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

 Marking criteria:

  • Correct stem i.e. hexane. ✔
  • All substituents (dichloro and methyl) correctly identified. ✔
  • IUPAC name completely correct including numbering, sequence, hyphens and commas. ✔

(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

phy sci p2 mem 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

Marking criteria
If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark. 

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

3.2

 Marking criteria

  • Identification of independent variable. ✔
  • Stating the relationship between dependent and independent variable. ✔

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

Marking criteria:

  • Compare structures. ✔
  • Compare the strength of intermolecular forces.✔
  • Compare the energy required to overcome intermolecular forces. ✔ 

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

phy sci p2 mem 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
    phy sci p2 mem 4.2.1
    ACCEPT
    As reactant: CH3(CH)2CH3 / CH3CH ═ CHCH3 / CH3 — CH ═ CH — CH3
    As product: CH3(CH2)2CH3 / CH3 — CH2 — CH2 — CH3 /CH3 — (CH2)2 — CH3 (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]

phy sci p2 mem 4.2.4
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

phy sci p2 mem 5.4.1

phy sci p2 mem 5.4.2
(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

phy sci p2 mem 6.2.3

OPTION 2

phy sci p2 mem 6.2.3b

 

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

phy sci p2 mem 6.2.3c

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 ✔

phy sci p2 mem 6.2.3d
(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

OPTION 1
pH = -log[H3O+] ✔
1,02 ✔= -log[H3O+]
[H3O+] = 0,0955 mol∙dm-3
Therefore/Dus
[HCℓ = 0,0955 mol∙dm-3
(0 0 6/0 1 mol∙dm-3)
OPTION 2
pH = -log[H3O+]
[H3O+] = 10-pH
= 10-1,02 ✔
= 0,0955 mol∙dm-3
Therefore/Dus
[HCℓ = 0,0955 mol∙dm-3
(0 0 6/0 1 mol∙dm-3)

(3)

✔Any one

7.2.2 POSITIVE MARKING FROM 7.2.1

Marking citeria:

  •  Formula: C = n/V   cₐvₐ    = nₐ/nb
                                   CbVb
  • 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

phy sci p2 mem 7.2.2a

phy sci p2 mem 7.2.2b

(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 ✔

phy sci p2 mem 8.2.0
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) + 2AgNO3(aq) ✔ → Cu(NO3)2(aq) + 2Ag(s) ✔ Bal ✔
    NOTE/LET WEL
  •  IF electrons are not cancelled – minus 1 mark
    (3)

8.5 OPTION 1
Eθcell =  Eθreduction - Eθoxidation
= 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            Eθ = 0,80 V ✔
Cu → Cu2+ + 2e              Eθ = - 0,34 V ✔
2Ag+ + Cu → 2Ag + Cu2+  Eθ= +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

phy sci p2 mem 9.4.0

[11]
TOTAL: 150

Last modified on Wednesday, 14 December 2022 06:42