Physical Sciences Grade 12 Paper 2 NSC November 2015 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) NOVEMBER 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 17 pages. Hierdie memorandum bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 2 DBE/November 2015 NSC/NSS – Memorandum QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 D  (2) 1.3 A  (2) 1.4 A  (2) 1.5 B  (2) 1.6 C  (2) 1.7 B  (2) 1.8 D  (2) 1.9 B  (2) 1.10 C  (2) [20] QUESTION 2/VRAAG 2 2.1 2.1.1 B (1) 2.1.2 O C H  (1) 2.1.3 CnH2n-2  (1) 2.1.4 4-ethyl-5-methylhept-2-yne / 4-ethyl-5-methyl-2-heptyne 4-etiel-5-metielhept-2-yn / 4-etiel-5-metiel-2-heptyn Marking criteria/Nasienriglyne:  4-ethyl / 4-etiel  OR/OF 4 ethyl / 4 etiel  5-methyl / 5-metiel  OR/OF 5 methyl / 5 metiel  hept-2-yne / 2-heptyne / hept-2-yn / 2-heptyn  OR/OF hept 2 yne / 2 heptyne / hept 2 yn / 2 heptyn IF/INDIEN: Any error e.g. hyphens omitted and/or incorrect sequence: Enige fout bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks. 2 3 (3) 2.1.5 Butan-2-one / 2-butanone / Butanone Butan-2-oon / 2-butanoon / Butanoon Marking criteria/Nasienriglyne:  Functional group / Funksionele groep   Whole name correct / Hele naam korrek  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 3 DBE/November 2015 NSC/NSS – Memorandum 2.2 2.2.1 Alkanes / Alkane  (1)   Notes/Aantekeninge: 2.2.2 2-methylpropane 2-metielpropaan IF/INDIEN: 2 methylpropane / 2 metielpropaan  1 2 OR/OF   IF sequence incorrect/INDIEN volgorde Methylpropane Metielpropaan verkeerd: Max./Maks. 1 2 Marking criteria structural formula: H Nasienriglyne struktuurformule: H C H  Three carbons in longest chain.  Drie koolstowwe in die langste ketting. H H  Methyl group on second carbon. H C C C H  Metielgroep op tweede koolstof.  Notes/Aantekeninge: H H H  One or more H atoms omitted: Een of meer H-atome uitgelaat: 1 2  Condensed or semi-structural formula: Gekondenseerde of semi-struktuur- formule: 1 2 (4) 2.2.3 Chain / Ketting  (1) 2.3 2.3.1 Haloalkanes / Alkyl halides  Haloalkane / Alkielhaliede (1) 2.3.2 Substitution / halogenation / bromonation  Substitusie / halogenering / halogenasie / bromonering (1) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 4 DBE/November 2015 NSC/NSS – Memorandum QUESTION 3/VRAAG 3 3.1 3.1.1 Esterification / Condensation  Esterifikasie / Verestering / Kondensasie (1) 3.1.2 O C O C  (1) 3.1.3 Propanoic acid / Propanoësuur  (1) 3.1.4 Dehydration / elimination  Dehidrasie / dehidratering / eliminasie (1) 3.1.5 (Concentrated) H2SO4 / sulphuric acid / H3PO4 / phosphoric acid  (Gekonsentreerde) H2SO4 / / swaelsuur / swawelsuur / H3PO4 / fosforsuur (1) 3.1.6 HH Notes/Aantekeninge  Functional group:  C C  Funksionele groep:  Whole structure correct:  H H Hele struktuur korrek: (2) 3.2 3.2.1 H H Notes/Aantekeninge  Functional group:  C C  Funksionele groep:  Whole structure correct:  H H Hele struktuur korrek: (2) 3.2.2 Addition / Addisie  (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 5 DBE/November 2015 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 A bond/an atom or a group of atoms  that determine(s) the (physical and chemical) properties of a group of organic compounds.  'n Binding of 'n atoom of 'n groep atome wat die (fisiese en chemiese) eienskappe van 'n groep organiese verbindings bepaal. (2) 4.2 4.2.1 D / ethanoic acid / etanoësuur  - Lowest vapour pressure.  Laagste dampdruk. (2) 4.2.2 A / butane / butaan  (1) 4.3  Between molecules of A / butane / alkanes are London / induced dipole / dispersion forces.  Tussen molekule van A / butaan / alkane is London / geïnduseerde dipole / dispersiekragte.  Between molecules of B / propan-2-one / ketones are dipole-dipole forces  in addition to London / induced dipole / dispersion forces. Tussen molekule van B / propan-2-oon / ketone is dipool-dipool-kragte tesame met London / geïnduseerde dipool /dispersiekragte.  Intermolecular forces in A are weaker than those in B. / Less energy is needed in A to break/overcome intermolecular forces.  Intermolekulêre kragte in A is swakker as die in B./ Minder energie word by A benodig om intermolekulêre kragte te breek/oorkom. - OR/OF Intermolecular forces in B are stronger than those in A. / More energy is needed in B to break/overcome intermolecular forces. Intermolekulêre kragte in B is sterker as die in A. / Meer energie word by B benodig om intermolekulêre kragte te breek/oorkom. OR/OF  Between molecules of A / butane / alkanes are weak London / induced dipole / dispersion forces. Tussen molekule van A / butaan/alkane is swak London / geïnduseerde dipool / dispersiekragte.  Between molecules of B /propan-2-one / ketone are strong(er) dipole- dipole forces in addition to London/induced dipole / dispersion forces. Tussen molekule van B / propan-2-oon / ketone is sterk(er) dipool- dipool/dispersiekragte. (3) 4.4 London forces/dispersion forces/induced dipole forces/dipole-dipole forces.  Londonkragte/dispersiekragte/geïnduseerde dipoolkragte/dipool-dipoolkragte. OR/OF A and B do not have hydrogen bonding./C and D have hydrogen bonding. A en B het nie waterstofbinding nie./C en D het waterstofbinding. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 6 DBE/November 2015 NSC/NSS – Memorandum 4.5 OPTION 1/OPSIE 1  D has more sites for hydrogen bonding than C / forms dimers / is more polar than C.  D het meer punte vir waterstofbinding as C / vorm dimere / is meer polêr as C.  D has stronger / more intermolecular forces / dipole-dipole forces.  D het sterker / meer intermolekulêre kragte / dipool-dipoolkragte. OR/OF D needs more energy to overcome/break the intermolecular forces. D het meer energie nodig om die intermolekulêre kragte te oorkom/breek. OPTION 2/OPSIE 2  C has less sites for hydrogen bonding than D. / C does not form dimers / C is less polar. C het minder plekke vir waterstofbinding as D. / C vorm nie dimere nie / C is minder polêr.  C has weaker / less intermolecular forces / dipole-dipole forces./ C needs less energy to overcome/break intermolecular forces / dipole-dipole forces. C het swakker / minder intermolekulêre kragte / dipool-dipoolkragte./ C benodig minder energie om intermolekulêre kragte / dipool-dipoolkragte te oorkom/breek. (2) 4.6 Marking criteria/Nasienriglyne  Mole ratio for V(CO2) correctly used. / Molverhouding vir V(CO2) korrek gebruik.  Mole ratio for V(H2O) correctly used. / Molverhouding vir V(H2O) korrek gebruik.  Mole ratio for V(O2 reacted) correctly used. / Molverhouding vir V(O2 reageer) korrek gebruik.  V(O2 excess/oormaat) = V(O2 initial/aanvanklik) – V(O2 change/verandering).  Vtot = 80 cm3 OPTION 1/OPSIE 1 V(O2 reacted/reageer): V(CO2) = 4V(C4H10) V(H2O) = 5V(C4H10) 13 = (4)(8)  = (5)(8)  V(O2) = V(C4H10) 2 = 32 cm3 = 40 cm3 13 = ( )(8)  = 52 cm3 2 V(O2 excess/oormaat): V(O2) = 60 – 52 = 8 cm3 Vtot = 32 + 40 + 8 = 80 cm3  OPTION 2/OPSIE 2 C4H10 O2 CO2 H2O 3 Initial V (cm ) 8 60 0 0 BeginV (cm3) Change in V (cm3) 8 52  32  40  Verandering V (cm3) Final V (cm3) 0 8 32 40 Finale V (cm3) Total/totale volume = 8 + 32 + 40 = 80 cm3  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 7 DBE/November 2015 NSC/NSS – Memorandum OPTION 3/OPSIE 3 C4H10 O2 CO2 H2O Initial V (dm3) 0,008 0,06 0 0 Begin V (dm3) Change in V (dm3) 0,008 0,052  0,032  0,04  Verandering V (dm3) Final V (dm3) 0 0,008  0,032 0,04 Finale V (dm3) Total/totale volume = 0,008 + 0,032 + 0,04 = 0,08 dm3  (5) [16] QUESTION 5/VRAAG 5 5.1 Time/Tyd: (Stop) watch / (Stop)horlosie  Volume: (Gas) syringe / Burette / Measuring cylinder / (Chemical) balance / Erlenmeyer flask / Graduated flask  (Gas)spuit / Buret / Maatsilinder / (Chemiese) balans / Erlenmeyer fles / Gegradueerde fles Notes/Aantekeninge  Only one mark per type of apparatus. / Slegs een punt per tipe apparaat. (2) 5.2 5.2.1 t1  (1) 5.2.2 t3  (1) 5.3 Between t1 and t2  Tussen t1 en t2 (1) 5.4 Volume (cm3) Exp. 2  Exp. 1   Exp. 3  Time (s) Marking criteria/Nasienriglyne Exp. Initial gradient higher than that of Exp.1. 2 Aanvanklike gradient groter as die van Eksp 1.  Curve reaches same constant volume as for Exp. 1 (but earlier). Kurwe bereik dieselfde konstante volume as in Eksp 1 (maar gouer).  Exp. Initial gradient lower than that of Exp.1. 3 Aanvanklike gradient kleiner as die van Eksp. 1.  Curve reaches a smaller constant volume as for Exp. 1 (at a later stage).  Kurwe bereik (later) 'n kleiner konstante volume as vir Eksp. 1. (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 8 DBE/November 2015 NSC/NSS – Memorandum 5.5.1 Marking criteria/Nasienriglyne  n(HCℓ) = (0,1)(100 x 10-3)  Use mole ratio/Gebruik molverhouding: n(Zn) = ½n(HCℓ) m  Substitute 65 into/ Vervang 65 in n = . M  n(Znfinal/finaal) = n(Zninitial/aanvanklik) – n(Znused/gebruik) m(Znfinal/finaal) = m(Zninitial/aanvanklik) – m(Znused/gebruik)  Final answer/Finale antwoord: Range/gebied: 0,33 g – 0,48 g OPTION/OPSIE 1 OPTION/OPSIE 2 n(HCℓ) = cV n(HCℓ) = cV = (0,1)(100 x 10-3)  = (0,1)(100 x 10-3)  = 0,01 mol = 0,01 mol n(Zn reacted/gereageer) = ½n(HCℓ) n(Zn reacted/gereageer): = ½(0,01)  n(Zn) = ½n(HCℓ) = 5 x 10-3 mol = ½(0,01)  m n(Zn)i = = 5 x 10-3 mol M 0,8 =  65 n(Zn reacted/gereageer): = 1,23 x 10-2 mol m(Zn) = (5 x 10-3)(65) = 0,325 g n(Zn)f = 1,23 x 10-2 – 5 x 10-3  = 7,3 x 10-3 mol m(Znf) = 0,8 – 0,325  = 0,48 g  (0,475 g) m(Zn) = nM = (7,3 x 10-3)(65) = 0,47 g  (5) 5.5.2 Smaller than / Kleiner as  (1) [15] QUESTION 6/VRAAG 6 6.1 Equal to / Gelyk aan  (1) 6.2 [X3 ]2 No KC expression, correct substitution /Geen Kc- uitdrukking, Kc = 3  korrekte substitusie: Max./Maks. 3 [X2 ] 4 (0,226 )2  Wrong KC expression /Verkeerde Kc-uitdrukking Max./Maks. 0 = 4 (0,06 )3  If one or more exponents are omitted in substitution step but = 236,46  correct answer obtained: Max 3 4 Indien een of meer eksponente uitgelaat by substitusie stap, maar korrekte antwoord verkry: Maks 3 (4) 4 6.3 (1) 6.3.1 Increases / Vermeerder  - 6.3.2  The increase in [X3] is opposed. / Change is opposed.  Die verhoging in [X3] word teengewerk. / Verandering word teenwerk.  The reverse reaction is favoured. / X3 is used / [X3] decreases.  Die terugwaartse reaksie word bevoordeel./X3 word gebruik / [X3] neem af. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 9 DBE/November 2015 NSC/NSS – Memorandum 6.4 Higher than / Hoër as  (1) 6.5 Exothermic / Eksotermies  -  The concentration of the product/X3(g) is lower / the concentration of the reactant / X2(g) is higher.  Die konsentrasie van die produkte/X3(g) is laer / die konsentrasie van die reaktans / X2(g) is hoër.  The increase in temperature favoured the reverse reaction.  Die toename in temperatuur het die terugwaartse reaksie bevoordeel.  According to Le Chatelier's principle an increase in temperature favours the endothermic reaction.  Volgens Le Chatelier se beginsel bevoordeel 'n toename in temperatuur die endotermiese reaksie. OR/OF Exothermic / Eksotermies  -  [X3] decreases and [X2] increases. / [X3] neem af en [X2] neem toe.   Kc decreases if temperature increases./Kc neem af as die temperatuur afneem.   Decrease in temperature favoured the forward reaction. / Verlaging in temperatuur het die voorwaartse reaksie bevoordeel.  (4) 6.6 Number of particles / 300 °C Aantal deeltjies 400 °C Kinetic energy /Kinetiese energie Marking criteria/Nasienriglyne Peak of curve at 400 °C lower than at 300 °C and shifted to the right. Piek van kurwe by 400 °C laer as by 300 °C en skuif na regs.  Curve at 400 °C has larger area at the higher Ek. Kurwe by 400 °C het groter oppervlak by hoë Ek.  (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 10 DBE/November 2015 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 7.1.1 Hydrolysis / Hidrolise  (1) 7.1.2 Acidic / Suur  - Forms H3O+ ions during hydrolysis./Vorm H3O+ ione gedurende hidrolise.  OR/OF Salt of strong acid and weak base./Sout van sterk suur en swak basis. OR/OF ( NH4 ) acts as proton donor. / ( NH4 ) tree op as ‘n protonskenker. (2) 7.2 7.2.1 n = cV  = (0,1)(0,1)  = 0,01 mol  (3) 7.2.2 POSITIVE MARKING FROM QUESTION 7.2.1. POSITIEWE NASIEN VAN VRAAG 7.2.1. Marking criteria/Nasienriglyne  Substitute volume and concentration to calculate n(HCℓ)  Vervang volume en konsentrasie om n(HCℓ) te bereken.  Use mole ratio/Gebruik molverhouding: n(NaOH) = n(HCℓ) = 1:1   n(NaOH) x 4 OR/OF V(HCℓ) x 4 OR/OF n(HCℓ) x 4   Subtraction/Aftrekking: n(NaOHinitial/aanvanklik) - n(NaOHexcess/oormaat)   Use mole ratio/Gebruik molverhouding: n(NaOH) = n(NH4Cℓ) = 1:1  m  Substitute/Vervang 53,5 g·mol-1 in n = . M  Percentage calculation/Persentasieberekening   Final answer/Finale antwoord: 0,11 g - 0,21 g  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 11 DBE/November 2015 NSC/NSS – Memorandum OPTION 1/OPSIE 1 n(HCℓ) = caVa = (0,11)(14,55 x 10-3)  = 1,6 x 10-3 mol n(NaOH) = n(HCℓ) = 1,6 x 10-3 mol  n(NaOH excess/oormaat) in 100 cm3 = 1,6 x 10-3 x 4  = 6,4 x 10-3 mol n(NaOH reacted/gereageer) = 0,01 - 6,4 x 10-3  = 3,6 x 10-3 mol n(NH4Cℓ) = n(NaOH) = 3,6 x 10-3 mol  (0,003598 mol) m(NH4Cℓ) = nM x m  n(NH4Cℓ) = = (3,6 x 10-3)(53,5) n(NH4Cℓ) = 0,92 53,5  53,5  = 0,193 g m  3,6 x 10-3 = x 53,5 92% : 0,193 g  3,6 x 10-3 = 0,92 100% : x 53,5 n(NH4Cℓ) = 0,192 g 0,193  100  x = 0,21 g  m(fertiliser/kunsmis): x =  0,192  100 92 m=  = 0,21 g  92 = 0,21 g  OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V(HCℓ) to neutralise 100 cm3 NaOH: n(HCℓ) to neutralise 100 cm3 NaOH: V(HCℓ) neutraliseer 100 cm3 NaOH: n(HCℓ) neutraliseer 100 cm3 NaOH: V(HCℓ) = 14,55 x 4  n(HCℓ) = cV = 58,2 cm3 = (0,11)(0,01455 x 4)  = 0,006402 mol (6,4 x 10-3 mol) n(HCℓ) = cV = (0,11)(0,0582)  n(NaOH excess/oormaat): = 0,006402 mol n(NaOH) = n(HCℓ) = 6,4 x 10-3 mol  n(NaOH) = n(HCℓ) = 0,006402 mol  n(NaOH reacted/gereageer): n(NaOH) = 0,01 - 0,006402  n(NaOH reacted/gereageer): = 0,003598 mol n(NaOH) = 0,01 - 0,006402  = 0,003598 mol n(NH4Cℓ) = n(NaOH) = 0,003598 mol  n(NH4Cℓ) = n(NaOH) = 0,003598 mol  m(NH4Cℓ) = nM m(NH4Cℓ) = nM = (0,003598)(53,5)  = (0,003598)(53,5)  = 0,192 g = 0,192 g 92% : 0,192 g 92% : 0,192 g 100 100% : 0,192 x  = 0,21 g  0,192  100 92 100% : = 0,21 g  92 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 12 DBE/November 2015 NSC/NSS – Memorandum OPTION 4/OPSIE 4 OPTION 5/OPSIE 5   c a Va na 0,11 x 14,55 1 c a Va na 0,11 x 14,55 1         c b Vb nb c b x 25 1 c b Vb nb c b x 25 1 cb = 0,064 mol·dm-3  cb = 0,064 mol·dm-3 n(NaOH in excess in 100 cm3): Δc(NaOH) = 0,1 - 0,064  n(NaOH in oormaat in 100 cm3): = 0,036 mol·dm-3 n(NaOH) = cV = (0,064)(0,1)  n(NaOH reacted/gereageer): n(NaOH) = cV = 6,4 x 10-3 mol = 0,036 x 0,1 n(NaOH reacted/gereageer): = 0,0036 mol n(NaOH) = 0,01 - 0,006402  = 0,003598 mol n(NH4Cℓ) = n(NaOH) = 0,0036 mol  m n(NH4Cℓ) = n(NaOH) n M = 0,003598 mol  92 x m(NH4Cℓ) = nM  0,0036 = 100  = (0,003598)(53,5)  53,5 = 0,192 g 0,0036(53,5) = 0,92x x = 0,21 g  92% : 0,192 g 100 100% : 0,192 x  = 0,21 g  92 (8) 7.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 [OH-] = [NaOH] = 0,5 mol∙dm-3 pOH = -log[OH-]  = -log(0,5)  Kw = [H3O+][OH-] = 0,301 1 x 10-14 = [H3O+]0,5   [H3O+] = 2 x 10-14 mol∙dm-3 pH + pOH = 14 pH = 14 – 0,301  pH = -log[H+]  = 13,7  (13,699) = -log(2 x 10-14)  = 13,7  Notes/Aantekeninge IF/INDIEN: Wrong formula/Verkeerde formule: pH = -log[OH−]; pOH = -log[NaOH] No marks for substitution and answer./Geen punte vir vervanging en antwoord. (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 13 DBE/November 2015 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1 Temperature/Temperatuur: 25 °C / 298 K  Pressure/Druk: 101,3 kPa / 1,013 x 105 Pa / 1 atm / 100 kPa  Concentration/Konsentrasie: 1 mol∙dm-3  (3) 8.2 8.2.1 Cd(s) / Cadmium / Kadmium / Cd∣Cd2+ / Cd2+∣Cd  Notes/Aantekeninge Ignore phases. / Ignoreer fases. (1)    8.2.2 Ecell  Ecathode  Eanode  Notes/Aantekeninge    Accept any other correct formula from the data 0,13  Ecathode  ( 0,40) sheet. / Aanvaar enige ander korrekte formule  vanaf gegewensblad. Ecathode  0,13  0,40  Any other formula using unconventional   0,27 ( V )  abbreviations, e.g. E°cell = E°OA - E°RA followed by correct substitutions: / Enige ander Q is Ni/nickel/nikkel  formule wat onkonvensionele afkortings gebruik bv. E°sel = E°OM - E°RM gevolg deur korrekte vervangings: 4 5 (5) 8.3 8.3.1 Cd(s) → Cd2+(aq) + 2e-  Ignore phases. / Ignoreer fases. Notes/Aantekeninge Cd2+ + 2e- ← Cd (2 ) Cd ⇌ Cd2+ + 2e- (1 ) 2 2 2+ Cd ← Cd + 2e - ( 0 ) Cd2+ + 2e- ⇌ Cd ( 0 ) 2 2 (2) 8.3.2 Pt/Platinum  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 14 DBE/November 2015 NSC/NSS – Memorandum 8.4 OPTION 1/OPSIE 1 Compare/Vergelyk Q2+ is reduced / Cd is oxidised and therefore Q2+ is a Q2+ & Cd2+ stronger oxidising agent than Cd2+. Q2+ word gereduseer / Cd word geoksideer, en dus is Q2+  'n sterker oksideermiddel as Cd2+. Compare/Vergelyk R2 is reduced / Cd is oxidised and therefore R2 is a R2 & Cd2+ stronger oxidising agent than Cd2+.  R2 word gereduseer / Cd word geoksideer, dus is R2 'n  sterker oksideermiddel as Cd2+. Compare/Vergelyk The cell potential of combination II is higher than that of R2 & Q2+ combination I, therefore R2 is a stronger oxidising agent than Q2+.  Die selpotensiaal van kombinasie II is hoër as dié van kombinasie I en dus is R2 'n sterker oksideermiddel as Q2+. Final answer/ Cd2+; Q2+;R2 OR/OF Cd2+ ; Ni2+ ; Cℓ2  Finale antwoord OPTION 2/OPSIE 2  The reduction potential of Cℓ-∣Cℓ2 = 1,36 V  because the cell potential of combination II is 1,76 V and the reduction potential of Cd∣Cd2+ is 0,4 V. Die reduksiepotensiaal van Cℓ-∣Cℓ2 = 1,36 V omdat die selpotensiaal van kombinasie II 1,76 V is en die reduksiepotensiaal van Cd∣Cd2+ 0,4 V is. OR/OF R2 is Cℓ2 because the cell potential of combination II is 1,76 V and the reduction potential of Cd∣Cd2+ is 0,4 V./ R2 is Cℓ2 omdat die selpotensiaal van kombinasie II 1,76 V is en die reduksiepotensiaal van Cd∣Cd2+ 0,4 V is.  Cd∣Cd2+ has the lowest reduction potential (-0,4 V) and therefore Cd2+ is the weakest oxidising agent. / Cd∣Cd2+ het die laagste reduksiepotensiaal (0,4 V) en dus is Cd2+ die swakste oksideermiddel.   Cℓ-∣Cℓ2 has the highest reduction potential and therefore Cℓ2 is the strongest oxidising agent. / Cℓ-∣Cℓ2 het die hoogste reduksiepotensiaal en dus is Cℓ2 die sterkste oksideermiddel.   Final answer/Finale antwoord: Cd2+; Q2+;R2  OR/OF Cd2+ ; Ni2+ ; Cℓ2 (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 15 DBE/November 2015 NSC/NSS – Memorandum QUESTION 9/VRAAG 9 9.1 ANY ONE/ENIGE EEN:  The chemical process in which electrical energy is converted to chemical energy.  Die chemiese proses waarin elektriese energie omgeskakel word na chemiese energie.  The use of electrical energy to produce a chemical change. Die gebruik van elektriese energie om 'n chemiese verandering te weeg te bring.  Decomposition of an ionic compound by means of electrical energy. Ontbinding van 'n ioniese verbinding met behulp van elektriese energie.  The process during which and electric current passes through a solution/ionic liquid/molten ionic compound. Die proses waardeur 'n elektriese stroom deur 'n oplossing/ioniese vloeistof/gesmelte ioniese verbinding beweeg. (2) 9.2 ANY ONE/ENIGE EEN:  To keep the polarity of the electrodes the same.  Om die polariteit van die elektrodes dieselfde te hou.  To prevent the anode and cathode from swopping. Om te verhoed dat die anode en katode omruil.  DC provides a one way flow of electrons ensuring that the same chemical reaction occurs all the time at the electrodes. GS verskaf 'n eenrigting vloei van elektrone en verseker dat dieselfde chemiese reaksie altyd by die elektrodes plaasvind.  If you use AC the polarity of the electrodes will keep changing. Wanneer jy WS gebruik word hou die polariteit van die elektrodes aan om te verander.  Pure copper deposited on only one electrode. Suiwer koper slaan slegs op een elektrode neer. (1) 9.3 Cu2+ (aq) + 2e-  Cu (s)  Ignore phases. / Ignoreer fases. Notes/Aantekeninge Cu2+ + 2e- ⇌ Cu ( 12 ) Cu ← Cu2+ + 2e- (22 ) Cu ⇌ Cu2+ + 2e- (02) Cu2+ + 2e- ← Cu (02) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 16 DBE/November 2015 NSC/NSS – Memorandum 9.4  Cu2+ is a stronger oxidising agent  than Zn2+.  Cu2+ is 'n sterker oksideermiddel as Zn2+.  Cu2+ will be reduced to Cu. / Cu2+ sal gereduseer word na Cu.  OR/OF  Zn is a stronger reducing agent than Cu. Zn is 'n sterker reduseermiddel as Cu.  Cu2+ will be reduced to Cu. / Cu2+ sal gereduseer word na Cu. OR/OF  The standard reduction potential of Cu2+|Cu is higher than that of Zn2+|Zn. Die standaard reduksie potensiaal van Cu2+|Cu is hoër as die van Zn2+|Zn.  Cu2+ will be reduced to Cu. / Cu2+ sal gereduseer word na Cu. OR/OF  The standard reduction potential of Zn2+|Zn is lower than that of Cu2+|Cu. Die standaard reduksie potensiaal van Zn2+|Zn is laer as die van Cu2+|Cu.  Cu2+ will be reduced to Cu. / Cu2+ sal gereduseer word na Cu. (3) 9.5 m Marking guidelines/Nasienriglyne n m M  Substitute 63,5 and 2,85 x 10-2  in n   m M 2,85 x 10 2  m 63,5  Vervang 63,5 en 2,85 x 10-2 in n  m  1,81 g M  Percentage purity.  Persentasie suiwerheid. 1,81 % purity   100  2  Final answer/Finale antwoord:  90,49 %  90,49%  (Accept/Aanvaar: 90,5%) (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om assbelief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 17 DBE/November 2015 NSC/NSS – Memorandum QUESTION 10/VRAAG 10 10.1 10.1.1 Haber (process) / Haber(proses)  (1) 10.1.2 N2 + 3H2 ⇌ 2NH3  bal  Notes/Aantekeninge  Reactants  Products  Balancing  Reaktanse  Produkte  Balansering   Ignore/Ignoreer → and phases / en fases  Marking rule 6.3.10/Nasienreël 6.3.10 (3) 10.1.3 Air / Lug  (1) 10.2 10.2.1 40%  (1) 10.2.2  High yield / percentage  Hoë opbrengs / persentasie  High rate due to higher concentration.  Hoë tempo weens hoër konsentrasie. (2) 10.2.3 Low reaction rate / Lae reaksietempo  (1) 10.3 Marking guidelines/Nasienriglyne 28  x 50  17,5 kg  80 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 28 28 % N in NH4NO3 =   100 m(N in NH4NO3) =  x 50  80 80 = 35% = 17,5 kg  m(N) in 50 kg: 35  50  = 17,5 kg  100 (3) TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou