QUESTION 1
1.1 D ✔✔ (2)
1.2 D ✔✔ (2)
1.3 A ✔✔ (2)
1.4 B ✔✔ (2)
1.5 D ✔✔ (2)
1.6 D ✔✔ (2)
1.7 C ✔✔ (2)
1.8 B ✔✔ (2)
1.9 A ✔✔ (2)
1.10 B ✔✔ (2)
[20]

QUESTION 2

2.1 A compound that contains a double bond/multiple bond/does NOT contain only single bonds (between C atoms). ✔✔ (2 or 0) (2)
2.2

2.2.1 B / E ✔ (1)
2.2.2 Carbonyl (group bonded to two C atoms) ✔

ACCEPT
Ketone (1)

2.2.3 F ✔✔ (2)
2.2.4 2,5-dichloro-3-methylhexane

(3)

2.2.5 CnH2n ✔ (1)

2.3 Compounds with the same molecular formula, ✔ but different functional groups/homologous series. ✔(2)
2.4

2.4.1 Carboxylic acids✔ (1)
2.4.2

(2)

2.5

2.5.1 Ethanol/✔ (1)
2.5.2 E ✔ ACCEPT : C2H4 (1)
2.5.3 (Concentrated) sulphuric acid/H2SO4/(concentrated) phosphoric acid/H3PO4 ✔ (1)

[18]

QUESTION 3
3.1

The temperature at which solid and liquid phases are in equilibrium. ✔✔(2)

3.2

As the chain length/number of C atoms/molecular mass/surface area/strength of the intermolecular forces ✔ increases, the melting points increase. ✔

OR

• As the chain length/ number of C atoms/molecular mass/surface area/strength of the intermolecular forces ✔ decreases, the melting points decrease. ✔(2)

3.3
London forces ✔
Londonkragte
ACCEPT
Dispersion forces/induced dipole forces  (1)

3.4

3.4.1 Liquid✔ (1)
3.4.2 Solid✔ (1)

3.5

3.5.1 Equal to ✔
Same molecular formula/Isomers/same number and types of atoms/same number of C and H atoms ✔ (2)
3.5.2 Lower than ✔ (1)
3.5.3

2,2-dimethylbutane:

• Structure:
  More branched/more compact/more spherical/smaller surface area (over which intermolecular forces act).✔
•  Intermolecular forces:
  Weaker/less intermolecular forces/Van der Waals forces/London forces/dispersion forces. ✔
•  Energy:
  Lesser energy needed to overcome or break intermolecular forces/Van der Waals forces. ✔
  
  OR
  Hexane
• Structure:
  Longer chain length/unbranched/less compact/less spherical/larger surface area (over which intermolecular forces act). ✔
•  Intermolecular forces:
  Stronger/more intermolecular forces/Van der Waals forces/London forces/dispersion forces. ✔
•  Energy:
  More energy needed to overcome or break intermolecular forces/Van der Waals forces. ✔

QUESTION 4

4.1
4.1.1 Substitution/Hydrolysis ✔

Substitusie/Hidrolise (1)

4.1.2 Primary (alcohol) ✔

ANY ONE:

• The C atom of the functional group is the terminal C atom.
• The C-atom bonded to the hydroxyl/-OH is bonded to (only) one other Catom. ✔
• The hydroxyl/-OH is bonded to a C-atom which is bonded to two hydrogen atoms.
• The hydroxyl/-OH is bonded to a primary C atom/terminal C atom/first C atom. ✔(2)

4.1.3

Marking criteria:

•  Four C atoms in longest chain. ✔
• One methyl substituent on C2. ✔
• Bromo substituent on C1. ✔
  
  IF
• Any error e.g. omission of H atoms, condensed or semi structural formula/Enige fout bv.
   Max:2/3 (3)

4.1.4 Elimination/dehydrohalogenation/dehydrobromination ✔(1)
4.1.5 Alkenes/Alkene ✔ (1)
4.1.6 Addition/Addisie ✔ (1)
4.1.7 2-bromo-2-methyl✔butane ✔

2-bromo-2-metiel✔butaan ✔ (2)

4.2
NOTE

• Penalise only once for the use of structural formulae or molecular formulae.

4.2.1 Marking criteria:

• Correct condensed structure for but-2-ene. ✔
• React but-2-ene with H2/H — H. ✔
•  Indicate the catalyst Pt/Ni/Pd on arrow/at the equation. ✔
• Correct condensed formula for butane as product. ✔
  IF: Any additional products or reactants - minus 1 mark
  
  ACCEPT
  As reactant: CH₃(CH)₂CH₃ / CH₃CH ═ CHCH₃ / CH₃ — CH ═ CH — CH₃
  As product: CH₃(CH₂)₂CH₃ / CH₃ — CH₂ — CH₂ — CH₃ /CH₃ — (CH₂)₂ — CH_{3 (}4)

4.2.2 Elimination/Cracking (1)
4.2.3 Propene/1-propene/prop-1-ene ✔✔ (2)
4.2.4

Marking criteria:

• Correct condensed formula for propene as reactant. ✔
• React (propene) with Br2/Br — Br ✔
• Correct condensed formula for 1,2-dibromopropane as product. ✔
  IF: Any additional products or reactants - minus 1 mark
  CH₃CHCH₂ ✔ + Br₂ ✔  → CH₃CHBrCH2Br ✔
  ACCEPT :
  As reactant CH₃CH ═ CH2 / CH₂ ═ CHCH₃
  As product CH₃CHBrCH₂Br / / BrCH₂CHBrCH₃ (3)
  [21]

QUESTION 5
5.1 NOTE
Give the mark for per unit time only if in context of reaction rate.

ANY ONE

•  Change in concentration ✔ of products/reactants per (unit) time. ✔
• Change in amount/number of moles/volume/mass of products or reactants per (unit) time.
•  Amount/number of moles/volume/mass of products formed/reactants used per (unit) time.
•  Rate of change in concentration/amount of moles/number of moles/volume/ mass. ✔✔ (2 or 0)
  (2)

5.2 Reaction rate decreases./Concentration of HCℓ decreases./Concentration of reactant decreases./Reactants are used up/Mass of CaCO3 decreases or is used up. ✔(1)

5.3
5.3.1 Exothermic/Eksotermies ✔ (1)
5.3.2

• Gradient increases/becomes steeper. / Curve becomes steeper. ✔
• Reaction rate increases/More (or larger volume) of CO2 is produced per unit time. ✔
• Temperature increases./Energy is released/Average kinetic energy of the molecules increases. ✔(3)

5.4 Marking criteria

• m(pure CaCO3) = 82,5/ 100 x 15 ✔ / V(CO2) = 82,5/ 100 x V(CO2) from15 g CaCO₃
• Divide by 100 g∙mol-1. ✔
•  Use mol ratio: n(CO2) = n(CaCO3). ✔
• Multiply n(CO2) by 24 000 cm3/24 dm3. ✔
•  Final answer: 2 976 cm3 ✔
•  Range: 2880 to 2970 cm3 / 2,88 to 2,97 dm3

(5)
5.5 Increases/Toeneem ✔ (1)
5.6 More (CaCO3) particles with correct orientation/exposed./ Greater (exposed) surface area. ✔

More effective collisions per unit time./Higher frequency of effective collisions. ✔

NOTE

•  If explanation in terms of CONCENTRATION: No mark for bullet 1.
•  Bullets are marked independently. (2)
  [15]

QUESTION 6

6.1 (The stage in a chemical reaction when the) rate of forward reaction equals the rate of reverse reaction. ✔✔ (2 or 0)

OR
(The stage in a chemical reaction when the) concentrations of reactants and products remain constant. (2 or 0)
(2)

6.2
6.2.1 Negative/Negatief ✔ (1)

6.2.2

• Increase in temperature favours an endothermic reaction. Accept: Decrease in temperature favours an exothermic. ✔
• Reverse reaction is favoured./Concentration of reactants increases./ Concentration of products decreases. ✔
• (Forward) reaction is exothermic.
  Accept: Reverse reaction is endothermic. ✔
  (3)

6.2.3 CALCULATIONS USING NUMBER OF MOLES

Marking criteria:

1. Initial n(P) and n(Q2) and n(PQ) from table. ✔
2. Change in n(P) = equilibrium n(P) – initial n(P).✔
3. USING ratio: P : Q2 : PQ = 2 : 1 : 2 ✔
4. Equilibrium n(Q2) = initial n(Q2) + change in n(Q2)
   Equilibrium n(PQ) = initial n(PQ) - change in n(PQ)
5. Divide equilibrium amounts of P and Q2 and PQ by 2 dm3. ✔
6. Correct Kc expression (formulae in square brackets). ✔
7. Substitution of equilibrium concentrations into Kc expression. ✔
8. Final answer: 10,889 ✔
   (3)

OPTION 1

OPTION 2

CALCULATIONS USING NUMBER OF MOLES
Marking criteria:

1. Initial n(P) = 4 mol and n(Q2) = 2,4 mol and n(PQ) = 0 ✔
2. Change in n(P) = equilibrium n(P) – initial n(P) = 2,8 mol.✔
3. USING ratio: P : Q2 : PQ = 2 : 1 : 2 ✔
4.  Equilibrium n(Q2) = initial n(Q2) + change in n(Q2)
   Equilibrium n(PQ) = initial n(PQ) - change in n(PQ)
5.  Divide equilibrium amounts of P and Q2 and PQ by 2 dm3. ✔
6. Correct Kc expression (formulae in square brackets). ✔
7. Substitution of equilibrium concentrations into Kc expression. ✔
8. Final answer: 10,89 / 10,889 ✔

OPTION 3

CALCULATIONS USING CONCENTRATION

Marking criteria:

1. Initial c(P) and c(Q2) and c(PQ) from table. ✔
2. Change in c(P) = equilibrium c(P) – initial c(P). ✔
3.  USING ratio: P : Q2 : PQ = 2 : 1 : 2 ✔
4. Equilibrium c(Q2) = initial c(Q2) + change in c(Q2)
   Equilibrium c(PQ) = initial c(PQ) - change in c(PQ)
5.  Divide initial amounts of P and Q2 and PQ by 2 dm3. ✔
6. Correct Kc expression (formulae in square brackets). ✔
7. Substitution of equilibrium concentrations into Kc expression. ✔
8. Final answer: 10,89 / 10,889 ✔
   
   

(8)

6.2.4 Remains the same/Bly dieselfde ✔

Only temperature can change Kc./Temperature remains constant. ✔(2)

6.3
6.3.1 Increases✔ (1)
6.3.2 Decreases✔ (1)
[18]

QUESTION 7

7.1
7.1.1 (It is a) proton/H₃O⁺ (ion)/H⁺ (ion) donor. ✔✔ (2)
7.1.2  HSO^-₄ /hydrogen sulphate ion/waterstofsulfaatioon ✔

ANY ONE:

•  It acts as base in reaction I and as acid in reaction II. ✔
• Acts as acid and base.(2)

7.1.3 HSO₄^- /Reaction (solution) II ✔

Smaller Ka value/weaker acid ✔
Lower ion concentration/Incompletely ionised. ✔
(3)

7.2
7.2.1

(3)

✔Any one

7.2.2 POSITIVE MARKING FROM 7.2.1

Marking citeria:

•  Formula: C = ⁿ/_V /    cₐvₐ    = nₐ/n_b
                                 C_bV_b
• Calculate n(Na₂CO₃): 0,075 x 0,025 ✔
• Calculate n(HCℓ): 0,0955 x 0,05 / 0,096 x 0,05 ✔
• Use ratios: n(HCℓ) = 2n(Na2CO3) ✔
•  n(HCℓ)_excess = n(HCℓ)_initial – n(HCℓ)_used = 0,00475 – 0,0038 ✔✔
• Substitute 0,075 dm³ in c = ⁿ/_v
• Final answer: 0,013 mol∙dm^-3 ✔ (1,3 x 10^-2 mol∙dm^-3)
  Range: 0,01 to 0,02 mol∙dm^-3

(8)
[18]

QUESTION 8
8.1 Chemical (energy) to electrical (energy) ✔ (1)
8.2 Marking criteria:

•  Any formula: c =    m  /c =    n/n = m/M
                                MV         V
•  Substitute 1 mol∙dm^-3.✔
• Substitute 170 g∙mol^-1 [or 108 + 14 + 3(16)] and 0,15 dm³ in correct formulae. ✔
• Final answer: 25,50 g ✔

8.3 ANY ONE:

• A substance that loses/donates electrons. ✔✔
• A substance that is oxidised.
• A substance whose oxidation number increases. (2)

8.4
8.4.1 Copper✔ (1)
8.4.2 Marking criteria/

• Reactants ✔ Products ✔ Balancing ✔
• Ignore double arrows.
• Ignore phase
• Marking rule 6.3.10.
  Cu_(s) + 2Ag⁺_(aq) ✔ → Cu²⁺_(aq) + 2Ag_(s) ✔ Bal ✔
  
  ACCEPT :
  Cu_(s) + 2AgNO_3(aq) ✔ → Cu(NO₃)_2(aq) + 2Ag_(s) ✔ Bal ✔
  NOTE/LET WEL
•  IF electrons are not cancelled – minus 1 mark
  (3)

8.5 OPTION 1
E^θ_cell =  E^θ_reduction - E^θ_oxidation ✔
= 0,80 ✔ – (0,34) ✔
= 0,46 V ✔
Notes

•  Accept any other correct formula from the data sheet. 
• Any other formula using unconventional
  abbreviations, e.g. E°cell = E°OA - E°RA followed by correct substitutions:

OPTION 2
2Ag⁺ + 2e^- → 2Ag            E^θ = 0,80 V ✔
Cu → Cu²⁺ + 2e^-               E^θ = - 0,34 V ✔
2Ag⁺ + Cu → 2Ag + Cu²⁺  E^θ= +0,46 V ✔
(4)

8.6 Decreases (1)
[16]

QUESTION 9
9.1 ANY ONE: (2 or 0)

• A substance whose (aqueous) solution contains ions. ✔✔
• Substance that dissolves in water to give a solution that conducts electricity.
• A substance that forms ions in water / when melted.
• A solution that conducts electricity through the movement of ions.
  (2)

9.2 Anode ✔

• Chromium is oxidised./Oxidation takes place (at the anode)./Chromium (it) loses electrons./Mass decreases./Cr →  Cr³⁺ + 3e^- ✔
  
  NOTE/LET WEL:
  If half-reaction is used, it must be correct/Indien halfreaksie gebruik word: Cr → Cr³⁺ + 3e^- (2)

9.3 Cr³⁺_(aq) + 3e^- → Cr_(s) ✔✔
Ignore phases.

Marking guidelines

•  Cr³⁺ + 3e^- ⇌ Cr  1/2
•  Cr ⇌ Cr³⁺ + 3e^-  0/2
• Cr ← Cr3⁺ + 3e^- 2/2
• Cr → Cr3⁺ + 3e^- 0/2
•  Ignore if charge omitted on electron.
• If charge (+) omitted on Cr³⁺Max: 1/2
  Example/Voorbeeld: Cr³ + 3e^- → Cr ✔
  (2)

9.4 Marking criteria:

• Substitute 52 g∙mol^-1 in ^m/_M/ratio ✔
• Use mol ratio: n(electrons): n(Cr) = 3 : 1. ✔
• Number of electrons = n x 6,02 x 10²³/No of Cr atoms = n x 6,02 x 10²³/ratio. ✔
• Total charge = number of electrons x 1,6 x 10-19/ratio. ✔
• Final answer: 11 113,85 C ✔
  Range: 11 076,8 to 11 580 C

[11]
TOTAL: 150