Chemistry Examination Practice © Keane Campbell 2016
21
Answers to CSEC® Chemistry Examination Practice
•
increase in plant and animal biomass
•
turbidity increase of water
•
rate of sedimentation increases.
16 a F [1]
It has a smaller atomic radius than Z [1] and therefore there
is less shielding effect. [1]
b X is potassium. [1]
c Y has electronic configuration 2, 8, 8, 2. [1]
d Y has greater reducing power. [1]
Y ionises more readily than Mg [1]
…because it loses its valence electrons more readily. [1]
e iThe product from X reacting with S will be stronger
than the product of Z and C. [1]
The product from X and S has an ionic bond [1], which
is stronger than the covalent bond in the product of Z
and C. [1]
ii
[1] for correct number of electrons in valence shells of
both Z and C; [1] for four pairs of electrons overlapping
between C and Z.
Z
Z
C
Z
Z
iii Chemical formula for compound of C and Z: CZ4 [1]
Chemical formula for compound of X and S: X2S [1]
17 a i
ii
Na2CO3(aq) + 2CH3COOH(aq) → 2CH3COONa(aq) +
CO2(g) + H2O(l) [2]
Calcium carbonate [1]
iii Molar mass of Na2CO3 = 106 g; mass used = 5 g
Amount of Na2CO3 used =
5g
106 g mol −1
= 4.72 × 10−2 mol [1]
iv Molar mass of CH3COOH = 60 g; mass used = 9.2 g
Amount of CH3COOH used =
v
9.2 g
60 g mol −1
= 1.53 × 10−1 mol [1]
Na2CO3 is the limiting reagent [1]
It has the smaller amount of moles of 4.72 × 10−2 mol [1]
22
CSEC® Chemistry Examination Practice © Keane Campbell 2016
section B
vi Mole ratio of Na2CO3 to CO2 is 1 : 1 [1]
4.72 × 10−2 mol of Na2CO3 will produce 4.72 × 10−2 mol
of CO2 [1]
1 mol of CO2 occupies 24 000 cm3 at rtp [1]
4.72 × 10−2 mol of CO2 will occupy
4.72 × 10 −2 mol × 24 000 cm 3
1 mol
bi
ii
= 1132.8 cm3 [1]
Alkanoic acids [1]
Cn H2n+1COOH [1]
iii Place blue litmus paper [1] in the milk, which should
turn red. [1]
18 a Presence of Ca(HCO3)2 [1] which is converted into CaCO3
(white deposits) when the water boils. [1]
b Water has a high specific heat capacity [1], therefore
the ocean will absorb a large amount of heat and yet its
temperature will rise only slowly, thereby allowing the
temperature of the ocean to remain relatively constant. [1]
c i
2NH4Cl(s) + Ca(OH)2(s) → CaCl2(s) + 2NH3(g) +
[1]
2H2O(l)
ii
[3]
ammonia
card cover
calcium hydroxide
and ammonium
chloride
calcium oxide to
dry the ammonia
heat
iii From the equation, 1 mol (53.5 g) of NH4Cl produces
1 mol (17 g) of NH3 [1]
7 g of NH4Cl will produce
7 g × 17 g
53.5 g
= 2.22 g of NH3 [1]
1 mol (53.5 g) of NH4Cl produces 1 mol (24 dm3) of NH3
gas at rtp [2]
7 g of NH4Cl will produce
7 g × 24 dm 3
53.5 g
= 3.14 dm3 of NH3 [1]
iv Place moist red litmus paper in contact with the gas, it
will change to blue. [1]
It will form dense white fumes with hydrogen chloride
gas. [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
23
Answers to CSEC® Chemistry Examination Practice
19 a i
ii
bi
ii
Al forms a protective layer of Al2O3 when exposed to
air [1] which prevents it from corroding. [1]
Acidic food reacts with the aluminium, resulting in
leaching of the metal in the food. [1] Acids also causes
discolouration or pitting. [1]
Haemoglobin [1]
Any two (for [1] mark each) from:
• fatigue
• weakness
• pale skin
• shortness of breath
• dizziness
• tongue swelling.
c i
ii
Acts as an acid [1]
Acts as a base [1]
iii Amphoteric [1]
di
ii
M > L > O [1]
M(s) + L2+(aq) → L(s) + M2+(aq) [2]
L(s) + 2O+(aq) → L2+(aq) + 2O(s) [2]
20 a i
ii
92
× 100% = 51.11% [1] for correct mass
Method A: 180
of desired reactant, which should be doubled, [1] for
correct answer, [1] for correct mass of reactants
× 100% = 100% [1] for correct answer,
Method B: 46
46
[1] for correct mass of reactants
Method A [1]
iii Global warming or thermal pollution [1]
iv Method A – fermentation [1]; method B – hydration [1]
v
bi
ii
Catalyst [1]
Decayed animal and vegetable matter [1]
Prevents soil from becoming hard, compact and cloddy. [1]
Provides nutrients for plants [1]
Improves water-absorbing capacity of soil [1]
Improves soil structure – preventing soil erosion [1]
21 a Heat a sample of one chemical in a test tube [1] connected
to another test tube containing calcium hydroxide via a
delivery tube. [1]
Heat the other sample in a second test tube [1] set up in the
same way.
24
CSEC® Chemistry Examination Practice © Keane Campbell 2016
section B
One sample will release a gas that turns the calcium hydroxide
to a white precipitate [1] and heat will have no effect on the
other. [1] The one that is affected is the PbCO3. [1]
bi
ii
K2CO3 – titration [1]
PbCO3 – ionic precipitation/double decomposition [1]
c Pb(NO3)2(aq) [1] and Na2CO3(aq)/K2CO3(aq) [1]
d Burette [1] and pipette [1]
e Delivery tube facing downward [1]; CO2 present in
collection tube [1]; rubber bung present to minimise the
escape of gas. [1]
delivery tube
rubber
bung
PbCO3 or K2CO3
loose-fitting bung
cooling water bath
CO2
heat
22 a i
Water has high surface tension. [1]
ii
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
O
Lipophilic tail
−
O
+
Na
[1]
iii When clothes are placed into soapy water, the nonpolar (lipophilic) tails of the detergent molecules attach
themselves to the grease [1] on the surface of the
material while the polar (hydrophilic) heads remain
dissolved in the water. [1] With constant agitation, the
grease and the dirt are dislodged from the surface. [1]
grease
fabric surface
bi
ii
[1]
materials [1]; do not cause pollution. [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
25
Answers to CSEC® Chemistry Examination Practice
c iBreaking of larger hydrocarbons into smaller
hydrocarbons in the presence of a catalyst [1]
ii A
H H H H H H H H H H H H
H
B
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
C
H
H
[1]
H
C
C
H
H
[1]
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
[1]
iii Compound B is ethene. React ethene with steam in the
presence of a phosphoric acid catalyst to form ethanol:
CH2=CH2 + H2O
H3PO4
300 C, 65atm
CH3–CH2–OH [2]
Use an oxidising agent H+/KMnO4 to convert the
ethanol to ethanoic acid, CH3COOH:
CH3–CH2–OH + [O] → CH3–COOH + H2O [1]
23 a i
Normal salts are salts which have no H+ ions present in
them. [1]
Acidic salts are salts which have some or all of the H+
ions [1] from the acid from which it was formed. [1]
ii
Normal salt: Na2SO4 [1]
Acidic salt: NaHSO4 [1]
iii Prepare two salt solutions and place blue and red
litmus papers in both of them. [1]
The one that changes blue to red is the acidic salt. [1]
bi
1) H3PO4: 3 [1]
2) H2SO4: 2 [1]
ii
NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l) [2]
c iDilute sodium chloride:
4OH−(aq) − 4e− → 2H2O(l) + O2(g)
or 4OH−(aq) → 2H2O(l) + O2(g) + 4e− [1]
Brine:
2Cl−(aq) − 2e− → Cl2(g)
or 2Cl−(aq) → Cl2(g) + 2e− [1]
ii
Concentration of the ions [1]
The greater the concentration of the anions, the more
likely it is to be discharged. [1]
24 a i
26
Basic oxide – compound that reacts with water to form
a base [1] or one that reacts with an acid to form a salt
and water [1], for example Na2O. [1]
CSEC® Chemistry Examination Practice © Keane Campbell 2016
Section B
ii
Acid anhydride – an acid oxide [1] that dissolves in
water to form an acid [1]; for example CO2(g) [1]
b SO2, CO2 and NO2 are all acid anhydrides which are emitted
from motor vehicle exhausts. [1]
A high concentration of these oxides lead to acid formation
when there is rainfall, which destroys statues and buildings
CO2(g) + H2O(l) → H2CO3(aq) [1] or
SO2(g) + H2O(l) → H2SO3(aq) [1]
High concentration of CO2 results in global warming which
results in rising sea levels due to melting of glaciers and ice
sheets [1].
CaO is used in the blast furnace to remove SiO2 impurities:
CaO + SiO2 → CaSiO3 [1]
c i
ii
Antacids [1]
Neutralisation reaction [1]
Antacids contain sodium hydrogencarbonate, which
neutralises the stomach acid. [1]
iii HCO3−(aq) + H+(aq) → H2O(l) + CO2(g) [1]
25 a i
ii
Atmospheric fixation [1]; industrial fixation [1];
biological fixation [1]
Atmospheric nitrogen is produced by lightning. The
electrical discharge in lightning provides sufficient
energy for nitrogen and oxygen to combine to form
nitrogen monoxide. [1]
N2(g) + O2(g) → 2NO(g) [1]
The nitrogen monoxide is further oxidised by oxygen in
the air to form nitrogen dioxide, which then reacts with
rainwater to produce nitric acid. [1]
2NO(g) + O2(g) → 2NO2(g) [1]
2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq) [1]
The nitric acid forms nitrates on reacting with soil
materials, which are incorporated in plant proteins. [1]
b Any three (for [1] mark each) from:
•
convert waste fats and oils into biodiesel
•
•
recycle aluminium beverage cans, which reduces by
about 5% the energy needed to extract aluminium
•
use light and water sensors for bathroom facilities
•
separate metals from waste and recycle them.
CSEC® Chemistry Examination Practice © Keane Campbell 2016
27
Answers to CSEC® Chemistry Examination Practice
c Water is most dense at about 4 °C and less dense at colder
temperatures – ice floats. So beneath the frozen surface of
water there is warmer water in which aquatic animals can
survive. [1]
Water is a universal solvent which allows for the transport
of nutrients vital to life in animals and plants. [1]
Water has a high specific heat capacity, which helps to
maintain the internal temperature of plants and animals. [1]
Multiple choice items
Item
Item
Item
Item
1
A
16
A
31
C
46
A
2
B
17
B
32
B
47
A
3
D
18
C
33
D
48
C
4
A
19
A
34
C
49
C
5
A
20
D
35
B
50
C
6
B
21
D
36
A
51
C
7
A
22
B
37
A
52
C
8
B
23
B
38
D
53
A
9
A
24
C
39
A
54
A
10
C
25
D
40
D
55
B
11
C
26
D
41
D
56
A
12
B
27
D
42
A
57
C
13
B
28
A
43
B
58
A
14
D
29
A
44
D
59
A
15
C
30
A
45
C
60
A
28
CSEC® Chemistry Examination Practice © Keane Campbell 2016
```
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