Physical Sciences Grade 12 P2 North West November 2017 Memo

Downloaded frpm hlayiso.com PhysivalSvienvewP2 NW/September'201%- 4 NSC — Momorandum QUESTION 1 (Start on a new page.) 110 Cv (2) 12 AW’ 2) 13° BYY 2) 14. Cv 2) 15 AVY (2) 16 Cry (2) 17 AW 2) 18 BYY 2) 19 Dv @) 110 Dvv (2) [29] QUESTION 2 {Start on a new page.} 24 24.1 EV Q) 241.2 FY - (1) 213 BY a) 214 DY (1) 22 Busan 2) 2.3 2.3.1 Propyl ethanoate. vv” (2) 2.3.2 aan —c—te—o w | I | | H H H H : . Marking criteria structural formula: * Three carbons in longest N.B. Accept condensed -OH chain * OH-group on terminal carbon Notes: « One or more H-atoms omitted: % « Condensed or semi- structural formuta:1/2 (2) uinpueiowiew — OSN toys i Pn bVOSHOCRURIES/ MAN OL ZqeBBoUsINgeo}shYg

Downloaded from hlayiso.com Physical Sciences/P2 NW/September 2017 NSC — Mermorendum QUESTION 3 (Start on a new page.) 3.4 3.1.1 Substitution(halogenation/bromination) (1) 3.1.2. Substitution(hydrolysis) ” ) 3.1.3. Elimination V e:.iif i OtiC (1) 3.2 CHsCHBrCHs + KOHY —CHsCHCH2z +H20+ KBr bal v’ 3) N.B. Max 1/3 if extra product or reactant is written 3.3 3.3.1 Ethanoic acid ¥% (®) 3.3.2 -—O—H ve wot You =p ie mp beet ty a ae ny Es 3) [N.B. Accept linear molecule of water. | 3.4 3.4.1 Heat/ Suntight/ UV light or rays/ Sun rays. (1) 3.4.2 Hydrogen bromide. v (1) 3.5 3.5.1 A polymer formed when two monomers combine through an addition reaction. vv (2) 3.5.2 iy ‘ Marking criteria structural formula: c= vv * Two carbons in chain | i ¢ Whole structure correct HOA Notes: « Qne or more H-atoms omitted % * Condensed or semi-structural formula 1/2 (2) 9) Pamerinht exnanind Dlanan tien mtine

Downloaded from hlayiso.com Physical Sciences/P2 NWiSeptember 2017 NSC - Memorandum QUESTION 4 (Start on a new page.) 44 4.1.1 Aseries of organic compounds that can be described by the same general formula. Vv“ (2) OR A series of organic compounds in which the members differ from the next with a —-CHz group. 42 Alcohol. v @) 43 : 4.3.1 Molecular mass(of different homologous series), (1) [N.B. Accept surface area, chain length/contact area. | 4.3.2 Vapour pressure. (1) 4.3.3. The larger the molecular mass, the lower the vapour pressure (2) OR The smaller the molecular mass, the higher the vapour pressure. [IN.B. Accept surface area/chain length/contact area’ | 44 Butane ¢ Weak” London forces between its molecules” (2) OR » London forces between its molecules” « less energy is needed for molecules to break out of liquid and form vapour 7 * Butan -1-o} « —alcohol, has strongerv hydrogen bonds between its molecules” = (2) OR * Hydrogen bonds between its molecules. v » lots of energy is needed for molecules to break out of liquid and form vapoury” 4.5 Compound Dv or (Butanol) « thas stronger hydrogen bonds between its molecules” « Therefore more energy is required to break the bonds between the molecules” Juries @) [N-B. Accept it has lowest vapour pressure. | [14] Dianna beim mune

Physical Sciences/P2 NSC - Memorandum QUESTION 5 (Start on a new page.} 5.1 5.2 5.3 5.4 . CO2(g) forms during the reaction. Downloaded from hlayiso.com NW/September 2017 (1) 4Qs. v The mass of the beaker and its contents remained 60,00g. (2) NCOz (reacted) = mae , = 200 a4 = 0,05 molY (C02) at STP = n(CO,) x Vmv = (0,05 x 22 ,4)¥v = 1,12 dmv" OR 1 mol = 22,4dm> 0,05mol - X X= 1,12 dm More particles per unit volume. ~ More HC£ molecules have enough kinetic energy and correct orientation. More effective collisions take place per second/ per unit time. [N-B. Accept: Higher frequency of effective collisions | Pantirint enamine @) (3) [12] Dianna dunn anne

Physical Sciences/P2 NSC — Memorandum QUESTION 6 (Start on a new page.) 6.1 A reversible reaction in which the rate of the forward reaction Downloaded from hlayiso.com Nw September 2017 equals to the rate of the reverse reaction. vv" (2) = 6.1.4 n(NHs) = = BS, 7 = 1,5 mol Nag | 3H2jg) | ZNHaig) Initial quantity(mol} 3 8 0 Change in (mol) 0,75 2,25 1,5v | Ratiov Quantity at equilibrium(mol) 2,25¥ | 5,75” 1,5 Equilibrium concentration (mol.dm?) Divide by 5v _ NEP v Wrong Ke expression Max 6/9 IN, IP No Ke expression Max 8/9 = 03" oy (0,45)01.15)° =0,13V (9) 6.1.2 | Smaller than v ¥ Ke Iv 2) 6.1.3 y Decrease” Increases in temperature favours the reverse reaction {endothermic reaction) v’ less product is formedv’, Ke value decreases. V (4) N.B. a) Accept: equilibrium position shifls to the left. b) The sign ¥ means negative marking in 6.1.2 & 6.1.3 above 17] Comurinht enennind Dtanan tim nuine

Downloaded from hlayiso.com Physical Sciences/P2 NW/September 2017 NSC ~Memorandum QUESTION 7 (Start on a new page.) 7A 7.1.1. A substance that can act as an acid and as a base. Vv (2) 7.12 Acid. It is a proton donor/it donates a proton. v (2) 7.4.3 PO*«(aq) “ (1) 7.2 7.24 Basic. v (1) 7.22 pH=-log[H] 13,3¥ =~ log [H*] [H*] = 5,01 x 104 mol.dm? 10 [#"] 10" * Fom0* = 0,2 mol.dm*v (5) [OH] =v = N.B Positi ing - 7.2.2 to 7.2.3. CBYb mn ositive marking - 7.2.2 to 7.2.3. Cal785) vy = 1y (0.2(25) 2 Ca = 0,14 moldmsv (5) 7.24 X+16+1=56 X = 39 g.mol!v X= K (potassium) ¥ {2)

Physical Sciences/P2 NWiSeptember 2017 NSC - Memorandum QUESTION 8 (Start on a new page.) 8.1. A solution/ liquid/ dissolved substance that conducts electricity through movement of ions. vv @) 8.2 Salt-bridge. v - completes the circuit! cell. - maintains the ceil neutrality. - Supplies a path through which ions can move to restore neutrality. (2) 8.3 BY Lead is a stronger reducing agent than than Phy’ and Mg will be oxidised to Mg? (3) 8.4 — Pb(NOs)2 or Pb?(aq)“or any saline solution with the corresponding ion as the cathode. (2) 85 8.5.1 Mg(s)“ —- Mg@Maq) + 2e “ (2) Notes: Mg(s) = Mg?*aq) + 2e° (1/2) Mg?(aq) + 2e° + Mg(s) (2/2) Mg(s) -- Mg*aq) +20" (0/2) Mg*faq) + 2e" = Ma(s) (0/2) 8.5.2. Mg(g) | Mg?*(aq)( (1 mol-dm) I! (1 mol-dmv3)H*(aq) | H2(g) Pt(s) (3) OR Ma(g) | Mg?aq) Il H*(aq) | He(g) Pt(s) OR Mg(g) | Mg?* (1 mol-dm) II H* (1 mol-div) | Ha(g) Pt(s) OR Mag(q) 1 Mg il H* t H2(g) Pi(s) 8.6 - It has Pt as inert/ does not react with the H* ions OR acid. - It has Pt as a conductor (of electricity) . (1) [15] N.B. QUESTION 8.6, subtract that 1 mark from the total (i.e 14 marks for it) and hence GRAND TOTAL will be 149 instead of 150 for the paper. Ranutinkt rannniae Dianna tere mune Downloaded fr bm hlayiso.com

° Downloaded from hlayiso.com Physical Sciences/P2 9 NW/September 2017 NSC — Memorandum QUESTION 9 (Start on a new page.) 9.1 Electrode where reduction takes place Vv“ (2) 9.2 Tv the negative electrode/ cathode “ (2) 93 Cus) > Cu* + 2ev¥ (2) Notes: Cus) = Cutag)+2e (1/2) CuPtag)+2e *-Cus) 2/2) Cu(s) ~ Cu?%ag)+2e (0/2) Cu*4aq) + 2e = Cu(s) (0/2) (2) 9.5 Ptand Ag are both weaker reducing agents than copper and will be oxidised to form ions. “¥ 2) OR Cu is a stronger reducing agent than Ag & Pt, so it will reduce Ag & Pt. 9.6 The rate at which copper is oxidised at the anode is equal to the rate at which copper ions are reduced at the cathode. vv (2) [10] QUESTION 10 (Start on a new page.) 10.1 Ostwald (process). “ (1) 10.2. 10.2.1 Pt(Platinum)~ (Accept Nicket (Ni) Temperature _+ 900°C Vv (2) 10.2.2 ANHs+ 502g) Vv -- 4NO(g) +6120 v7’ bal (3) 10.2.3 Catalytic oxidation. “ (1) 10.2.4 Nitrogen dioxide/ Nitrogen(IV)oxide (1) 10.2.5 20 or Water a) 10.3 10.3.1 KNOs vv @) 10.3.2 m(N)= Bx 6.35 ¥ , = 3,58 kg (4) [15] Panvminint rannntnd GRAND TOTAL: 149 Dianna tien nune