PHYSICAL SCIENCES
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS JUNE 2019

MEMORANDUM 

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

QUESTION 2
2.1 CONCENTRATED √ (1)
2.2 To prevent reagents escaping √ /To smell the ester/Acts as a condenser (1)
2.3 H₂O √ (1)
2.4 propan-1-ol √√ Accept 1-propanol propanol (1/2) (2)
2.5

• n= 54,55/12 √ = 4,55 n = 9,1/1 √= 9,1 n =(100-54.55-9.1) √ /16 √ = 2.27
                         = 2         = 4
  • Empirical formula C₂H₄O √
  • Molar mass(R) = 130 + 18 - 60= 88 √
  • MMolar Mass (Empirical formula) = 2 x 12 + 4 x 1 + 1 x 16 = 44 √
  • Molecular formula  = C₄H₈O₂ √ (8)  [13]

QUESTION 3
3.1
3.1.1 

1. Esters √  (1)
2.   (2)

Marking Criteria

• Whole structure correct2/2
• Only functional group correct ½   

3.1.2 Same molecular mass √√ /Same molar mass (2)
3.1.3 B has two sites for hydrogen bonding √. A has one site for hydrogen bonding √ (2)
3.2
3.2.1 Yes √

• Same molecular formula √ but different structural formulae √ (3)

3.2.2 Y √

• D has a larger surface area/chain length than E √
• London forces/Induced-dipole forces √/Dispersion forces stronger√ in D than in E
• More energy needed to break/overcome forces in √
  OR

Y √

• E has a smaller surface area/chainlength than D √
• London forces/Induced-dipole forces √/Dispersion forces weaker in E than in D
• Less energy needed to break forces in E √(4)[14]

QUESTION 4
4.1
4.1.1

1. Addition √ /Halogenation/Bromination √ (1)
2. Elimination √/Dehydrohalogenation √ (1)

4.1.2 

1. (3)
   • Marking criteria
     • Whole structure correct 3/3
     • Two Br atoms in structure 1/3
2. Chlorine √ (1)

4.1.3 

1. (4)
   • Marking criteria
     • Whole structure correct 2/2
     • Only functional group correct ½
2. Strong heat √ OR Concentrated strong base (1)

4.2
4.2.1 ADDITION √ (1)
4.2.2 n = 1000 √ (1)
4.2.3 Monomer √ (1)
4.2.4 Make plastics√/ (Any other correct answer) (1)
4.3
4.3.1 Breaking down of a long chain √ /hydrocarbon/alkane into more useful shorter chains √ (2)
4.3.2 THERMAL CRACKING √
TERMIESE KRAKING (1)
4.3.3
 Hexane √√ (4)
Marking criteria

• Whole structure correct 2/2
•  If one or more hydrogens are omitted ½[22]

QUESTION 5
5.1 Temperature√/Concentration √ (of H2O2)/Add a catalyst √(3)
5.2 Change in concentration per unit time/Rate of change of concentration √√ OR change in amount/volume/mass of reactant/product per unit time. (2)
5.3
5.3.1 Average rate

• = - ∆c/∆t= - (1,45-1,9) √ /(15-0) √
  = 0,03√ (mol·dm-3·min^-1) (3)

5.3.2 High concentration √√ ( of H₂O₂ initially)(2)
5.4
5.4.1 ENDOTHERMIC √ (1)
5.4.2 Catalyst increases rate of reaction√

• By lowering activation energy /Deur aktiveringsenergie te verlaag √
• More particles have sufficient Ek to react √/More particles have Ek greater or equal to Ea (4)

5.5
5.5.1 Experiment 1 √ : More particles have higher Ek √(2)
5.5.2 EQUAL TO √Same amount of H₂O₂ used in both experiments √ (2)
5.6

• nO₂ = V/V_m = 0,2/24,8 √ = 8,065 x 10^-3 mol
  nH₂O₂ = 2 √ x 8,065 x 10^-3
  = 0,061 mol
  m H₂O₂ = nM √= 0,0161 x (2+32) √
  = 0,547g√ (0,55 g ) (5)

5.7
5.7.1 Q √ (1)
5.7.2 R √ (1)
5.7.3 P √ (1)  [27]

QUESTION 6
6.1 Stage reached by a chemical reaction where the rate of forward reaction equals the rate of reverse reaction √√ (2)
6.2 6.2.1 REMAINS THE SAME √ (1)
6.2.2 INCREASES √ (1)
6.2.3 INCREASES √ (1)
6.3 Decreasing pressure is opposed √Reaction which produces more gas moles is favoured √ Forward reaction is favoured √ (3)
6.4 6.4.1 REVERSE √(1)
6.4.2 Catalyst √√ (added)(2)

OPTION 1
6.5 CALCULATIONS USING NUMBER OF MOLES

• Divide by  44 in n = m/M √
•  Divide nCO_{2 equilm} & nCO _equilm by 2 √
•  ∆n(CO₂) = n_initial -n_eq √
•  Ratio  CO₂ : CO 1 : 2 √
•  n_equil CO = nCO_initial+ ∆n(CO) √
• Kc expression √
• Substituting c_equilmCO en c_equilimCO₂ √
• Final answer √

OPTION 1
n_initial CO₂= m/M = 104,72/44 √ = 2,38 mol

• Kc = [CO]²/[CO₂] √
  = 3,842/7,6 √
  = 1,94 √

OPTION 2
CALCULATIONS USING CONCENTRATION

• Substitute into  C=m/MV
• Substitute into c=n/V√
• ∆c(CO₂) = c_initial -c_eq √
• Ratio CO₂ : CO 1:2 √
• cequil CO = cCO_initial+ ∆c(CO) √
• Kc expression√
• Substituting  c_equilmCO and c_equilimCO₂ √
• Final answer  √

• Kc = [CO]²/[CO₂] √
  = 3,842/7,6 √
  = 1,94 √ (8)

6.6
6.6.1 Low √ Yield/
Kc is low √/ Kc is less than 1/ [REACTANTS] > [PRODUCTS]2)
6.6.2 Exothermic √ 

• As temperature decreases, Kc decreases, [Products]decreases √OR [Reactants increases]
• Reverse reaction is favoured √
• Decrease in temperature favours exothermic reaction √ (4) [25]

QUESTION 7
7.1
7.1.1 Reaction of a salt with water √√ (2)
7.2
7.1.2 Can act as acid and as a base √√/Can donate H⁺ and accept H⁺(2)
7.1.3 HCO₃^- √ (1)
7.1.4 H₂CO₃ √ (1)
7.2 7.2.1 Standard √ (solution) (1)
7.2.2 CH₃COOH √√ (2)
7.2.3

• c₁V₁=c₂V₂                                                   n=cV=(0,2)(0.02) √ =0,004 mol
  0,2 x 20 √ = 0,16 x V₂ √                        V=n/c= (0,004)/(0,16) √
  V₂ = 25 cm³√ OR 0,025 dm³                V=25 cm³ √ OR 0,025 dm³ (3)

7.3 7.3.1 Neutralisation √ (1)
7.3.2 6,8 to 7,2 √ Titration between strong base and a strong acid √ Solution at endpoint is neutral √ (3)

TOTAL: 150