PHYSICAL SCIENCES: CHEMISTRY PAPER 2
GRADE 12
MEMORANDUM
SENIOR CERTIFICATE EXAMINATIONS
MAY/JUNE 2017

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

QUESTION 2
2.1 A bond / an atom / a group of atoms ✓ that determine(s) the (physical and chemical) properties of a group of organic compounds. ✓ (2)
2.2
2.2.1 2.21✓ (1)
2.2.2 Carboxyl (group)✓ (1)
2.3
2.3.1 Ketones ✓ (1)
2.3.22.32
Marking criteria:

  • Functional group ✓
  • Two methyl substituents ✓
  • Whole structure correct: 3 
                                           3    (3)

2.4
2.4.1 5-bromo-4-ethyl-2,2-dimethylhexane
Marking criteria:

  • Correct stem i.e. hexane.✓
  • All substituents (bromo, ethyl and dimethyl) correctly identified.✓
  • IUPAC name completely correct including numbering, sequence, hyphens and commas. ✓(3)

2.4.2 4-methylpent-2-yne ✓✓
OR
4-methyl-2-pentyne (2)

NOTE
4-methyl  ✓
pent-2-yne  ✓
IUPAC name correct but hyphens omitted: ½


[13]

QUESTION 3
3.1
3.1.13.11

Marking criteria:

  • Functional group ✓
  • Methyl substituent on C-2✓
  • Whole structure correct:
                                           3

Accept:

  • OH as condensed

(3)

3.1.2 D ✓
Accept:
butan-1-ol (1)
3.1.3 G ✓
Accept
2-methylpropan-2-ol(1)
3.2
3.2.1 (Increase in) chain length / molecular size / molecular mass/ number of C-atoms/ surface area / contact area / number of electrons✓(1)
3.2.2 London forces / dispersion forces / induced dipole forces ✓(1)
3.3
3.3.1 108 (°C) ✓ (1)
3.3.2 Compare compound F with compounds C and D:

  • Compound F has a larger molecular mass / molecular size / surface area/contact area / number of C-atoms / number of electrons / than compound C. ✓
  • Compound F is more branched than compound D. ✓
  • Intermolecular forces in compound F are stronger than in compound C and weaker than in compound D. ✓
  • More energy needed to overcome intermolecular forces in compound F than in compound C and less energy needed to overcome (break) intermolecular forces in compound F than in compound D. ✓(4)

3.43.4

Marking criteria:
At least one structural formula of methanol as shown.✓
Dotted line drawn from O atom on one molecule to H atom bonded to an O atom in the second molecule. (H atom should be between two O atoms.)
Accept:

  • OH as condensed✓

(2)

3.5
3.5.1 Esterification / Condensation ✓(1)
3.5.23.52

Marking criteria:

  • Functional group✓
  • Whole structure correct 2/2(2)

[17]

QUESTION 4
4.1
4.1.1 Addition / hydrogenation ✓(1)
4.1.2 Substitution / halogenation / chlorination ✓(1)
4.1.3 Elimination / dehydration ✓(1)
4.2 2-bromopropane ✓
Note:
IF:
Bromopropane ½
2-bromo✓
propane✓ (2)
4.3
4.3.1 Dehydrohalogenation / Dehydrobromination ✓(1)
4.3.2 Hot✓ ethanolic strong base✓

  • Concentrated strong base / NaOH / KOH✓
    OR
    Strong base with no water.
    OR
    Strong base in (pure) ethanol as solvent.
  • Strongly heated or hot base
    OR
    High temperature/heat strongly(2)

4.4
4.4.1 H2O / NaOH / KOH ✓ (1)
4.4.24.42

Notes

  • Ignore ⇌
  • Accept HCℓ and H2O as condensed.
  • Any additional reactants and/or products: Max.¾
  • Accept coefficients that are multiples.
  • Incorrect balancing: Max.¾
  • Molecular/condensed formulae: Max.2/4
    Accept: -OH as condensed

(4)
[13]

QUESTION 5
5.1
5.1.1 To measure volume / amount ✓ of gas/oxygen produced.(1)
5.1.2 Catalyst / Speeds up the reaction. / Increases reaction rate. ✓ (1)
5.2 No gas / bubbles produced.✓
OR
Volume of gas in syringe remains constant. / The plunger stops moving.(1)
5.3 CuO / Copper(II) oxide / catalyst ✓ (1)
5.4

  • A catalyst provides an alternative pathway of lower activation energy. ✓
  • More molecules have sufficient / enough kinetic energy. ✓
    OR
  • More molecules have kinetic energy equal to or greater than the activation energy.
    More effective collisions per unit time. / Frequency of effective collisions increases. ✓(3)

5.5
5.5.1 Released ✓
Products at lower (potential) energy than reactant. / Reaction is exothermic / ΔH < 0 ✓ (2)
5.5.2 B ✓ (1)
5.6
5.6
(6)
[16]

QUESTION 6
6.1
6.1.1 Products can be converted back to reactants. ✓
OR
Both forward and reverse reactions can take place.(1)
6.1.2 Endothermic ✓ (1)
6.1.3

  • Kc increases with increase in temperature. ✓
  • Forward reaction is favoured. / Concentration of products increases. / Concentration of reactants decreases. ✓
  • Increase in temperature favours an endothermic reaction. ✓(3)

6.1.4 Increases  ✓ (1)
6.1.5 Remains the same  ✓ (1)
6.2
6.2.1

Marking criteria:

  1. Use 48 g·mol-1 to calculate n(Ti). ✓
  2. Use 71 g·mol-1 to calculate n(Cℓ2). ✓
  3. Use mole ratio/Gebruik molverhouding: n(Cℓ2) = 2n(Ti). ✓
  4. n(Cℓ2)equilibrium = n(Cℓ2)initial – n(Cℓ2)reacted. ✓
  5. Divide by volume✓
  6. Kc expression ✓
  7. Substitution of c(Cℓ2) in Kc. ✓
  8. Final answer: 0,25 ✓

 

OPTION 1
  Cℓ2  Ti 
Initial quantity (mol)  6 ✓(b)  7 ✓(a) 
Change (mol)  2  1
Quantity at equilibrium (mol) 4 ✓(d)  6 (a) 
Equilibrium concentration (mol∙dm-3)✓  2  

Use mole ratio
Divide by 2

No KC expression: Max. 
                                         8
Wrong KC expression: Max.
                                              8

Kc =   1    ✓(f)
      [Cl12]2 
= 1  ✓(g)
  (2)2
= 0,25 ✓(h)

OPTION 2
n(Ti)reacted = m = 336 - 288− = 1 mol
                     M         48
n(Cℓ2)reacted = 2n(Ti) = 2(1) ✓ = 2 mol
n(Cℓ2)initial = = 426 = 6 mol
                     M      71
n(Cℓ2)equilibrium = 6 – 2 ✓ = 4 mol
c = =
      V    2
= 2 mol·dm-3
Kc =  1   
     [Cl12]2
= 1  ✓
  (2)2
= 0,25 ✓
(8)

No KC expression: Max.
                                        8
Wrong KC expression: Max. 5 
                                              8

6.2.2 Remains the same✓ (1)
[16]

QUESTION 7
7.1
7.1.1 Weak (acid)✓ (1)
7.1.2 pH = -log[H3O+] ✓
4 ✓ = -log[H3O+]
[H3O+] = 1 x 10-4 mol∙dm-3 ✓
(3)
7.2
7.2.1 A substance that produces hydroxide ions / OH- in water. ✓✓
NOTE:
If water is omitted: ½ (2)
7.2.2

Marking guidelines:

  • Formula: c =  /n = cV/ ca × Va  = na
                          V               Cb × Vb × nb
  • Use mol ratio: 1:1 ✓
  • Substitution of : 0,16 x 25 /0,16 x 0,025 ✓
  • Use Vb = 12,5 cm3/0,0125 dm3 ✓
  • Use 56 g·mol-1 ✓
  • Substitute V = 0,25 dm3 ✓
  • Final answer: 4,48 g ✓
 7.22a
 7.22b

(7)
7.2.3 Greater than 7 ✓ (1)
7.2.4 7.24
Due to formation of (OH-), the solution is basic / alkaline✓

Notes

  • Reactants✓ Products ✓ Ignore balancing
  • Ignore/Ignoreer → and phases.
  • Marking rule 6.3.10

(3)
[17]

QUESTION 8
8.1
8.1.1 Emf ✓(1)
8.1.2 Voltmeter ✓ (1)
8.1.3 Salt bridge ✓ (1)
8.1.4 Temperature : 25 °C / 298 K ✓
Concentration : 1 mol·dm-3 ✓ (2)
8.2

Marking criteria   
Dependent and independent variables correctly identified.  ✓ 
Relationship between the independent and dependent variables correctly stated

Examples
Emf increases as concentration (of oxidising agent) increases.
NOTE:
IF:Emf is directly proportional to concentration.½ (2)
8.3

OPTION 1
Eθcell = Eθreduction - Eθoxidation
1,11 ✓ = Eθ+x/x2+ - (- 0,76) ✓
Eθ+x/x2+ = 0,35 (V) ✓
X = Copper ✓
Accept:
Cu/Cu2+ half reaction 

Notes

  • Accept any other correct formula from the data sheet./Aanvaar enige ander korrekte formule vanaf gegewensblad.
  • Any other formula using unconventional abbreviations, e.g. Eºcell = EºOA - EºRMfollowed by correct substitutions:Max;¾
 
OPTION 2
X2+(aq) + 2e- → X(s)                          0,35 (V) ✓
Zn(s) → Zn2+(aq) + 2e-                       0,76 (V) ✓
X2+(aq) + Zn(s) ✓ X(s) +Zn2+(aq)       1,11 (V) ✓
X = Copper/Cu ✓
Accept:
Cu/Cu2+ half reaction

(5)

8.4 Cu2+(aq) + Zn(s) ✓ → Zn2+(aq) + Cu(s) ✓ Bal. ✓
Accept:

  • X2+(aq) + Zn(s) → Zn2+(aq) + X(s)
  • Any metal identified in QUESTION 8.3 of which the ion has a +2 charge.

Notes

  • Reactants ✓ Products ✓ Balancing ✓
  • Ignore phases..
  • Ignore double arrows..
  • Marking rule 6.3.10.

(3)
[15]

QUESTION 9
9.1 Electrolytic (cell) ✓ (1)
9.2 P ✓ (1)
9.3
9.3.1 Au(s) → Au3+(aq) + 3e- ✓✓
Ignore phases 

Notes
Au3+ + 3e- ← Au (2) Au3+ + 3e- ⇌ Au (0/2)
Au ⇌ Au3+ + 3e- (21) Au3+ + 3e → Au (0/2)

(2)
9.3.2 (+)3 ✓ (1)
9.3.3 Electrical energy (is converted) to chemical energy. ✓(1)
9.3.4 Becomes smaller / thinner / eroded / decrease in mass. ✓(1)
9.4 ANY ONE

  • Increase in value. ✓
  • Protection against rust. (1)

9.5 ANY ONE

  • Replace Au3+(aq) / electrolyte with Ag+(aq) / silver(I) solution / use a silver solution
  • Replace P / anode / gold with Ag(s) / silver (1)

[9]

QUESTION 10
10.1
10.1.1 B/air ✓ & C/methane ✓ (2)
10.1.2 Nitric acid / HNO3 ✓ (1)
10.1.3 A / Sulphur / S ✓ (1)
10.1.4 2NH3(g) + H2SO4 ✓ → (NH4)2SO4 ✓ Bal. ✓ (3)

Notes

  • Reactants ✓ Products ✓ Balancing ✓
  • Ignore ⇌ and phases
  • Marking rule 6.3.10

10.1.5 D / potassium chloride ✓ (1)
10.2
10.2.1

OPTION 1:
%P = 3 ✓x 22%
         7
= 9,43%
∴m(P) = 9.43 × 2kg✓
              100
= 0,19 kg ✓ 
OPTION 2:
∴m(P)=✓(0,44) ✓( 22 x 2 = 0,44)
          7                    100
= 0,19 kg ✓

(3)
10.2.2

OPTION 1
m(fertiliser) = 22  x 2
                     100
= 0,44 kg
m(filler/bindstof) = 2 – 0,44
= 1,56 kg ✓
OPTION 2
%filler = 100 – 22 ✓
= 78%
m(filler) =   78  x 2
                 100
= 1,56 kg ✓

(3)
[14]

TOTAL: 150

Last modified on Wednesday, 30 June 2021 09:48