Physical Sciences Grade 12 Paper 2 Gauteng Mock 2020 Memo

Downloaded from hlayiso.com So) GAUTENG PROVINCE FPCAT ION 2 REPUBLIC OF SOUTH AFRKA GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2020 MARKING GUIDELINES NASIENRIGLYNE 10842 PHYSICAL SCIENCES: CHEMISTRY FISIESE WETENSKAPPE: CHEMIE PAPERI/IVRAESTEL 2 12 pagesibladsye P.T.O.

Downloaded from hlayiso.com 10842/20 QUESTIONIVRAAG 1 11 Dvv (2) 12 AvY (2) 1.3 AYY (2) 14. Cv (2) 1.5 D¥vv¥ (2) 16 Avv (2) 170 Cvv (2) 18 ByY (2) 19 Avy (2) 110 Dvv (2) [20] QUESTION/VRAAG 2 2.1.1 A orlofC ¥ (1) 212 BY (1) 2.1.3 Fv (1) 2.1.4 DY (1) 2.2.1 4,5-dimethyl “hex-2-ene v /4,5-dimetietheks-2-een Marking criteria 2,.2,1-2.2.3/Nasienriglyne: Only functional group correct/Slegs funksionele groep korrek: Max/Maks: %B Whole structure correct/Hele struktuur korrek: % (2) 2.2.2 2,3-dibromo-5-methyl “heptane ~ 2,3-dibromo-5-metietheptaan (2) 2.2.3 4-methyl “’pent-2-yne ¥ 4-metielpent-2-yn {2)

Downloaded from hlayiso.com 10842/20 EET H ; c—c i ck he aan: vv H H H H H HoH (2) Marking criteria/Nasienriglyne ¢ Only functional group correct/Sfegs funksionele groep korrek: Max/Maks: Ne % * Whole structure correct:/hele struktuur korrek 72 2.3.2 Propanoic acidv ¥ Propanoésuur (1) 2.3.3 Sulphuric acid/H2SO4 ¥ Swawelsuur/ H2SO4 (1) [14] QUESTION/VRAAG 3 3.1 The temperature at which the solid and liquid phase are at equilibrium vv” Die temperatuur waarby die vaste stof en die vioeistof fase in ewewig is. ““ (2) 3.2 C Vv It has a lower vapour pressure ¥ loer: ¥ Cv Dithet ndampdruk ¥, Sonche oun © (2) 3.3.1. Organic compounds with the same molecular formula, but different structural formulae. ¥v ¥ 2 oF zero Organiese verbindings met dieselfde molekulére formule, maar verskillende struktuurformules. “ “ 2 of nul (2)

3.3.2 3.4 Downloaded from hlayiso.com 10842/20 Compound B heptane is less branched/has a longer chain/less compact/less sphericai/has a larger surface area. Stronger intermolecular forces/Van der Waals forces/dispersion forces/London forces/Induced dipole- induced dipole forces. “ More energy needed to overcome intermolecular forces” OR Compound A/2-methyhexane is more branched/more compact/more spherical/has shorter chain/has a smaller surface area. “ It has weaker intermolecular forces/Van der Waals forces/dispersion forces/London forces/Induced dipole- induced dipole forces. V Less energy needed to overcome intermolecular forces. “ Verbinding B, heptaan, is minder vertak/het jn langer ketting/minder kompak/minder sferies/ het 'n groter opperviak area. Sterker intermolekulére kragte/Van der Waals kragte/dispersiekragte/_ondonkragte/geinduseerde kragte. “ Meer energie benodig om intermolekulére kragte te oorkom._ “ OF Verbinding A/2-metielheksaan is meer vertak/meer kompak/meer sferies/het 'n korter kettinglengte/et 'n kleiner opperviak area. “ Dit het 'n swakker intermolekulére krag/Van der Waals kragte/dispersiekragte/Londonkragte/geinduseerde dipoolkragte. ~“ Minder energie word benodig om die intermolekulére kragte te oorkom. _“ (3) Both compounds C and D have hydrogen bonding between molecules. “Compound C has one site for hydrogen bonding whilst compound D has two sites for hydrogen bending “ OR Compound C has less sites for hydrogen bondingAweaker hydrogen bonding than compound D. More eneray needed to overcome intermolecular forces in compound Dfless energy needed to overcome intermolecular forces in compound E. v Beide verbindings C en D het waterstofbindings tussen die molekule. ” Verbinding C het een plek vir waterstofbinding terwy! verbinding D twee plekke het vir waterstofbinding. “ OF Verbinding C het minder plekke vir waterstofbinding/swakker waterstofbinding as verbinding D. Meer energie word benodig om die intermolekulére kragte te oorkom in verbinding D/minder energie word benodig om die intermolekulére kragte in verbinding Ete oorkom. ~ (3) [12]

Downloaded from hlayiso.com 10842/20 QUESTION/VRAAG 4 4.1.1 It contains a double bond/C=C between two carbon atoms in its hydrocarbon chain/All carbon atoms not bonded to the maximum number of atoms/four atoms¥ Daar is dubbelbindings/C=C tussen twee koolstofatome in die koolwaterstof ketting/Alle koolstofatome is nie gebind aan die maksimum aantal atome/vier atome nie. “ (1) 4.1.2 (a) Addition/Bromination ¥ Addisie/Broominasie (1) (b) Substitution Substitusie (1) 4.1.3 Heat/sunlight/ultraviolet light “ Hitte/sonlig/ultraviolet lig (1) 4.1.4 Butanev/butaan (1) 4.1.5 Hydrogen bromide/HBrY Waterstofbromied/ HBr (1) 4.2.1 Elimination” Eliminasie (1) 4.2.2 Butv-2-enev But-2-een (2) 423 v v i! iy v Hf lew ¥ Heo H+ NaOH ——* H-G— E7o-H + NaCt HH H HH H (6) [15]

Downloaded from hlayiso.com 10842/20 QUESTION/VRAAG 5 5.1 Carbon dioxide/CO2 ¥ Koolstofdioksied/CO2 (1) 5.2.1. The decrease in concentration of hydrochloric acid per unit time. “” Die afname in konsentrasie van waterstofchloried per eenheid tyd. wv (2) 5.2.2 Concentration (of the acid) ¥ Konsenirasie (van die suur) (1) 5.2.3 Fora fair test/comparison “ Vir n regverdige toets/of vergelyking “ (1) 5.3.1. e Higher acid concentration in experiment 2 means more particles/molecules per unit volume” e More particles have kinetic eneray equal to or greater than activation energy/More particles have enough kinetic energy ¥ e More effective collisions per unit time/frequency of effective collisions increases ¢ Rate of effective collisions increases. “ Hoér konsentrasie suur in eksperiment 2 beteken meer deeltjies/molekule per eenheid volume” Meer deeltjies het kinetiese energie gelyk of groter as die aktiveringsenergie/Meer deeltjies het genoeg kinetiese energie. v” Meer effektiewe botsings per eenheid tyd/frekwensie van effektiewe botsings verhoog. Tempo van effektiewe botsings verhoog. (4) 5.3.2 n(CaCOs) = mM Marking Criteria * Subsiituting 100 into formula m/M e Using ratio 2:1 e Substituting the number of moles into formula c= nv = 4/100 = 0,04 mol n(HCl) : n(CaCOs) = 2: 1 n(HCl) = 2(0,04) ¥ = 0,08mol f e Final answer — i Nasienriglyne: ¢(HCl) = nv | © Vervang 100 in formule m/M © Gebruik verhouding 2:1 0.4 = 0,08/VY e Vervang geial mol in formule c=n/V e Finale antwoord V=0,2 dm? = 200 cem3v | (4)

5.4 Downloaded from hlayiso.com 10842/20 Activated complex Geaktiveerda kompleks vom v Shape Potential energy Potensi#le energie Reaksle kodrdinate Reaction coordinates QUESTIONIVRAAG 6 6.1 6.2 6.3 6.3.1 6.3.2 The reaction is reversible. “ Die reaksie is omkeerbaar. “ This reaction is in contact with a catalyst, and therefore is known as the contact process. ¥ Hierdie reaksie is in kontak met 'n katalisator en daarom staan dit bekend as die Kontakproses. “ If the temperature is too low the reaction rate will be too slow for the yield to be enough. Vv Indien die temperatuur te laag is sal die reaksietempo te stadiq wees vir die opbrengs om genoeg te wees, ““ A high temperature favours an endothermic reaction ¥ If the temperature is high the reverse reaction will be favoured ” Less product will be formed which is not favourable. in Hoé temperatuur bevoordeel die endotermiese reaksie. Indien die temperatuur hoog is word die teruqwaarise_reaksie bevoordee!, “” Minder produkte sal vorm en dit is nie voordelig nie. “ (4) [17] (1) () (2) (3)

6.4 65 Downloaded from hlayiso.com 10842/20 Marking guideline/Nasienriglyne: e Use the Ratio/Gebruik ratio © N(SO2)eqewe = n(SO2)initiatbegin — An(SO2) N(O2)eqewe = x — An(O2) Divide equilibrium moles by 200 dm3/Deel ewewigsmol deur 200 dm3V v ° e Correct Ke expression (formulae in square brackets)“ Korrekte Ke uitdrukking(formules in blokhakies) * Substitution of values in expression/Vervanging van waardes in uitdrukking ¥ e Substitution of/Vervanging van 32 g:mol' in m = nM ¥ e Final answer/Finale antwoord: 891,179 “ e Range/Gebied: 891,167 - 891,2 SOz2 O2 SOz Initial amount(moles) Aanvangs hoeveelheid 50 x 9 Change in amount(moles) 22 io’ Verandering in hoeveelheid 22 " Ratio Equilibrium amount(moies) Hoeveelheid 28 xo 22 Equilibrium concentration(mol-dm) (x11) Divide by Ewewigkonsentrasie(mol-dm*) 0,14 200 011 200 dm3v’ _ _ [so3]2 ¥ a isogl* 0 ] 0,11 7,328 = me m=nxM 14) oa) = (278,849)(32)v x = 27,849 mol = 891,167g ¥ a ne! (7) Forward ¥ “/ Voorwaarts (2) [16]

Downloaded from hlayiso.com QUESTION/VRAAG 7 7.1 7.1.4 7.1.2 7.13 7.14 7.156 Acid-base reaction/neutralisation/protolysis Suur-basis reaksie/ neutralisasie/protoliese ~~ Barium hydroxide Bariumhidroksied X-Burette “ Buret A white precipitate in a yellow solution that gradually turns orange Vv ‘n Wit neerslag in n geel oplossing wat geleidelik oranje word. v~ (a) ¥ Remains constant Barium Hydroxide is a strong base that will dissociatefionise from the start Bly konstant. Bariumhidroksied is in sterk basis wat sal dissosieer/ioniseer van die begin af. (b) Decrease “ Verminder (c} Decrease “ Verminder Marking quidelines/Nasienriglyne e Any formulae/Enige formule: “ Division by Volume ~/ dee/ deur volume Use mole ratio “ gebruik molverhouding Use M = 233 g:mol* inm = nM “ Answer = 1,165 g “ 10842/20 (1) (1) (1) (3) OPTION 1: OPSIE 1 OPTION 2: OPSIE 2 Cae v Ma v c= CoVp My cqa(30) 1 O1= == v , 1)G60) 1 0,005 mol Cq = 0,167mol « dm-3 n c=- |ast= S oS oT Vv n 0,167 = d03 n = 0,005 mol n(BaSO,) =n(Ba(OH), v = 0,005 mol m=nxM = (0,005)(233) =1,1659 “ (5)

Downloaded from hlayiso.com 7.2 7.2.1 NHa*(aq) + H20(é) > NHaY + Hs0* ¥ 7.2.2 CHsCHOO-(aq) + H2O(é) - CHsCOOHY + OHY QUESTION/IVRAAG 8 8.1 A galvanic/voltaic cellv Galvaniese sel/voltaiese sel 82 Ags) | Ae?*(aq)(1 mol-dm) V1 Cu2*(aq)(1 mol-dm) | Cu(s) “ Accept Marking criteria: _Nasienriglyne; i e Oxidation “ Oksidasie Atl Ags [| Cu2* ICu i ° Saltbridge” soutbrug e Reduction’ reduksie 10842/20 (2) (2) [17] (1) (3) 83 AétoCu” Aéna Cu” (1) 8.4 AL> AS + 3a Vv Marking quidelines/Nasienrigiyne 1 0 © Af 2AB+3e° Ms AG +3e 2 Ae % ie} Ags +3e° Al % Ag +3e AL % e Ignore if charge omitted on electron./ignoreer indien lading weggelaat op elektron. e If charge (+) omitted on Aé* /indien lading (+) weggelaat op Aé*: Example/Voorbeeld: A¢é > AG +3eV¥ Max./Maks: he _ ; . — my i Notes (2) _ e Accept any other correct formula from the 8.5 E°cat = EP reduction = E°oxidation 7 data sheet./Aanvaar enige ander korrekte = 0,34 — (-1,66) ¥ formule vanaf gegewensbiad. =2Vv i (4) £ j e Any other formula using unconventional ' abbreviations, e.g. E°cay = E°oa- E°ra j followed by correct substitutions:/Enige i ander formule wat onkanvensionele i afkortings gebruik bv. E°sei =.3/4 i e E*ou- E°rmgevolg deur korrekte ' ] i vervangings: ¥%, 10

Downloaded from hlayiso.com 10842/20 8.6 8.6.1 Decrease” Verminder (1) 8.6.2 NEGATIVE MARKING FROM 8.6.1/NEGATIEWE NASIEN VANAF 8.6.7 The concentration of Aé*? will increase. According to Le Chatelier the system needs to decrease the concentration. ¥ The reverse reaction will be favoured. ¥ Therefore the reading will decrease. ¥ Die konsentrasie van AZ sal vermeerder. Volgens Le Chatelier sal die sisteem die konsentrasie moet verlaag. ~ Die terugwaartse reaksie sal bevoordeel word. “” Dus sal die lesing verminder. ~ (3) 8.7 Chemical energy to electrical energy” Chemiese potensiéle energie na elektriese energie. “ (1) [16] QUESTION/VRAAG 9 91 Av (1) 92 Cu%+ 2e >CuvY (2) A 93 Marking quidelines/Nasienriglyne « Multiply nx2 or mx2 “ = Vermenigvuldig nx2 of mx2 © Maeposite = 31,759 © Miostat anode = (31,75 + 15,8)g9 % e Answer/antwoord 66,77% ¥ n= > M ‘Tost at anode v Ze, 1,75 (0.25% 2)" se = 2 e100 = 3175 (31,75 + 15,8) Ce = 66,67% “ 0,25 mol Cu formed in 1 800 s so in 3 600s (2x0,25) mol formed 0,25 mol Cu vorm in 7 800 s dus in 3 600 s (2x0,25) moi gevorm (4) ue Ypurity = —~seeeste4 5100 9.4,1 Electrolyte is a solution/liquid/dissolved substance that conducts electricity through the movement of ions (2 or 0) Elektroliete is 1n oplossing/vioeistof/opgeloste stof wat elektrisiteit gelei deur die beweging van ione (2 of 0) (2) 9.4.2 The rate of oxidation is equal to the rate of reduction. ” Die tempo waarteen oksidasie plaasvind is gelyk aan die tempo van reduksie. “ (1) [10] 11

Downloaded from hlayiso.com 10842/20 QUESTION/VRAAG 10 10.1 10.1.1 10.1.2 10.1.3 10.1.4 10.2 Nitrogen ¥ stikstof Phosphorus ” fosfor Potassium ” kalium Acceptlaanvaar N:P:K (3) Potassium/KY Kalium/ K (1) Phosphates/P “ Fosfate/P (1) Haber orfof Ostwald Vv (1) Eutrophication: The process by which an ecosystem eg. A river or dam, becomes enriched with inorganic nutrients”, especially phosphorus and nitrogen, resulting in excessive plant growth ” Eutrofikasie: Die proses waarby 'n ekosisteem bv. 'n rivier of dam, verryk word met anorganiese voedingstowwe v, veral fosfor en stikstof, wat 'n oorgroot toename in plantegroei tot gevoighet. ~“ (2) Marking quidelines/Nasienriglyne * m(N) = 2x36 ¥ ¢ Double up on Fertilizer BY verdubbel kunsmis B « Fertilizer A m(N) = 24kg, m(P)=m(K) =6 kg * Fertilizer B m(N) = 19,09kg, m(P)=m(K) = 5,45 kg « Answer Fertilizer BY Fertilizer Al/kunsmis A + Fertilizer Bilkunsmis B In 100kg 4.1.1(36) In 50 kg 7.2.2(15) m(N) = £236 + In 100kg 14.4.4(30) =24kg “ - m(N) = #4230 m(P)=m(K) = 2x36 v = 19,09 kg =6kg op m(P}=m(k) = Sx30 = 5,45 kg v Therefore Fertilizer B will be the best “ Kunsmis B sal die beter een wees. (5) [13] TOTAL/ITOTAAL: 150 12