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1CSEC® Chemistry Examination Practice © Keane Campbell 2016

Answers to CSEC® Chemistry Examination PracticeSection A2 a i Readings 1 2 3

Final volume (cm3) 12.60 23.80 26.90Initial volume (cm3) 0.70 12.20 15.30Volume used (cm3) 11.90 11.60 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

5

1. × − g

= 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].)

20 4 6 81 53 7

Time (min)

Tem

per

atu

re (

°C)

40

60

80

100100

0

20

Cooling curve of naphthalene

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

3 a 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].)

100 20 25 305 15Time (min)

Mass of silver deposited on twowatch casings at different currents

Mas

s o

f si

lver

dep

osi

ted

(g

)

2

3

4

5

6

0

1

Key

1.5 A 3.0 A

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.

Test Observation InferenceA small amount of solid M was heated in a dry test tube.

A brown gas evolves which turns moist blue litmus red.Another gas evolves which rekindles a glowing splint.

NO2(g) [1]O2(g) [1]

To a sample of a solution of M, aqueous sodium hydroxide is added until in excess.

Pale green precipitate [1]Precipitate remains in excess [1]

Fe2+ ion present

To another sample of solution M, a few copper turnings are added, followed by concentrated sulfuric acid.

Brown fumes evolved which turn moist blue litmus red.

NO3− ion present [1]

section A


y = 5400C 107.87 g

96500C×

[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 Ammeter read incorrectly.

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 Observation InferencesTo a sample of solution FA1, aqueous ammonia was added until in excess.

No precipitate. Ca2+ ion present. [1]

Barium chloride followed by dilute nitric acid was added to another sample of solution FA1.

White precipitate. [1]Precipitate dissolves. [1]

SO32− ion present.

Potassium permanganate was added to another sample of solution FA1.

Colour change from purple to colourless. [1]

SO32− ion present.

4 a i 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]



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])

heat

+ –

cathodeanode

molten sodiumchloride

Cl– Na+

c

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 0 51000

3

3

cm molcm

× . = 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

Test Observations InferencesX 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 acidifi ed potassium manganate(VII).

• Choking odour gas evolves which turns moist blue litmus red.

• Colour change from purple to colourless. [1]

SO2 present. [1]Gas is acidic. [1]Reducing agent present. [1]

To 1 ml of solution X is added silver nitrate followed by aqueous ammonia.

No observable change. No Br− , Cl− or I − ions. [3]

Aqueous sodium hydroxide is added to 1 ml of solution X and heated.

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]

section A


1000 cm3 H2SO4 solution contains 0.5 mol of H2SO4 [1]

25 cm3 of H2SO4 solution will contain

25 0 51000

3

3

cm molcm

× . = 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 = −× × −

19532 1 25 10 2

Jmol.

= −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.

0 Time from start of reaction

Tota

l vo

lum

e o

f p

rod

uct

0.25 mol dm−3

0.5 mol dm−3

e i 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]

6 a 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.



• 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]

v 1) Sketch graph for [2] marks.

Mas

s o

f p

reci

pit

ate

Volume of aqueous NH3 added

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]

section A


So 100 g of CaCO3 will produce 24 dm3 of CO2 [1]

22.4 g of CaCO3 will produce 22 4 24100

3. g dmg

× = 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 5 38 44

24

3

3

. dm gdm

×

= 9.86 g [1]

c i 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]

8 a 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.

Volume of acid added (cm3)

Temperature of solution (°C)

0 26.0 5 27.510 28.515 31.520 34.525 36.030 39.035 36.040 35.5

iii A neutralisation reaction is a reaction between a base and an acid to form a salt and water. [1]



iv Graph plotted for [2] marks.

100 20 30 405 15 25 35 45

Volume of acid added (cm3)

10

20

30

40

0

5

15

Tem

per

atu

re o

f so

luti

on

(°C

)Thermometric titration of slaked lime and

an acidic soil solution

25

35

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]

9 a 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.

section A


ii Completed table for [5] marks.

Temperature after reaction (°C) 36.0Temperature before reaction (°C) 26.0Temperature change (°C) 10.0Total volume of solution (cm3) 75.0Total mass of solution (g) 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

5 0 10 2

Jmol.

[1]

Heat of neutralisation = −62 700 J mol−1 [1]

b i The dryness in Robyn’s hair is due to the presence of sulfate compounds in her shampoo. [2]

ii Aqueous Ba(NO3)2 or BaCl2 [1]; aqueous HCl or HNO3 [1]

iii CO32− ions [1]; SO3

2− ions [1]

iv Add a sample of the shampoo to H+/K2Cr2O7; if it turns from orange to green then it contains SO3

2− ions. [1]

Heat a sample of the shampoo and place lime water above it; if it forms a white precipitate, then it contains CO3

2− ions. [1]

10 a i 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]

ii Osmosis [1]

b i Gas syringe [1]; correct set up [1].

gas

gas syringe

reactants

ii In the gas syringe [1]



iii Stop watch [1]

iv Graph for [3] marks.

1000 200 250 30050 350150

Time (s)

Reaction of 4.1976g of sodium carbonate and31.68cm3 of 2.5moldm−3 hyrodochloric acid

Vo

lum

e o

f C

O2

pro

du

ced

(cm

3 )

40

60

80

100

120

0

20

1) The initial stage of the graph [1]

2) Rate = change in volumeof producttime taken for thechange

[1]

This is the gradient of the curve at time = 60 s. Drawing a tangent gives:

rate = 108215

3cms

[1]

rate = 0.50 cm3 s−1 [1]

1000 200 250 30050 350150

Time (s)

Reaction of 4.1976g of sodium carbonate and31.68cm3 of 2.5moldm−3 hyrodochloric acid

Vo

lum

e o

f C

O2

pro

du

ced

(cm

3 )

40

60

80

100

120

0

20

section A


v Curve added to graph as shown [1].

1000 200 250 30050 350150

Time (s)

Reaction of 4.1976g of sodium carbonate and0.5moldm−3 and 2.5moldm−3 hyrodochloric acid

Vo

lum

e o

f C

O2

pro

du

ced

(cm

3 )

40

60

80

100

120

0

20

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 Observation InferenceA small amount of solid T was placed in a test tube and heated over a Bunsen flame. The gas was bubbled through lime water.

White precipitate. [1] CO23− ions were present.

Sodium hydroxide was added to a solution of T until in excess.

A white precipitate was formed that dissolved in excess.

Pb2+, Al3+ or Zn2+ ions are present. [1]

A solution of potassium iodide was added to a separate portion solution of T.

No observable change. Pb2+ ion not present. [1]

To a third portion of solution T, excess aqueous ammonia was added.

Precipitate did not dissolve. [1] Al3+ ions were present.



Section B2 a The structural formula shows the arrangement of atoms in

the molecule of a compound. [1]

b [2]

OH N

H

C

O

CH3

c i Alcohol [1]

ii 1) Upward delivery [1]; because the gas is lighter than air. [1]

2) Covalent [1]

3) H H× [1]

4) Use a lighted splint. [1] A popping sound identifi es 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]

g Tylenol/Baralgin/Panadol [1]

h HO N

H

C

O

CH3

or HO N

H

C

O

CH3

[1]

3 a i Note that a crucible is used and not a beaker because it is molten NaCl. [2]

heat

+ –

cathodeanode

molten sodiumchloride

Cl– Na+

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]

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 = 3000 242 96500

3C dm×× C = 0.373 dm3 [1]

b i 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]

4 a 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 g of oxide are used, which is 1500260 mol

So amount of gas produced = 5 × 1500260 mol [1]

At stp 1 mol of gas occupies 22.4 dm3



So volume of CO2 gas produced = 5 × 1500260 × 22.4 dm3 = 646 dm3 [1]

iii From the equation, each mole of oxide produces 2 mol of R.

So 1500 g of oxide will produce 1500260 × (2 × 90) g of R = 1040 g [1]

5 a i 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

• design for degradation

• 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

section B


tillering in small grains; low protein content in seed and vegetative parts; fewer leaves; higher susceptibility to weather stress, pests and diseases.

6 a 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

Diamond Graphite [1]

Diamond Graphite [1]

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 O

AI

O O

AI AI3+ AI3+

O2-

O2-

O2- [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

C

H

CC

O

O

CH

H

H

H

H H [1]



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 Ionic equation for the preferential reaction at

Changes in electrolyte

Cathode Anodei Concentrated hydrochloric acid using

inert electrodesFrom H2O: H+, OH−

From HCl: H+, Cl−

2H+(aq) + 2e− → H2(g)

2Cl− (aq) − 2e− → Cl2(g) or2Cl− (aq) → Cl2(g) + 2e−

Electrolyte becomes more dilute.

ii Aqueous copper sulfate using active electrode

From H2O: H+, OH−

From CuSO4: Cu2+, SO42−

Cu2+(aq) + 2e− → Cu(s)

Cu(s) − 2e− → Cu2+(aq)orCu(s) → Cu2+(aq) + 2e−

No change in the electrolyte.

iii Aqueous copper sulfate using inert electrodes

From H2O: H+, OH−

From CuSO4: Cu2+, SO42−

Cu2+(aq) + 2e− → Cu(s)

4OH− (aq) − 4e− → 2H2O(l) + O2(g)or4OH− (aq) → 2H2O(l) + O2(g) + 4e−

Electrolyte becomes acidic.

b i Number of moles of HCl used:

1000 cm3 HCl(aq) contains 0.5 mol HCl

50 cm3 HCl contains 50 0 51000

3

3

cm molcm

× . = 0.025 mol HCl [1]

Number of moles of NaOH used:

1000 cm3 NaOH(aq) contains 0.25 mol NaOH

25 cm3 NaOH contains 25 0 25

1000

3

3

cm molcm

× .

= 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]

section B


So heat of neutralisation = −1724 250 00625

..

Jmol

= −275 880 J mol−1 or −275.88 kJ mol−1 [1]

ii

Pote

nti

al e

ner

gy

Reaction progress

Activationenergy

Energyreleased

NaOH(aq) + HCl(aq)

NaCl(aq) + H2O(l)

[2]

9 a i Electrolysis [1]

ii 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 10 273 96500

9. × ××

C gC

= 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



• 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 i Inhalation (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 Global warming [1]

ii Any two (for [1] mark each) from: planting of trees; using renewable energy; using hydrogen-powered vehicles

11 a i

13 p13 n

[2]

ii Electron configuration of W is 2, 8, 3 [1]

iii Group: 3 [1]; Period: 3 [1]

iv Neon [1]

v Yes [1]; isotopes [1]

b i 2W(s) + 6HCl(aq) → 2WCl3(aq) + 3H2(g)

Balancing [1]; correct formula [1]

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 3 22 42 26

3g dmg

× ××

. = 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 C C C H

H

H

H

H

H

H [1]

H C C C OH

H

H

H

H

H

H [1]

B C

H C C C

H

H

H

H

O

O H [1]

H C C C

H

H

H

H

O

O

C C C H

H

H

H

H

H

H [1]

E F

ii

H C C C H

H

H

H

OH H

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

C C

CH3

H

H

H n [1]

b i Moles of C in 1 mol of G = 24

12 1

ggmol− = 2 mol [1]

ii 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 = 6

1 1

ggmol− = 6 mol [1]

iii

H C C O H

H

H

H

H [1]



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]

0 Time from start of reaction

Tota

l vo

lum

e o

f p

rod

uct

0.5 mole dm−3

1.5 mole dm−3

g Ca(s) + 2HCl(aq) → CaCl2 + H2(g) [1]

h i 1) Calcium turnings: amount in mol = 5 5

40 1

. ggmol−

= 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 Hydrochloric acid [1]

i Temperature or surface area [1]

14 a i Neutralise the acidity [1]; precipitate the impurities [1]

ii 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]

b i Brine [1]

ii The OH− ion will be preferentially discharged [1] because it is lower in the electrochemical series than Cl−. [1]

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 Acidic oxide [1]; reducing agent [1]

ii 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]

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

→heat



• 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 i The 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 ZC

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 Na2CO3(aq) + 2CH3COOH(aq) → 2CH3COONa(aq) + CO2(g) + H2O(l) [2]

ii Calcium carbonate [1]

iii Molar mass of Na2CO3 = 106 g; mass used = 5 g

Amount of Na2CO3 used = 5

106 1

ggmol−

= 4.72 × 10−2 mol [1]

iv Molar mass of CH3COOH = 60 g; mass used = 9.2 g

Amount of CH3COOH used = 9 2

60 1

. ggmol−

= 1.53 × 10−1 mol [1]

v Na2CO3 is the limiting reagent [1]

It has the smaller amount of moles of 4.72 × 10−2 mol [1]

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 24000

1

2 3. × ×− mol cmmol

= 1132.8 cm3 [1]

b i Alkanoic acids [1]

ii 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 specifi c 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) + 2H2O(l) [1]

ii

[3]

ammonia

card cover

calcium oxide todry the ammonia

calcium hydroxideand ammoniumchloride

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 1753 5g g

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 2453 5

3g dmg

×.

= 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]



19 a i Al forms a protective layer of Al2O3 when exposed to air [1] which prevents it from corroding. [1]

ii Acidic food reacts with the aluminium, resulting in leaching of the metal in the food. [1] Acids also causes discolouration or pitting. [1]

b i Haemoglobin [1]

ii Any two (for [1] mark each) from:

• fatigue

• weakness

• pale skin

• shortness of breath

• dizziness

• tongue swelling.

c i Acts as an acid [1]

ii Acts as a base [1]

iii Amphoteric [1]

d i M > L > O [1]

ii M(s) + L2+(aq) → L(s) + M2+(aq) [2]

L(s) + 2O+(aq) → L2+(aq) + 2O(s) [2]

20 a i Method A: 92180 × 100% = 51.11% [1] for correct mass

of desired reactant, which should be doubled, [1] for correct answer, [1] for correct mass of reactants

Method B: 4646 × 100% = 100% [1] for correct answer, [1] for correct mass of reactants

ii Method A [1]

iii Global warming or thermal pollution [1]

iv Method A – fermentation [1]; method B – hydration [1]

v Catalyst [1]

b i Decayed animal and vegetable matter [1]

ii 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.

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]

b i K2CO3 – titration [1]

ii 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]

PbCO3 or K2CO3

heat

CO2

delivery tuberubberbung

cooling water bath

loose-fitting bung

22 a i Water has high surface tension. [1]

ii

H C C C C C C C C C C C C C C C C O Na

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Lipophilic tail Polar head

H

H

H

H O

+−

[1]

iii When clothes are placed into soapy water, the non-polar (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]

fabric surface

grease

[1]

b i Any one from: biodegradable [1]; made from renewable materials [1]; do not cause pollution. [1]

ii Any one from: leads to pollution [1]; may be non-biodegradable. [1]



c i Breaking of larger hydrocarbons into smaller hydrocarbons in the presence of a catalyst [1]

ii A

H 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

H

H

H

H

H

H

H

H

H

H [1]

B

C C

H

H

H

H [1]

C

H 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

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 � ��

H PO

300 C, 65atm3 4 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]

b i 1) H3PO4: 3 [1]

2) H2SO4: 2 [1]

ii NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l) [2]

c i Dilute 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 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]

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 made of limestone. [1]

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 Antacids [1]

ii 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 Atmospheric fixation [1]; industrial fixation [1]; biological fixation [1]

ii 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

• use biodegradable garbage disposal bags

• 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.



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 itemsItem Answer Item Answer Item Answer Item Answer

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

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