Physical Sciences Grade 12 North West Mock 2020 Memo

o4o40000 Lefapha la Thuto la Bokone Bophirima Noordwes Departement van Onderwys North West Depariment of Education ~” NORTH WEST PROVINGE NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 > o PHYSICAL SCIENCES: CHEMISTRY (P2) ic - FISIESE WETENSKAPPE: CHEMIE (V2) a fi u : SEPTEMBER 2020 F a a t MARKING GUIDELINES/NASIENRIGLYNE a Lay 7 MARKS/PUNTE: 150 These marking guidelines consist of 12 pages and 3 pages with the cognitive grid. Hierdie nasienriglyne bestaan uit 14 bladsye en 3 bladsye met die kognitiewe tabel. Copyright reserved/Kopiereg voorbehou : Please turn over/Blaai om asseblief

Physical Sciences Downloaded. from hlayiso.GMPemver 2020 QUESTION 1/VRAAG 7 A 1.2 1.3 1.4 1.5 16 1.7 1.8 1.9 1.10 Copyright reserved/Kopiereg voorbehou B NSC Marking guidelines /NSS Nasienriglyne (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [20] Please turn over/Blaai om asseblief

og040000 ~ Physical Sciences Dawnleaded, f ro m h lay so . GR Baotember 2020 NSC Marking guidelines /VSS Nasienriglyne QUESTION 2/QUESTION 2 21 2.2 2.3 24 Molecules with the same molecular formulae, but different structural formulae. Vv ¥ Molekule met dieselfde molekulére formule, maar verskillende struktuurformules. 2.21 AVY 222 Dv 22:3 BY 2.3.1 Esterification/Esterifikasie Vv (2) (1) (2) (1) 2.3.2 Acts as a catalyst/a substance that speeds up the rate of the reaction/dehydrating agent. ¥ Tree op as ‘n katalisator/'n stof wat die tempo van ‘n reaksie verhoog/dehidrateermidde! 2.3.3 H | O H Cc c Cc H cl oH | | c—c feel H Hjn Vv Copyright reserved/Kopiereg voorbehou vv (1) (2) (2) [12] Please turn over/Blaai om asseblief

Physical Sciences Pawnloaded f ram h layiso 6 QMember 2020 NSC Marking guidelines /VSS Nasienriglyne QUESTION 3/VRAAG 3 3.4 3.2 3.3 3.4 The temperature at which the vapour pressure of a substance equals atmospheric/external pressure. Vv“ Die temperatuur waar die dampdruk van die stof gelyk is aan die atmosferiese/eksterne druk (2) AY Lowest boiling point/Shortest chain lengthY Laagste kookpunt/ Kortste keitinglengte (2) 3.3.1. Ethyl ethanoate. vv /Etieletanoaat (2) 3.3.2 H H vv (2) Marking guidelines/nasienriglyne e BOTH have hydrogen bonding” /BEIDE het waterstofbinding e Compare number of sites for hydrogen bondingY /Vergelyk aantal plekke vir waterstofbinding Compare strength of IMFs” /Vergelyk sterkte van IMKs © Compare energy required /Vergelyk energie benodig Both compounds/C and D have (in addition to London forces and dipole- dipole forces) hydrogen bonding. ¥ Beide verbindings/C en D het (in addisie tot Londonkragte en dipoo!- dipoolkragte) waterstofbinding Compound C/Ethanol/alcohol has one site for hydrogen bonding and compound D/Ethanoic acid/ carboxylic acid has two/more sites for hydrogen bonding. ” Verbinding C/etanol! /alkoho! het een plek vir waterstofbinding en verbinding D/etanoésuur/karboksielsuur het twee/meer plekke vir waterstofbinding. Intermolecular forces in compound D/Ethanoic acid/ carboxylic acid are stronger than intermolecular forces in compound C/Ethanol/alcohol. ” Intermolekulére kragte in verbinding D/etanoésuur /karboksielsuur is sterker as die intermolekulére kragte in verbinding C/etanol/alkohol. Copyright reserved/Kopiereg voorbehou : Please turn over/Blaai om asseblief

UcUUUUU Physical Sciences Dawnlaaded, f ro 4) h lay so 2 GR otember 2020 NSC Marking guidelines /NSS Nasienriglyne OR/OF intermolecular forces in compound C/Ethanol/alcohol are weaker than intermolecular forces in compound D/Ethancic acid/ carboxylic acid. Intermolekulére kragte in verbinding C/etanol/alkohol is swakker as die intermolekulére kragte in verbinding D/etanoésuur/karboksielsuur More energy is needed to overcome/break intermolecular forces in compound D/Ethanoic acid/ carboxylic acid than in compound C/ Ethanol/alcohol. ¥ Meer energie word benodig om die intermolekulére kragte in verbinding D /etanoésuur /karboksielsuur te breek/oorkom as in verbinding C/etanol alkohol OR/OF Less energy is needed to overcome/break intermolecular forces in compound C/Ethanol/alcohol than in compound D/Ethanoic acid/ carboxylic acid. Minder energie word benodig om die intermolekulére kragte in verbinding C/etanol /alcohol te breek/oorkom as in verbinding D/Ethanoic acid/ carboxylic acid. (4) [12] QUESTION 4/VRAAG 4 441 4.1.1 Addition v /Addisie (1) 4.1.2 2C2Hs +702 — 4COQ2 + 6H20 Notes/Nota’s e Reactants /Reaktantev Products/Produkte “ Balancing/Balansering v e Ignore/ignoreer = and phases/en fases e Marking rule/Nasienreél 6.3.10 (3) 4.2 4.2.1 1,2-dichloroethaney Vv /1,2-dichloroetaan (2) 4.2.2 Halogenation/Chlorination Vv /Halogenering/Chlorinering (1) 4.3 4.3.1 Hydrolysis ¥ /Hidrolise (1) 4.3.2 el eer H H ¥ a | | +NaOH/KOH A [4 + NaCt/KCe H—C—C—H — oe ae ce dy H OH (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences eLaw nleaded fram h layiso RS Gbember 2020 NSC Marking guidelines /NSS Nasienriglyne 44 4.4.1. Hydration v /Hidrasie (1) 4.4.2 -ANY ONE /ENIGE EEN - Excess water Vv /Oormaat water - Concentrated H2SOz /Gekonsentreerde H2SO4 (1) [14] QUESTION 5/VRAAG 5 5.1 The influence/the effect of a catalyst V /Invoed/effek van katalisator (1) 52 LY : (1) 5.3 No-catalystv’“/Geen kaitalisator (2) 5.4 Change in concentration of products/ reactantsv’ per (unit ) time” Verandering in konsentrasie van produkte/reaktante per tydseenheid (2) OR/IOF Change in amount/number of moles/volume/mass of products or reactants per (unit) time Verandering in die hoeveelheid/aantal molvolume/massa van produkte/reaktante per tydseenheid A [H20 5.5 Average rate = - AT202) At _ _ 0,0131-0,020 v 400-(0) v = 1,73 x 10 > mol-dm? stv (3) 56 Copyright reserved/Kopiereg voorbehou MARK ALLOCATION/PUNTETOEKENNING =e =a, =e sicre or/of ¢ = = or/of n a v e Substitute/vervang (0,0200 — 0,0106) and/en 50 x 10° v e n(O2) = %(H202) ¥ e Using/gebruik M = 32 in m = Mn or/of n = cMV or/of ratio calculation/verhouding berekening “ e Answer/aniwoord: 7,52 x 10° g/ 0,008 g/ 0,01 g ¥ Please turn over/Biaai om asseblief ve aie amt

Physical Sciences D winloa led f ro m h lay | so . oA Sela 2020 NSC Marking guidelines /NSS Nasienriglyne OPTION 4/OPSIE 1 OPTION 2/OPSIE 2 cay Ac [H202] = 0,0200-0,0106 <Is = 0,094 (0,0200 — 0, 0106) = n ¥ in Ac(O2) = 1/2(H202) =1/2(0,0094) ¥ n(Qz) = 4/2 n(H2O2) = 1/2(4,7 x 10“) ¥ = 2,35 x10 mol =0,0047 ¥ m n=— M Gs MV 2,35 x 104 = m/32 Vv Am(O2) = cMV m= 7, 52x 10%9 v = (0,0047)(32) v (50X10) =7,52x10-3g = 0,008 g = 0.019” (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences Pawnloaded fram h layiso fe AMember 2020 NSC Marking guidelines /NSS Nasienriglyne QUESTION 6/VRAAG 6 6.1. The stage in a chemical reaction when the rate of the forward reaction is equals to the rate of the reverse reaction. V“” Stadium in ‘n chemiese reaksie waar die tempo van die voorwaartse reaksie elyk is aan die tempo van die terugwaartse reaksie. OR/OF The stage in a chemical reaction when the concentration of reactants and products remain constant. Stadium in ‘n chemiese reaksie waar die konsentrasie van reaktante en produkte konstant bly. (2) 6.2. STAY THE SAME. v¥ BLY DIESELFDE At equilibruim there are equal number of moles of gaseous reactants and products. vv By ewewig is daar gelyke aanial mol gasagtige reaktante en produkte. (3) 6.3 6.3.1 A very small amount of NO(g) is formed. VW ‘n Baie klein hoeveelheid NO(g) is gevorm. (2) 6:3'2 Na(g) O2(g) | 2NO() Initial moles/Aanvangsmol 7 2 0 Moles reacted/mol gereageer 0,2 0,2 047 Moles at equilibrium/mol by ewewig ie 0,4v 6,8 Equilibrium concentration/ 3,4 0,9 0,2 Ewewigskonsentrasie_(mol-div) Divide by 2/deel met 2)v __{[Noy? ~ [No] [02] 0,04 “(3.4)(0.9) v = 0,013 ¥ (8) c v Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences Dawnleaded.f rom h layiso GO aptember 2020 NSC Marking guidelines /NSS Nasienriglyne OPTION 2/OPSIE 2 Nez Oz nM 2NO Initial concentration _ Ls 3,5 224 Aanvangskonsentrasie 2 2 0 (mol-dim) Change/verandering 0,1v 0,1 Of ov (mol-dm) 2° Equilibrium concentration 3,4 0,9 02Vv /konsentrasie by ewewig (mol-dim) __INO}? *"[NallO,] __ (02 ~ 3,4)(0,9) = 0.04 , 3,06 = 0,013V 63.3 e When the system is heated, Ke increased’, which implies that more NO(g) was formed. Met toename in temperatuur het die K- verhoog, wat impliseer dat meer NO¢qg) gevorm het. e The forward reaction is favoured. ¥ Die voorwaartse reaksie is bevoordeel. e Increase in temperature favours the reaction which absorbs energy” / endothermic reaction. Verhoging in temperatuur bevoordee! die reaksie wat energie absorbeer / endotermiese reaksie. (4) [19] 01040000 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences pReanleadead fram hlayiso.GQMrember 2020 NSC Marking guidelines /NSS Nasienriglyne QUESTION 7/VRAAG 7 7.1 7.1.1 An acid that ionises completely in water. VW ‘n Suur wat volledig ioniseer in water. (2) 7.1.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 PH = - log [HsO*] ¥ pH+pOH =14 3 +pOH =14 3 =- log [H30*] ¥ pOH =- log [OH] 11. =-log [OH] [H30*] =0,001 mol-dm?v (3) [OH] =1x10-1 Kw = [H30*] [OH] =14 [H3O*] = 0,001 mol-dm®? 7.1.3 No /Nee (1) 7.1.4. The concentration of the hydrogen ions in the HX solution is lower than’ that of HC? of the same concentration . “ Die konsentrasie van die waterstofione in die HX_oplossing is laer as die van HC? met dieselfde konsentrasie. (2) OR/OF The pH of HX is higher than that of HC£ of the same concentration Die Ph van HX is hoér as die van HCt met dieselfde konsentrasie 2 7.2.1 A substance that can react as an acid or a base in a chemical reaction. Vv Vv ‘n Stof wat beide as ‘n suur en ‘n basis kan optree in ‘n chemiese reaksie (2) 7.2.2 HSOs ¥ (1) 7.2.3. Sulphite ion Y Sulfietioon (1) 7.3 7.3.1 ¢(H2SOx) = 2v Vv n 02=—v 0,1 n = 0,02 molv (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences Downleaded from hlayiSo.GQptencer 2020 NSC Marking guidelines /NSS Nasienriglyne 7.3.2 n[Mg(OH)2]= cV = (0,2) x (0,02) ¥ = 0,004 mol n(H2SO« in excess/oormaat) = 0,004 mol v (mol ratio) n(H2SQa4) reacts with/reageer met Na2zCOz = 0,02 -0,004” = 0,016 mol n(Na2CQs3) = n(H2S04) = 0,016 mol” = (mol ratio) m(Na2zCO3) = nM = 0,016 x 106V” = 1,696 Mass of impurity/massa onsuiwerheid = 5 — 1,696V = 3,304g V (7) [22] QUESTION 8/VRAAG 8 8.1. The substance/ species which gains electrons. vv Stof wat elektrone bykry. (2) 82 BiAgtY Silver ions (Ag*) has a stronger oxidising ability than cobalt (Co)v’, and silver ions (Ag*) will be reduced to silver v. Silwer ione (Ag*) het ‘n sterker oksideervermoé as kobalt (Co), en silwer (Ag*) sal gereduseer word na silwer (3) 8.3 Co(s) — Co* + 2e 2Ag* +2e => 2Aq(s Co(s) + 2Ag* == Co + 2Ag(s)V¥v (bal)” (3) 8.4 E® cei = E®cathodelkatode - E®anode ¥ Accept any other correct formula E® sel = E®katode/katode - E®anode from the data sheet. _ Aanvaar enige ander korrekte = +0,80 v - (-0,28) formule vanaf gegewensblad. = 1,08 vv E®cetser<1,5 V, thus the bulb will not glow or light up/ gloeilamp sal nie brand.¥ (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences pRawnleadead from h layiso GOP tember 2020 NSC Marking guidelines /NSS Nasienriglyne QUESTION 9/VRAAG 9 9.1 Electrical energy is converted to chemical energy. v~“ Elektriese energie word omgeskakel in chemiese energie. (2) 9.2 Cathode/katode V (1) 9.3.1 Cryolite/kriolietv (NasAIFe) Vv (2) 9.3.2 AlzOs(.2H20) VY (2) 9.4 Carbon in the electrodes reacts with O2 gas produced, forming CO2 ¥ carbon is used upY causing electrodes to disintegrate/corrode away. Koolstof in die elektrodes reageer met O2 gas wat vorm en vorm CO2/ Koolstof word opgebruik wat veroorsaak dat elektrodes disintegreer/ weggevreet word (2) [9] QUESTION 10/VRAAG 10 10.1 10.1.1 Combustion (process) “ Verbrandings(proses) (1) 10.1.2 4NH3s +502 Y> 4NO + 6GH20V balv (3) 10.2 10.2.1 Neutralisation (Acid-Base) V /neutralisasie (Suur-basis) (1) 10.2.2 HNOs+NH3sv—> NH4NOs3 v bal v (3) 10.3 10.3.1 Haber ¥ (1) 10.3.2 Feliron/yster ¥ (1) 10.4 10.4.1 The ratio of nitrogen, phosphorus and potassium in the fertiliser” v'IDie verhouding stikstof, suurstof fosfor en kalium in die kunsmis. (2) 10.4.2 Pure fertiliser/Suiwer kunsmis = 10 kg x 23 V 100 ¥ =23kgv (3) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou