Physical Sciences Grade 12 P2 KwaZulu Natal Mock 2019 Memo

Downloaded from hlayiso.com Physical Sciences / P2 1 19 NSC-Memorandum PHYSICAL SCIENCES P2 (CHEMISTRY) PREPARATORY EXAMINATION SEPTEMBER 2019 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS : 150 This marking guideline consists of 8 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 2 Preparatory Examination September 2019 NSC-Memorandum QUESTION 1 1.1 D  (2) 1.2 C (2) 1.3 A (2) 1.4 C (2) 1.5 D (2) 1.6 D (2) 1.7 D (2) 1.8 C (2) 1.9 C (2) 1.10 B (2) [20] QUESTION 2 2.1.1 hexan-3-one  (2) 2.1.2 carboxyl (group)  (1) 2.1.3 H H H │ │ │  Whole structure correct: 2/2 H ― C ― C ≡ C ― C ― C ― H  Only functional group correct 1/2 │ │  More than one functional H H group 0/2 H ― C ― H │ H (2) 2.1.4 addition polymerisation (1) 2.1.5 H H │ │  Whole structure correct  ― C ― C ―  Name  │ │ H H n polyethene (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 3 Preparatory Examination September 2019 NSC-Memorandum 2.1.6 G (1) 2.2.1 esters/alkyl alkanoate (1) 2.2.2 ethyl propanoate (2) 2.2.3 H H O │ │ ║ H ― C ― C ― C ― O ― H │ │ H H  Whole structure correct: 2/2  Only functional group correct 1/2  More than one functional (2) group 0/2 2.2.4 acts as a catalyst/speeds up the reaction.  or acts as a dehydrating agent.  (1) 2.3.1 Compounds that have the same molecular formula but different functional groups (2) 2.3.2 pentanoic acid (2) [20] QUESTION 3 3.1 the pressure exerted by a vapour at equilibrium with its liquid in a closed system.  (2 or 0) (2) 3.2.1 length of carbon chain/surface area/branching (1) 3.2.2 number of carbon atoms/molecular mass (1) 3.3.1 GREATER THAN (1) 3.3.2 M has a longer carbon chain/greater surface area than N/ M has more sites for intermolecular forces Intermolecular forces between molecules of M are stronger than between molecules of N More energy is required to overcome the intermolecular forces between molecules of M (3) 3.4 N (1) [9] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 4 Preparatory Examination September 2019 NSC-Memorandum QUESTION 4 4.1.1 warm/mild heat dilute KOH/warm dilute strong base (2) 4.1.2 hot KOH concentrated Base(KOH) (2) 4.2 substitution (1) 4.3 H H H H H H H H │ │ │ │ │ │ │ │ H ─ C ─ C ─ C ─ C ─ H + KOH H ─ C ─ C ═ C ─ C ─ H + KBr + H2O │ │ │ │ │ │ H Br H H H H   left hand side   for organic product   balancing (4) 4.4 unsaturated contains a double bond/multiple bond between atoms of carbon (3) 4.5 hydration (2) [14] QUESTION 5 5.1 calcium carbonate there is some unreacted CaCO3 at the end of the reaction (time 60s)  (2) 5.2.1 ANY ONE  The change in concentration of reactants/products per unit time.   Rate of change in concentration of reactants or products.   Change in amount/number of moles/volume/mass of reactants/products per (unit) time.  Amount/number of moles/volume/mass of products formed OR reactants used per (unit) time. (2) change in mass of CaCO3 Marking criteria 5.2.2 rate     Equation(accept if negative sign t is omitted)  54  X   Substitute 54 – X in equation  1,07 = - 30  0   Substitute 30 – 0 in equation  Substitute 1,07for rate = 86,10 g   Final answer: X = 86,10 g  (if answer is negative minus 1 mark) (5) 5.3 0 (cm3).  (1) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 5 Preparatory Examination September 2019 NSC-Memorandum 5.4 A decrease in concentration of reactants decreases the number of molecules per unit volume.  Fewer number of collisions per unit time A fewer number of effective collisions occur per unit time/lower frequency of effective collisions.  (3) 5.5 REMAINS THE SAME (1) [14] QUESTION 6 6.1 When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance.  (2 or 0) (2) 6.2 the reaction has reached a state of (dynamic) equilibrium/the rate of the forward reaction is equal to the rate of the reverse reaction. (2 or 0) (2) 6.3 Marking criteria:  Indicating that the number of mols of CO equilibrium is 0,6  Correct mol ratio  Calculating the quantity(mol) at equilibrium of all three substances  n  Substitute V = 2 dm3 in c  to determine concentration at equilibrium of all the V substances.  Kc expression  Substitution of concentrations in Kc expression   Final answer: 0,456  No Kc expression, correct substitution: Max. 6 7 Wrong Kc expression : Max. 4 7 CH4 H2O CO H2 Initial quantity(mol) 2,2 1,8 0 0 Change(mol) -0,6 -0,6 +0,6 + 1,8 Ratio  Quantity at  1,6 1,2 0,6 1,8 equilibrium(mol) Equilibrium 0,8 0,6 0,3 0,9 Divide by 2 concentration(mol.dm-3) Kc = [CO] [H2 ]3 = 0,3 0,93  = 0,456  (7) [CH4 ] [H2O] 0,8 0,6 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 6 Preparatory Examination September 2019 NSC-Memorandum 6.4  Graph starts at 2,2 Number of moles CH4 2,2 Decreases to 1,6  Correct shape  curve  1,6 horizontal ignore labels on axes Time in minutes (4) 6.5.1 INCREASES (1) 6.5.2 REMAINS THE SAME (1) 6.6 An increase in the number of moles of CH4 increases the concentration of CH4(reactant). According to Le Chateliers Principle an increase in the concentration of the reactants  favours the reaction that decreases the concentration of the reactants In this case the forward reaction is favoured (3) [20] QUESTION 7 7.1.1 An acid is a substance that produces hydrogen ions(H+)/hydronium ions(H3O+) when it dissolves in water.  (2) 7.1.2 strong it ionises completely in water  (ACCEPT: dissociates) (3) 7.2.1 pH = - log [H3O+]   Formula pH = - log [H3O+]  0,65 = - log [H3O+]  Substitute 0,65 for pH  [H3O+] = 0,224 mol.dm-3  c((H2SO4) = ½c(H3O+)  c((H2SO4) = 0,112 mol.dm-3 c((H2SO4) = ½c(H3O+)  = ½(0,224) = 0,112 mol.dm-3 (4) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 7 Preparatory Examination September 2019 NSC-Memorandum 7.2.2 POSITIVE MARKING FROM QUESTION 7.2.1: concentration of H2SO4 Marking guidelines/Nasienriglyne: n  Formulae: c  /n = cV/  V  Calculate initial number of moles of H2SO4  Calculate number of moles of H2SO4 that reacted  Calculate number of moles of H2SO4 in excess  Calculate number of moles of NaOH that reacted  Ratio of NaOH to H2SO4  Final answer cm3 or dm3 n(H2SO4)initial = cV = (0,25)(0,024)  = 6 x 10-3 mols n(H2SO4)excess = cV = (0,112)(X + 24)  1000 (0,112)(X + 24) n(H2SO4)reacted = 6 x 10-3 -  1000 n(NaOH)reacted = cV 0,15(X)  1000 n(NaOH)reacted = 2 (n(H2SO4)reacted) 0,15(X) (0,112)(X + 24) = 2( (6 x 10-3 - ) 1000 1000 X = 17,71 cm3 0,01771 dm3 (7) [16] QUESTION 8 8.1 a solution/liquid/dissolved substance that conducts electricity through the movement of ions. (2) 8.2 Zn(s) Zn 2+(aq) + 2e- Notes  Zn2+ + 2e- ← Zn (2 ) Zn2+ + 2e- ⇌ Zn (0 ) 2 2 Zn ⇌ Zn2+ + 2e- (1 ) Zn2+ + 2e → Zn ( 0 ) 2 2  Ignore if charge on electron is omitted.  If a charge of an ion is omitted e.g. Zn → Zn2 + 2e- Max.: 1 2 (2) 8.3.1 Temperature of 25 ºC/298K Concentration of the electrolytes equals 1 mol.dm-3.  (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 8 Preparatory Examination September 2019 NSC-Memorandum 8.3.2 EΘcell = EΘcathode - EΘanode  Notes  Accept any other correct formula from the 0,63 = EΘcathode - (-0,76)  data sheet. EΘcathode = - 0,13 V   Any other formula using unconventional X is lead(Pb)  abbreviations, e.g. E°cell = E°OA - E°RA followed by correct substitutions: E°sel = E°OM - E°RM Max/: 4 5 (5) 8.4 Zn(s)/Zn2+(aq) //Pb2+(aq)/Pb(s)  Zn(s)/Zn2+(1 mol.dm-3) //Pb2+(1 mol.dm-3)/Pb(s) Accept Zn/Zn2+//Pb2+/Pb (3) 8.5 0(V)  (1) [15] QUESTION 9 9.1 from electrical energy to chemical energy (1) 9.2 P (1) - 9.3 Cu2+ + 2 e → Cu Notes  Cu2+ + 2e- ← Cu (0 ) Cu2+ + 2e- ⇌ Cu (1 ) 2 2 Cu ⇌ Cu2+ + 2e- (0 ) Cu2+ + 2e → Cu ( 2 ) 2 2  Ignore if charge on electron is omitted.  If a charge of an ion is omitted e.g. Cu → Cu2 + 2e- Max.: 1 2 (2) 9.4.1 Q will break down/become eroded/surface becomes rough and eroded ACCEPT Q will be oxidised.  (2) 9.4.2 Cu/electrode Q is a stronger reducing agent than the Cℓ- ions. Cu/Q will be oxidised/loses electrons resulting in the electrode becoming eroded OR The Cℓ- ion is a weaker reducing agent than Cu(Q) and will therefore not be oxidised.  (3) 9.4.3 P (1) [10] QUESTION 10 10.1.1 Haber process (1) 10.1.2 nitric oxide NO (1) 10.1.3 platinum Notes: (1) 10.1.4 HNO3 + NH3 → (NH4)NO3  Reactants  Products  Balancing   Ignore double arrows. (3)  Marking rule 6.3.10. 10.2 Mass P = 31/149 x11,95 = 2,486 kg 2/8 x Z/100 x 20 = 2,486 Z = 49,72% (6) TOTAL MARKS: 150 Copyright Reserved Please Turn Over