Physical Sciences Grade 12 Paper 2 EC June 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2019 PHYSICAL SCIENCES P2 MARKING GUIDELINE/ FISIESE WETENSKAPPE V2 NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 10 pages. Hierdie nasienriglyn bestaan uit 10 bladsye.

Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P2 (EC/JUNE 2019) QUESTION 1/VRAAG 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/VRAAG 2 2.1 CONCENTRATED √ GEKONSENTREERD (1) 2.2 To prevent reagents escaping √ /To smell the ester/Acts as a condenser Om te voorkom dat reagense ontsnap / Om die esters te ruik√ / Dit tree as ʼn kondensor op (1) 2.3 H2O √ (1) 2.4 propan-1-ol √√ Accept 1-propanol propanol (1/2) propaan-1-ol Aanvaar 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/Empiriese formule C2H4O √ Molar mass(R)/Molêre massa(R) = 130 + 18 – 60= 88 √ MMolar Mass (Empirical formula)/Molêre massa (Empiriese formule)= 2 x 12 + 4 x 1 + 1 x 16 = 44 √ Molecular formula/Molekulêre formule= C4H8O2 √ (8) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V2 3 QUESTION 3/VRAAG 3 3.1 3.1.1 (a) Esters √ (1) (b) H O H H Marking Criteria Whole structure correct H C O C C C H 2/2 H H H Volle struktuur 2/2 Only functional group correct ½ Slegs funksionele groep korrek ½ (2) 3.1.2 Same molecular mass √√ /Same molar mass Dieselfde molekulêre massa / Dieselfde molêre massa (2) 3.1.3 B has two sites for hydrogen bonding √. A has one site for hydrogen bonding √ B het twee plekke vir waterstofbinding. √ A het een plek vir waterstofbinding (2) 3.2 3.2.1 Yes √ Ja Same molecular formula √ but different structural formulae √ Dieselfde molekulêre formule maar verskillende struktuurformules (3) 3.2.2 Y√ D has a larger surface area/chain length than E √ D het ʼn groter oppervlakte/ kettinglengte as E London forces/Induced-dipole forces √/Dispersion forces stronger√ in D than in E Londonkragte/geїnduseerde-dipool kragte/Dispersie-kragte is sterker in D as in E More energy needed to break/overcome forces in D√ Meer energie word benodig om die kragte te breek/oorkom in D OR/OF Y√ E has a smaller surface area/chainlength than D √ E het ‘n kleiner oppervlakte/ kettinglengte as D London forces/Induced-dipole forces √/Dispersion forces weaker in E than in D Londonkragte/geїnduseerde-dipool kragte/Dispersie-kragte is swakker in E as in D Less energy needed to break forces in E √ Minder energie word benodig om intermolekulȇre kragte te breek/oorkom in E (4) [14] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P2 (EC/JUNE 2019) QUESTION 4/VRAAG 4 4.1 4.1.1 (a) Addition √ /Halogenation/Bromination √ Addisie / Halogenasie √/Brominasie (1) (b) Elimination √/Dehydrohalogenation √ Eliminasie / Dehidrohalogenering (1) 4.1.2 (a) H Cℓ Cℓ H C H H√√ H C C C C H H H H H Marking criteria/Nasienriglyne  Whole structure correct 3/3  Volle struktuur korrek 3/3  Two Br atoms in structure 1/3  Twee Br atome in struktuur 1/3 (3) (b) Chlorine √ Choor (1) 4.1.3 (a) H H H H √ R H C C = C C H√ but-2-ene√ √ ` OR 2-butene H H but-2-een OF 2-buteen Marking criteria/Nasienriglyne Whole structure correct 2/2 Vollestruktuur korrek 2/2 Only functional group correct ½ Slegs een funksionele groep korrek ½ (4) (b) Strong heat √ OR Concentrated strong base Vurige hitte OF Gekonsentreerde sterk basis (1) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V2 5 4.2 4.2.1 ADDITION √ ADDISIE (1) 4.2.2 n = 1000 √ (1) 4.2.3 Monomer √ Monomeer (1) 4.2.4 Make plastics√/ (Any other correct answer) Vervaardiging van plastiek / (Enige ander korrekte antwoord) (1) 4.3 4.3.1 Breaking down of a long chain √ /hydrocarbon/alkane into more useful shorter chains √ Breek van ʼn lang ketting / koolwaterstof / alkaan in nuttiger korter kettings (2) 4.3.2 THERMAL CRACKING √ TERMIESE KRAKING (1) 4.3.3 H H H H H H H C C C C C C H √√ Hexane √√ Heksaan H H H H H H Marking criteria/Nasienriglyne  Whole structure correct 2/2  Volle struktuur korrek 2/2  If one or more hydrogens are omitted ½  Een of meer waterstofatoome uitgesluit ½ (4) [22] QUESTION 5/VRAAG 5 5.1 Temperature√/Concentration √ (of H2O2)/Add a catalyst √ Temperatuur / Konsentrasie (van H2O2 ) / addisie van ʼn katalisator (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. Verandering in konsentrasie per eenheidstyd / Tempo van verandering van konsentrasie OF ʼn verandering in hoeveelheid / volume / massa van reaktant / produk per eenheidstyd (2) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P2 (EC/JUNE 2019) 5.3 5.3.1 Average rate/Gemiddelde tempo = - ∆c/∆t= - (1,45-1,9) √ /(15-0) √ = 0,03√ (mol·dm-3·min-1) (3) 5.3.2 High concentration √√ ( of H2O2 initially) Hoë konsentrasie ( van H2O2 oorspronklik) (2) 5.4 5.4.1 ENDOTHERMIC √ ENDOTERMIES (1) 5.4.2 Catalyst increases rate of reaction√/Katalisator verhoog reaksietempo  By lowering activation energy /Deur aktiveringsenergie te verlaag √  More particles have sufficient Ek to react √/More particles have Ek greater or equal to Ea/Meer deeltjies het genoeg Ek om te reageer √ / meer deeltjies het ʼn Ek groter of gelyk aan Ea √  More effective collisions per unit time √/Meer effektiewe botsings per eenheidstyd (4) 5.5 5.5.1 Experiment 1 √ : More particles have higher Ek √ Eksperiment 1 : Meer deeltjies het hoër Ek (2) 5.5.2 EQUAL TO √ GELYK AAN Same amount of H2O2 used in both experiments √ Dieselfde hoeveelheid H2 O2 word in beide eksperimente gebruik (2) 5.6 nO2 = V/Vm = 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] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V2 7 QUESTION 6/VRAAG 6 6.1 Stage reached by a chemical reaction where the rate of forward reaction equals the rate of reverse reaction √√ Fase word bereik deur ʼn chemiese reaksie waar die tempo van voorwaartse reaksie gelyk is aan die tempo van terugwaartse reaksie (2) 6.2 6.2.1 REMAINS THE SAME √ BLY DIESELFDE (1) 6.2.2 INCREASES √ TOENEEM (1) 6.2.3 INCREASES √ TOENEEM (1) 6.3 Decreasing pressure is opposed √ Reaction which produces more gas moles is favoured √ Forward reaction is favoured √ Afnemende druk is word teengestaan Reaksie wat meer gasmol produseer, word bevoordeel Voorwaartse reaksie word bevoordeel (3) 6.4 6.4.1 REVERSE √ TERUGWAARTS (1) 6.4.2 Catalyst √√ (added) Katalisator (bygevoeg) (2) OPTION 1/OPSIE 1 6.5 CALCULATIONS USING NUMBER OF MOLES BEREKENINGE MET DIE GEBRUIK VAN AANTAL MOL  Divide by/Deel deur 44 in n = m/M √  Divide nCO2 equilm & nCO equilm by 2 √  ∆n(CO2) = ninitial -neq √  Ratio verhouding CO2 : CO 1 : 2 √  nequil CO = nCOinitial+ ∆n(CO) √  Kc expression/Uitdrukking √  Substituting/Substitusie cequilmCO en cequilimCO2 √  Final answer /Finale antwoord √ Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P2 (EC/JUNE 2019) OPTION 1/OPSIE 1 ninitial CO2= m/M = 104,72/44 √ = 2,38 mol CO2 C CO ninitial 2,38 0 ∆n 0,48 √ 0,96√(Ratio) n equilm 1,9 0,96√ cequilm 7,6 3,84√(Division by 0,25) Kc = [CO]2/[CO2] √ = 3,842/7,6 √ = 1,94 √ OPTION 2/OPSIE 2 CALCULATIONS USING CONCENTRATION BEREKENINGE MET DIE GEBRUIK VAN KONSENTRASIE  Substitute into / Substitusie in C=m/MV Substitute into/ Substitusie in c=n/V√  ∆c(CO2) = cinitial -ceq √  Ratio/Verhouding CO2 : CO 1:2 √  cequil CO = cCOinitial+ ∆c(CO) √  Kc expression /Uitdrukking √  Substituting /Substitusie cequilmCO and cequilimCO2 √  Final answer /Finale antwoord √ Initial Concentration of CO2 Equilibrium concentration of CO2 Aanvanklike konsentrasie van Ekwilibrium konsentrasie van CO2 CO2 m n c c MV v 104,72 1,9 c √ c √ (44)(0,25) 0,25 c  9,52mol  dm 3 c  3,84mol  dm 3 CO2 C CO Cinitial 9,52 - 0 ∆c 1,90√ - 3,84√(Ratio) c equilm 7,60 - 3,84√ Kc = [CO]2/[CO2] √ = 3,842/7,6 √ = 1,94 √ (8) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNIE 2019) FISIESE WETENSKAPPE V2 9 6.6 6.6.1 Low √ Yield/ Lae Opbrengs Kc is low √/ Kc is less than 1/ [REACTANTS] > [PRODUCTS] Kc is laag / Kc is minder as 1 / [REAKTANTE]> [PRODUKTE] (2) 6.6.2 Exothermic √ Eksotermies  As temperature decreases, Kc decreases, [Products] decreases √ OR [Reactants increases]  Reverse reaction is favoured √  Decrease in temperature favours exothermic reaction √  Soos temperatuur afneem, verminder Kc, verminder [Produkte] √ OF [Reaktante neem toe]  Omgekeerde reaksie word bevoordeel √  Afname in temperatuur bevoordeel eksotermiese reaksie √ (4) [25] QUESTION 7 7.1 7.1.1 Reaction of a salt with water √√ Reaksie van ʼn sout met water (2) 7.2 7.1.2 Can act as acid and as a base √√/Can donate H+ and accept H+ Kan as suur en as ʼn basis optree / kan H+ skenk en H+ aanvaar (2) 7.1.3 HCO3 √ (1) 7.1.4 H2CO3 √ (1) 7.2 7.2.1 Standard √ (solution) Standaard (oplossing) (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/OF 0,025 dm3 (3) 7.3 7.3.1 Neutralisation √ Neutralisasie (1) 7.3.2 6,8 to 7,2 √ Titration between strong base and a strong acid √ Solution at endpoint is neutral √ Titrasie tussen sterk basis en ʼn sterk suur Oplossing by eindpunt is neutraal (3) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P2 (EC/JUNE 2019) 7.3.3 nNaOH reacting = cV = 0,1 x 30/1000 √= 0,003 mol nHCl reacting = 0,003 mol √ cHCl = n/V = 0,003/(25/1000) √ = 0,12 mol.dm-3 √ OR cAVA/cBVB= nA/nB cA x 25/0,1 x30 = 1/1 cA = 0,12 mol.dm-3 (4) 7.3.4 Point where an indicator changes colour √√ Punt waar ʼn indikator van kleur verander (2) 7.3.5 POSITIVE MARKING FROM 7.3.3 nHCl excess = cV√ = 0,12 x 8/1000 √ =9,6 x 10-4 mol cHCl= n/V = 9,6 x 10-4/(30 + 25 + 8) √/1 000 √ = 0,015 mol.dm-3 pH = -log[H3O+] √ = - log (0,15) √ = 0,82 √ (7) [29] TOTAL /TOTAAL: 150 Copyright reserved Please turn over

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