Physical Sciences Grade 12 P2 Western Cape Mock 2025 Memo

Page 1 of 15 Western Cape Government Education METRO CENTRAL EDUCATION DISTRICT GRADE 12 : PHYSICAL SCIENCES: PHYSICS (P2)— MARKING GUIDELINE : SEPTEMBER 2025 MARKS: 150 TIME: 3 hours

Page 2 of 15 QUESTION 1 14 Do vv (2) 12 Bo vv (2) 13 CC vv (2) 14 D vv (2) 15 D vv (2) 16 DO vv (2) 17 A vv (2) 18 DO vv (2) 19 A vv (2) 110 D vv (2) [20]

Page 3 of 15 QUESTION 2 2.1.1 Av (1) 2.1.2 BY (1) 2.2.1 CnH2n2 ¥ (1) 22.2 H i ii | | i H | Cc c=>=c Cc i | H H HH sv 7 Ccchain with all bonds and H H —¥ —H H—c—H v Alkyne functional group on carbon 3 = Y Dimethyl on carbon 2,6 and ethyl on 5 H—G—H i (3) 2.3. Pentanal Pentan 7 al’ (2) 2.4.1 Butan-2-ol Butan VY 2-ol ¥ ACCEPT: 2-butanol (2) 2.4.2 H | fe) H | H H I | 1 = 7 a a : Y 4C chain with all bonds and H H | H H Y¥ OH and methyl group on carbon 2 H-C-H | H (2) 2.5 CsHs + 502 > 3CO2 + 4H20 v Reactant, VY product, ¥ Balanced eq. (3) Note: No balancing mark, if reactants and products are incorrect. [15]

Page 4 of 15 QUESTION 3 34 3.2 3.3 3.4 3.5 The temperature at which the vapour pressure of a substance equals atmospheric pressure. vv (2) 3.2.1. Carbon chain length/Molecular mass/Number of carbon atoms/ number of London forces.(sites) ¥ (1) 3.2.2 Primary Alcohols/ Straight chains alcohol ACCEPT: Functional group/ Homologous series/ type of IMF/ atmospheric pressureY (1) FROM Ato D e Chain length increases / larger surface area / higher molecular mass. W e The number of London forces increases OR Increasing strength of the Intermolecular forces. e More energy needed to overcome the increase in intermolecular forces. ~ (If learners write, “break bonds”, only penalise once.) Boling point increases (3) NO Y Two independent variables ACCEPT: molar mass/type of IMF/ different functional groups). ¥ (2) OPTION 1 e Compound D/ Pentan-1-ol has hydrogen bonding (and London forces) between its molecules, whereas Compound E / Pentane has London forces between its molecules. ~ e Hydrogen bonding forces are stronger than London forces OR Intermolecular forces in Compound D / Pentan-1-ol are stronger than the intermolecular forces in Compound E / Pentane. “ e More energy is needed to overcome the intermolecular forces in Compound D / Pentan-1-ol than in Compound E / Pentane.. -. Boiling point of Compound D / Pentan-1-ol is higher than compound E /Pentane. “ OPTION 2 * Compound E / Pentane has London forces between its molecules whereas Compound D / Pentan-1-ol has hydrogen bonding (and London forces) between its molecules. ¥ e London forces are weaker than hydrogen bonding OR Intermolecular forces in Compound E / Pentane are weaker than the intermolecular forces in Compound D / Pentan-1-ol. v e Less energy is needed to overcome the intermolecular forces in Compound E / Pentane than in Compound D / Pentan-1-ol. Boiling point of Compound D / Pentan-1-ol is higher than Compound E / Pentane v (3)

3.6 Compound Z (Ester) v Compound Z has the weakest intermolecular forces (dipole-dipole) of the two. ¥ (while compound F has stronger hydrogen bonds between the molecule. Or Compound F has the stronger intermolecular forces (Hydrogen bonding) of the two. (while v SHESTION 4 ¥ 4-C chain with all H 44 H H Hy H . Hp CH y v Br on second C H H b, H 4.2. 4.2.1 Elimination / dehydrohalogenation / dehydrobromination ¥ 4.2.2 H oH Hv ¥ 4-C chain with all H _— i Ce GH v ¥ double bond on first C H oH H 4.3. CHsCH2zCHBrCH3 + NaOH ¥ ——» CHsCH2CHOHCH3vY + NaBrv (3) [If molecular formula used > 1 mark for NaBr] 44 e Add HBr or NaBr or KBr ¥ e Absence of water or heat Vv 4.5 Hydration Vv 4.6.1 Heat or Catalyst Ni/Pd/Pt v¥ 4.6.2 CnHoane2 v Page 5 of 15 (2) [14] (2) (2) (1) (1) (1) [13]

Page 6 of 15 QUESTION 5 5.1.1. The unstable transition statev from reactants to products ¥ (2) 51.2) AH = Eabsorbed — Ereteased - 203 = 2 144-Y Y = 2 347 (kJ-mol'!) vv (2) 5.1.3 Chemical potential energy, E,. {kJ-mot't) Course of reaction Y correct energy of reactants and product AND lower activated complex Y label C (2) 5.2.1 The change in concentration of reactants or products per unit time v.[2 or 0] (2) 5.2.2 RV e Ris at a higher temperature and therefore there are more particles with sufficient kinetic energy or Ek > Ea. v More collisions per unit time. More effective collisions per unit time. v e Higher/greater/faster rate of reaction. NB. If P selected and the following explanation given [maz 2 marks]: e Fewer particles with sufficient kinetic energy above Ea ¥ e Fewer effective collisions per unit time “ 1 5.3.1 Rate =- “Sv At 1,73K10-8V = — 6-080 y (200-0) C = 6,54 x 10-3 VY (mol:dm*) > unit not necessary Or C= 0,00654 = 0,007 accept (4)

5.3.2 (a) 1m (H2O2) = cV 25 = = vv (—— (0,01 - 0,0053) v (—) = 1,175 x 104% moles (b) — n(H2O2) : n(O2) 2:1 v ratio 4,175 x 104 : 6875 x 10) (c) n(O2) =v x me = 7 25,7 V=1,510x 10% dmv —-0,00151 dm* Page 7 of 15 (6) [23]

Page 8 of 15 QUESTION 6 6.1. It is that stage of a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction VY [2 or 0] (2) 6.2. Marking criteria Assume QS (2x) reacts and Q2 (x) and RSs (x) produced. Equilibrium moles Equilibrium concentration Ke equation Substitution of Ke Substitution of conc of reactants and products Answer for x (= 0,127 moles). OPTION 1 Reaction | Q2 RS3 2QS RS(s) Initial moles 0,6 0,6 0,4 0,4 Change in +x +x - 2x v moles Equilibri 0,6 + 6+x |04-2 v a an rium +x 06+x Xx ’ Divide by By Equilibrium[] | 06+x | 0,6+x | 04—2x 0,6an mol-dm? 0,5 0,5 0,5 oe. [Qs]? [Q2][RS3] (ae = 0.5 0,04 v= =e ¥ 0,5 0,5 0,4-2x 0,2 = 0,6+x x = 0,127 moles = change in moles of Q2 ¥ (7)

ACCEPT [max 5 marks if correct conclusion made > then 6 marks] Reaction Qo RS3 2QSs RS(s) Initial moles 0,6 0,6 0,4 0,4 Change in -X -~X + 2x moles Equilibrium 0,6 -x 06-x |0,4+ 2x mol Equilibrium [ ] 06—x | 06-—x | 04+ 2x mol-dm? 0,5 0,5 0,5 Ke = 181? [Q2][RS3] pee y 0,04 v 0,5 v x 0,44+2x 0,6—-x - 0,127 moles No mark v t+ .. change in moles of Qe = - (-0,127) = 0,127 mole V¥ Page 9 of 15 Divide by 0,5 dm°v

OPTION Concentration route Reaction Q2 RSs 2QS RS Initial moles 0,6 0,6 0,4 0,4 mol Initial conc 1,2 1,2 0,8 Change in conc | +x +x - 2x Equillbens 1,2+x 1,2+x 0,8-2x mol-dm S 2 ge aS [Q2][RS3] (0,8-2x)? 0,04 v = ————— v (1,2+x)(1,2+2) 0,8-2x 0,2 = 1,2+% 0,2(1,2+x) = 0,8-2x 0,24 +0,2x = 0,8 — 2x 2,2x = 0,8—0,24 X = 0,25455 mol-dm? .. change in moles of Q = cV (0,25455)(0,5) 0,127 moles ¥ Page 10 of 15 v— Divide by 0,5 dm°v

6.3 6.4 6.5 6.6 Page 11 of 15 POSITIVE ¥ Mark Explanation independently « An increase in temperature favours an endothermic reaction, v the reactants are being decreased indicating the forward reaction is favoured e Therefore the forward reaction is endothermic. ~ (3) Concentration of Reactants increases / Q2 and/or RSs added ¥ (1) e Increase in concentration of reactants results in an increase in the rate of the forward reaction. ¥ e As reactants are used, the rate of forward reaction decreases. V As products are formed, the rate of the reverse reaction increases. V Equilibrium is obtained after 20 s when the rates of the forward and reverse reactions are equal. (3) a v =Sece=e Criteria 2 is Y both vertical line a \ Y both horizontal lines c 1 2 Esaoaoed 1 Bs} oO ® ow (2) [18] Time (s)

Page 12 of 15 QUESTION 7 7.1.1. An acid is a substance when ionized in water produces (H*) /(H30*) hydronium ions OR Acids produces hydrogen ions (H*) /(H3O*) hydronium ions Y in an aqueous solution. vv [2 or 0] 7.1.2 HSO4 / hydrogen sulphate ion ¥ 7.1.3 Hs0* ¥ me pH = -log[HsO*] ¥ = -log (0,6) v = 0,22 v 7.2.1 pH>7v 7.2.2 OPT 1: COs * (aq) + H2O (2) ¥ — HCOs~ (aq) + OH (aq) ¥ OR OPT 2: COsz * (aq) + 2H20 (£) ¥ — H2COs (aq) + 20H (aq) (OH was formed .. pH > 7) ACCEPT: NazCO3 + 2H20 ¥ + 2NaOH + H2CO3 VY [ 7.3.1. Initial mol of H2SO, (a) pH = - log [H3O*] v 0,2 ¥ =- log [H30*] [HsO*) = 0,63096 mol.dm? (b) [H30*] : [H2SO4] 2:1 0,63096 : 0,31548 mol.dm [H2SO4] [ACCEPT: 0,32 mol-dm*] ¥ 7.3.2 POSITIVE MARKING FROM 7.3.1 Marking criteria Ninitial = C.V formula ¥ Substitution into formula ¥ Initial mol of NaOH substitution ¥ Mol of H2SOs4 reacted with Na2COs subtraction Vv m=n.M formula ¥ Substitution into formula ¥ Percentage purity answer “ (2) (1) (1) (3) (1) (2) (3)

(a) Ninitiaa = CV % = (0,31548)(0,1) ¥ 0,031548 mol Initial mol of NaOH/ (b) — Ninitiaa = CV i = (0,4)(0,05) ¥ = 0,005 mol Mol of H2SO, reacted with NaOH (c) — n(NaOH) ;'n(H2SOa) 1 } :2 0,008 (Q,0025 mol of HzSO. Mol of H,SO, (reacted with Na,CO; (d) — Ninitian- Nreacted(vaoH) 0,031548) -(0,0028) v = 0,029048 mol Mass of pure!Na,CO3 (e) — n(H2SQs) : n(NazCOs) ' iat 0,029048 : 0,029048 mol of Na2zCOs (f) n(NazCOs) = 7 v mm 0,029048 )= v 106 m = 3,079 g Percentage purity of Na,CO3 (g) | Percentage purity = “a x = =61,58% Vv Page 13 of 15 [Range: 61,48% — 61,59%] (7) [20]

Page 14 of 15 QUESTION 8 8.1 Chemical energy to electrical energy Vv (1) 8.2 Mg (s) — Mg?* (aq) +2e Vv (2) 8.3 8.4 8.5 8.6 8.7 8.8 Double arrow = 1 mark Accept: Mg?* (aq) + 2e <— Mg (s) VY Double arrow = no marks X** is the oxidising agent because it gains electrons and its oxidation number decreases from +3 to0. “ (1) e Ptis inert or Unreactive. OR e Very good conductor of electricity ANY ONE v (1) e Temperature: 25°C or 298K “ « Pressure: tatm or 101,3 kPa ¥ [Must give values for temperature and pressure] (2) Marking Criteria ¢ anode (reducing agent | oxidised species) “ e salt bridge || ¥ e cathode (oxidising agent | reduced species) ¥ v vo v X(s)| X**(aq) | | Cte(g) | Ce (aq) | Pt(s) (1 mol.dns) (1 mol.dm-$) [NOTE: NO deductions if 1 mol.dm* or state of matter omitted] (3) E® cet = E® reduction - E®° oxidation % 2,10 V =+ 1,36 - (X)v X=- 0,74 (V) v X is Cr / Chromium ¥ (5) Increases Vv Nett Cell Reaction: 2Cr + 3Cf2 > 2Cr* + 3Cf Adding AgNOs will precipitate AgC£ v leading to a decrease in chloride ion (Cf) concentration. Y Therefore forward reaction is favoured. (3) [18]

Page 15 of 15 QUESTION 9 9.1 9.2 9.3 9.4 9.5 9.6 An aqueous solution that conducts electricityY through the movement ions ¥ OR A substance of which the aqueous solution V contains ions V OR A substance that dissolves in water v'to give a solution that conducts electricityv .(2) Cu(NO3)2 / CuNOsY (1) {ron rod ¥ Mark reason below independently Reduction takes place (on the iron rod) OR Cu?* (or Cu*) ions gain electrons to form Cu v (2) Cu(s) > Cu?*(aq) + 2e Vw ACCEPT: Cu(s} — Cu*(aq) +e vv (2) Copper ions ¥/copper(Il) ions / copper(l) ions [ACCEPT: Cu?* /Cu* ] Silver ions ¥ [ACCEPT: Ag* ] (2) Mark independently from 9.5 Since Ag is connected to the anode it will undergo oxidation to produce Ag* ions. In the solution there will now be Ag* ions and Cu?* ions. Ag‘/silver (I) ions is a stronger oxidising agent V than Cu2*/copper (II) ions V and will be reduced more readily to form silver/Ag on the iron rod. OR Cu?*/copper (II) ions is a weaker oxidising agent V than Ag*/silver ions and will not be reduced readily to form copper/Cu on the iron rod. ACCEPT: Ag* +e — Ag (one mark) (2) [11] TOTAL 150 MARKS