Physical Sciences Grade 12 P2 2019 Memo

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 1 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienrigl ————————————————————— DEPARTMENT OF EDUCATION ————————————————————— DEPARTEMENT VAN ONDERWYS ————————————————————— LEFAPHA LA THUTO ————————————————————— ISEBE LEZEMFUNDO ————————————————————— PROVINCIAL PREPARATORY EXAMINATION PROVINSIALE VOORBEREIDINGSEKSAMEN/ GRADE/GRAAD 12 PHYSICAL SCIENCES P2: CHEMISTRY FISIESE WETENSKAPPE V2: CHEMIE MARKING GUIDELINES/NASIENRIGLYNE SEPTEMBER 2019 MARKS/PUNTE: 150 TIME/TYD: 3 hours/uur This memorandum consists of 17 pages./ Hierdie memorandum bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 2 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 C (2) 1.2 B (2) 1.3 A (2) 1.4 C (2) 1.5 D (2) 1.6 D (2) 1.7 D (2) 1.8 A (2) 1.9 C (2) 1.10 B (2) [20] QUESTION 2/VRAAG 2 2.1 2.1.1 E (1) 2.1.2 A and C  A en C (1) 2.2 2.2.1 Propanone  ACCEPT: Propan-2-one; 2-propanone Propanoon AANVAAR: Propan-2-oon; 2-propanoon (2) 2.2.2 Ethyl propanoate Etielpropanoaat (2) 2.2.3 Methanoic acid Metanoësuur (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 3 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 2.3 H  H H H Marking Criteria/Nasien kriteria  H C C C C C C H Functional group Funksionelegroep H H H Ethyl substituent on the fourth carbon  Etielsytak op die vierde koolstof Whole structure correct H C H  Totale struktuur korrek H C H H (3) 2.4.1 H H C C H H (1) 2.4.2 Addition Addissie (1) 2.5 3n CnH2n + ( ) O2 → nCO2 + nH2O Marking Criteria/Nasien kriteria: 2 Balancing reactants  m 192 Balansering reaktante n(O 2 ) = = =6 Balancing Products M 32 Balansering produkte 3n = 6 Dividing by 32 2 Deel deur 32 n=4 3n ∴ CnH2n = C4H8 Equating/Gelykstelling = 6 2 Final answer Finale antwoord (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 4 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 Boiling point increases with an increase in the number of carbons. Kookpunt neem toe met 'n toename in die aantal koolstof atome. OR/OF As the number of carbons increases, the boiling point also increases.  Indien die aantal koolstofatome toeneem, sal die kookpunt ook toeneem. (2) 3.2 3.2.1 Higher than Hoër as (1) 3.2.2 ANY TWO: ENIGE TWEE: Hydrogen bond Waterstofbinding Dipole-dipole force Dipool-dipool krag London force/induced dipole-induced dipole force/dispersion force. Londonkrag/geïnduseerde dipool-geïnduseerde dipool krag/dispersiekrag (2) 3.2.3  Butan-1-ol has strong hydrogen bonds in addition to the weak london forces. Butan-1-ol het sterk waterstofbindings saam met swak Londonkragte.  Butane has weak London forces.  Butaan het swak Londonkragte  More energy needed to overcome/break the hydrogen bonds/ IMF in butan-1-ol than the London forces in butane.  Meer energie is nodig om die waterstofbindings te oorkom/te breek/IMK in butan-1-ol te breek/te oorkom as die Londonkragte in butaan. OR/OF  Butane has London forces/induced dipole- induced dipole forces/ dispersion forces Butaan het Londonkragte/geïnduseerde dipool-geïnduseerde dipool krag/dispersiekrag  Butan-1-ol has strong hydrogen bonds in addition to the weak london forces. Butan-1-ol het sterk waterstofbindings saam met swak Londonkragte.  Less energy needed to overcome/break the London forces/ IMF in butane than the hydrogen bonds in butan-1-ol.  Minder energie is nodig om die Londonkragte te oorkom/te breek/IMK in butaan te oorkom/te breek as die waterstofbindings in butan-1-ol. (3) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 5 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 4.1.1 Hydrogenation Hidrogenasie (1) 4.1.2 Platinum, nickel/nikkel or/of palladium/Pt, Ni ,Pd (1) 4.2 H H H Marking Criteria/Nasien kriteria: H H H Functional group on second C-2  H C C C H Funksionele groep op die tweede C O Whole structure correct  H H Totale struktuur korrek  H Propan-2-ol/2- propanol (3) 4.3 Secondary  Sekondêre The carbon atom bonded with -OH (hydroxyl) group is bonded to two other carbon atoms.  Die koolstofatoom gebind aan die -OH (hidroksiel) groep is gebind aan twee ander koolstofatome. (2) 4.4 4.4.1 Concentrated sulphuric acid/H2SO4/phosphoric acid/H3PO4/heat  Gekonsentreerde swawelsuur/H2SO4 /fosforsuur/H3PO4/hitte (1) 4.4.2 Prop-1-ene/Propene Prop-1-een/Propeen (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 6 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 4.5 H H H  H H  HH H  H C C C H + NaOH H C C C H + NaCℓ + H2O  H Cℓ H  H H Marking Criteria/Nasien kriteria: Reactants/Reaktante: 2-chloropropane  2-chloropropaan NaOH Products/Produkte: Double bond on prop-1-ene Dubbelbinding op prop-1-een Whole structure of prop-1-ene correct  Totale struktuur van prop-1-een korrek NaCℓ + H2O Any additional reactants and/or products/Addisionele reaktante en/produkte: Max. 4/5 Molecular/condensed formulae/ Molekulêre/gekondenseerde formules: Max. 4/5 Missing arrow/Sonder pyl: Max. 4/5 Reactants only /Slegs reaktante:Max. 2/5 Products only/Slegs produkte: Max. 3/5 (5) 4.6 4.6.1 Substitution/hydrolysis Substitusie/hidrolise (1) 4.6.2 Dilute strong base Verdunde sterk basis Mild heat  (2) Matige hitte [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 7 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 ANY ONE/ ENIGE EEN  Change in concentration of reactant/product per (unit) time. Verandering in konsentrasie van reaktant/produk per (eenheid) tyd.  Rate of change in concentration. Tempo van verandering in konsentrasie  Change in amount/number of moles/volume/mass of products or reactants per (unit) time Verandering in hoeveelheid/aantal mol/volume/massa van produkte of reaktante per (eenheid) tyd  Amount/number of moles/volume/mass of products formed or reactants used per (unit) time. Hoeveelheid/aantal mol/volume/massa van produkte gevorm of reaktante opgebruik per (eenheid) tyd. (2) 5.2 5.2.1 Surface area/state of division Oppervlak area/toestand van verdeling (1) 5.2.2 ANY ONE/ ENIGE EEN Concentration of the acid Konsentrasie van die suur Temperature of the acid Temperatuur van die suur Mass of the sodium carbonate Massa van die natriumkarbonaat (1) 5.3 Carbon dioxide/CO2  Koolstofdioksied/CO2 (1) 5.4 90 seconds  90 sekondes (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 8 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 5.5 Greater than  Groter as  In experiment 2, the sodium carbonate has a greater surface area which means more reaction sites.  In eksperiment 2 het die natriumkarbonaat 'n groter oppervlak wat meer plekke waar reaksies kan plaasvind beteken. OR/OF More particles with correct orientation/more contact points Meer deeltjies met die regte oriëntering/meer kontakpunte.  There will be more effective collisions per unit time in experiment 2 than in experiment 1.  Daar sal meer effektiewe botsings per tydseenheid wees in eksperiment 2 as in eksperiment 1. (3) 5.6 5.6.1 Mass of CO2 formed/Massa CO2 gevorm = 150 -145,60  = 4,4 g m n= M 4,4 n= 44  n = 0,1moles/mol n (CO2) : n (Na2CO3) = 1:1 m (Na2CO3) = n x M = (0,1)(106) = 10,6 g (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 9 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 5.6.2 150,00 Experiment 2 mass/g Eksperiment 2 145,60 time/s Marking Criteria/Nasien kriteria:  Graph for experiment 2 should have a steeper gradient than experiment 1.  Grafiek vir eksperiment 2 moet 'n steiler helling hê as eksperiment 1.  Time for reaction to come to completion must be shorter.  Tyd vir reaksie om te voltooi moet korter wees. (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 10 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 A stage in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction.  'n Tydstip tydens 'n omkeerbare reaksie waar die tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. OR/OF The stage in a chemical reaction when the concentration/amounts of reactants and products remain constant. Die tydstip in 'n chemiese reaksie wanneer die konsentrase/hoeveelheid van die reaktante en produkte onveranderd bly. (2) 6.2 CALCULATIONS USING NUMBER OF MOLES BEREKENINGE WAT AANTAL MOL GEBRUIK Mark allocation/Punte toekenning:  Change n(H2)  Verandering n(H2)  Ratio to determine n(CO) reacted and n(CH3OH) formed = 1 : 1  Verhouding om te bepaal n(CO) gereageer en n (CH3OH) gevorm=1:1  n(CO) & n(CH3OH) at equilibrium  n(CO) & n(CH3OH) tydens ewewig  Divide three equilibrium amounts by 2 (calculation of concentration)  Verdeel drie ewewighoeveelhede deur 2 (berekening van konsentrasie)  Kc expression KC uitdrukking  Substitution into Kc expression Vervanging in KC uitdrukking  Answer : X = 0,8 mol  Antwoord: X= 0,8 mol  Substitution into m = n x M Vervanging in m= n x M  Final answer m = 22,4 g Finale antwoord m = 22,4 g OPTION 1/OPSIE 1: CO (g) + 2 H2(g) → CH3OH(g) 1 mole/mol 2 moles/mol 1 mole/mol Start/begin (mol) X 1,5 0 Reacted (mol) 0,375 0,75  0,375 ratio Equilibrium (mol) X-0,375 0,75 0,375 Equilibrium X - 0,375 0,75 0,375 n  Concentration c = 2 2 2 V (mol∙dm ) -3 =0,375 =0,1875 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 11 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne [CH3 OH] Kc =  [H2 ] 2 [CO] 0,1875 6,27 =  2 X - 0,375 [0,375] [ ] 2 x = 0,8 mol m n= M m = (0,8)(28) m = 22,4 g  OPTION 2 (USING CONCENTRATIONS)/OPSIE 2 (GEBRUIK VAN KONSENTRASIES) CO (g) + 2 H2(g) → CH3OH(g) Ratio/Verhouding 1 2 1 Initial concentration/Aanvanklike X 1,5 konsentrasie = 0,5 X = 0,75 0  2 2 (mol∙dm-3) Change in 0,1875 concentration/Verandering in 0,1875 konsentrasie 0,375  ratio (mol∙dm-3) Equilibrium 0,5X - 0,75 Concentration/Ewewigskonsentrasie = 0,375 0,1875 0,1875 2 (mol∙dm-3) [CH3 OH] Kc =  [H2 ] 2 [CO] [0,1875] Kc = 2  [0,375] [0,5 X 0,1875] x = 0,8 mol Initial mass of CO injected in the container/Aanvanklike massa CO toegevoeg tot houer (m) = nxM m = (0,8)(28) m = 22,4g  (9) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 12 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 1 6.3 Kc( reverse reaction/terugwaartse reaksie) = Kc ( forward reaction) 1  Kc( reverse reaction/terugwaartse reaksie) = 6,27 Kc(reverse reaction/terugwaartse reaksie ) = 0,16 NOTE Give 2 marks if the learners write only the correct answer. NB Gee 2 punte as die leerders slegs die korrekte antwoord gegee het. (2) 6.4 6.4.1 Remains the same  Bly dieselfde The temperature is constant, Kc value is only affected by a change in temperature. Die temperatuur is konstant, KC se waarde word slegs beïnvloed deur ’n verandering in temperatuur. (2) 6.4.2 Decreases  Neem af. A larger volume (results in a lower pressure which) will favour the reaction producing a larger number of moles/reverse reaction is favoured. 'n Groter volume (het tot gevolg 'n verlaagde druk) sal die reaksie bevoordeel wat 'n groter aantal mol sal produseer/terugwaartse reaksie word bevoordeel. (2) 6.5 Decreases  Neem af A decrease in temperature favours the exothermic reaction. The reverse reaction will be favoured. 'n Afname in temperatuur bevoordeel die eksotermiese reaksie. Die terugwaartse reaksie word bevoordeel. (3) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 13 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 An acid is a proton donor.  'n Suur is 'n protonskenker. (2) 7.2 NaOH → Na+ +OH- 0,75 mol∙dm-3 (NaOH) gives/gee 0,75 mol∙dm-3 OH- ( Ratio/Verhouding 1:1)  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 pOH = - log [OH-]  [H+][OH-] =10-14 pOH = - log[0,75 ]  [H+][0,75] =10-14  pOH = 0,1249 [H+] = 1,3333x10-14 mol∙dm-3  pH + pOH = 14 pH = - log [H+]  pH + 0,1249 = 14 pH = - log [1,3333x10-14] pH = 13,88 pH = 13,88 (5) 7.3 ninitial/aanvanklik (HCℓ) n c=  V n initial/aanvanklik = cV = (0,5) (0,075) = 0,0375 moles For the reaction between HCℓ (excess) and NaOH/Vir die reaksie tussen HCℓ (oormaat) en NaOH n(NaOH) reacted/gereageer n = cV n = 0,75X 0,022 n = 0,0165 moles n(NaOH) : n(HCℓ) = 1:1 nexcess/oormaat (HCℓ) = 0,0165 moles(ratio/verhouding) nreacted/gereageer (HCℓ) with/met CaCO3 n= 0,0375-0,0165 n=0,021moles/mol 1 nreacted/gereageer(CaCO3) = × 0,021 = 0,0105 moles/mol 2 m=nxM m = (0,0105)(100) m =1,05 g 1,05 % CaCO3 = × 100  5 (8) = 21 % [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 14 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Salt bridge. Soutbrug. (1) 8.2 Standard hydrogen electrode. (1) Standaard waterstofelektrode. 8.3 Provides a surface for the transfer of electrons/Allows conduction. (1) Voorsien 'n oppervlakte vir die oordrag van elektrone/Laat geleiding toe. 8.4 Mg (s) | Mg2+ (aq)  || I2 (g) | I- (aq) |Pt (3) 8.5 Eθcell  Eθcathode  Eθanode  E0sel = E0katode- E0anode Ecell/sel = 0,54- (-2,36) Ecel/sel = 2,90 V Ecell is less than 3 V so the bulb will not glow to its maximum brightness.  Esel is minder as 3 V en die gloeilamp sal nie gloei teen sy maksimum helderheid nie. (5) 8.6 The bulb will switch off/it will no longer glow at all. Die gloeilamp sal afskakel/dit sal nie langer brand nie. The cell potential will be 0 V. Die selpotensiaal sal 0 V wees. (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 15 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Electrical energy to chemical energy. (1) Elektriese energie na chemiese energie. 9.2. 2O2-(ℓ)→O2 (g) + 4e- OR/OF 2O2- (ℓ )+ C(s)→O2(g) + 4e (2) 9.3 They undergo a combustion reaction and get used up. Hulle ondergaan 'n verbrandingsreaksie en word opgebruik. ACCEPT/AANVAAR: They get burnt. ( since they react with oxygen at a high temperature) Hulle verbrand. (aangesien hulle met suurstof reageer by hoë temperatuur) (2) 9.4 To reduce the working temperature  and hence save on extraction cost. Om die werkende temperatuur te verlaag en om sodoende ekstraksiekoste te bespaar. (1) 9.5 ANY TWO/ENIGE TWEE (Consider any other reasonable answers from learners/Oorweeg enige ander redelike antwoord van die leerders.)  Excessive CO2 (which is a green house gas) is formed during the extraction process and this causes global warming. Oormaat CO2 (wat 'n kweekhuisgas is) word gevorm gedurende die ekstraksieproses en dit veroorsaak Aardverwarming.  The CO2 formed can also react with water when it rains and this causes acid rain which corrodes buildings and damages plants.  Die gevormde CO2 kan ook met water reageer wanneer dit reën en dit veroorsaak suurreën wat geboue verweer en plante beskadig.  The electricity used in the extraction process is generated from the burning of coal which pollutes the environment and depletes a non renewable energy source. Die elektrisiteit wat gebruik word gedurende die ekstraksieproses word opgewek deur die verbranding van steenkool wat besoedeling van die omgewing veroorsaak en 'n nie-hernubare energiebron uitput.  The plant requires a very large landscape so people might be moved from their places of origin OR massive deforestation can occur in order to accommodate the plant. Die aanleg benodig 'n baie groot gebied en mense word sodoende verskuif van hulle oorspronklike blyplekke OF grootskaalse ontbossing kan gebeur om die aanleg te akkomodeer.  Red mud used in the extraction process can contaminate underground water reserves. Rooi modder wat gebruik word gedurende die ekstraksie kan die (2) ondergrondse waterreserwes besoedel. [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 16 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 Nitrogen (gas)  Stikstof (gas) (1) 10.1.2 Haber process Haberproses (1) 10.1.3 Ammonium sulphate/(NH4)2SO4  Ammoniumsulfaat/(NH4)2SO4 (1) 10.1.4 Nitrogen dioxide/NO2 Stikstofdioksied/NO2 (1) 10.2 3NO2 + H2O  → 2HNO3 + NO  (balancing/balansering) (3) 10.3 10.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Using NPK ratios: P:K = 2:1 Using NPK ratios Gebruik NPK verhoudings Gebruik NPK verhoudings 2 NP: K = 9:1 m (P) = × 1,2 9 1 m ( NP)= × 1,2  m(P) =2,4 kg 1 Any One m ( NP) =10,8 kg 7 m(N) = × 1,2 m (NPK) =10,8 + 1,2 1 m (fertiliser/kunsmis) = 12 kg m (N) =8,4 kg m (fertiliser/kunsmis) = 1,2 + 2,4 12 % fertiliser/kunsmis = ×100  +8,4 50 = 12 kg 12 % fertiliser/kunsmis = 24% % fertiliser/kunsmis = ×100  50 X = 24 % fertiliser = 24%: X = 24 OPTION 3/OPSIE 3 If 100% fertiliser/Indien 100% kunsmis 1 m (K)= × 50  10 m (K) =5 kg 1,2 m (K)= × 100  5 X = 24% (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Grade 12 17 September 2019 Fisiese Wetenskappe V2/Graad 12 Marking Guidelines/Nasienriglyne 10.3.2 No. Nee The chosen fertilizer contains too much nitrogen. Die gekose kunsmis bevat te veel stikstof. This will make the leaves of the maize plant grow big and green and not promote the actual growth of the maize crop. Dit sal veroorsaak dat die blare van die mielieplant groot en groen groei en sal nie die eintlike groei van die mielie-oes bevoordeel nie. (3) 10.4 Use organic fertilisers like manure in place of inorganic fertilisers.  Gebruik organiese misstowwe soos beesmis in plaas van kunsmis. Avoid putting too much or excessive inorganic fertilisers.  Moet nie te veel anorganiese kunsmisstowwe gebruik nie. (2) [16] GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief