Physical Sciences Grade 12 Limpopo November 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES P2: CHEMISTRY FISIESE WETENSKAPPE VR 2: CHEMIE SEPTEMBER 2016 MEMORANDUM MARKS: 150

Downloaded from hlayiso.com 2 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION?VRAAG 1 1.1 A (2) 1.2 D (2) 1.3 B (2) 1.4 C (2) 1.5 D (2) 1.6 C  (2) 1.7 A (2) 1.8 A (2) 1.9 C (2) 1.10 B (2) [20] QUESTION/VRAAG 2 2.1 2.1.1 Esters  (1) 2.1.2 Carboxyl group/karboksiel groep (1) 2.1.3 A (1) 2.1.4 5-bromo-2,4-dimethylheptane // (3) 5-broom–2,4–dimetielheptaan of / 5-bromo–2,4– dimetielheptaan 2.1.5 C n H 2n  (1) 2.1.6 H Notes H C O H  ∙ Functional group  H ∙ Whole structure correct  (2) 2.1.7 propanoic acid  / propanoësuur (1) 2.2 2.2.1 Polyethene/ Polythene/ polieteen/ politeen/ polietileen Accept: Polyethylene (1) 2.2.2 Addition Polymerisation  addisie polimerisasie (1) 2.2.3 (a) CH 2  (1) (b) CH 2  (1) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com 3 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION/VRAAG 3 3.1 A series of organic compounds that can be described by the same general formula.  / ‘n reeks organiese verbindings wat deur ‘n algemene formule beskryf kan word OR: A series of organic compounds in which one member differs from the next with a CH 2 group./ ‘n reeks organiese verbindings waar een van die ander met CH 2 groep/eenheid verskil NOTES: 2 or 0 3.2 Ketones / ketone (1) 3.3 Butanoic acid/ butanoöesuur (1) 3.4.1 Boiling point/ kookpunt (1) 3.4.2 Molecular mass / molekulêre massa (1) 3.5 The boiling point of carboxylic acids > b.p of alcohols > b.p of aldehydes (1) Kookpunt van karboksielsure > k.pt alkohole > k.pt van aldehiede 3.6 ∙ Both compounds A and B are hydrogen bonded / beide A en B het waterstofbinding IM kragte ∙ Compound A has two (more) sites for hydrogen bonding /A forms dimers/A is more polar/ B has less/one sites for hydrogen bonding // A het twee plekke vir waterstofbinding (B het slegs 1 plek vir H-bd) ∙ Therefore A has stronger intermolecular forces than B/B has weaker Intermolecular forces than A  / A het sterke IM kragte (3) 3.7 The stronger/weaker the intermolecular forces, the lower/higher the vapour pressure. // hoe sterker die IM kragte hoe swakker is die dampdruk / h (1) swakker die IM kragte hoe hoer is die dampdruk [11] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION/VRAAG 4 4.1.1 Addition/hydrobromination / Hydrohalogeantion / Addisie/ hidrohalogenasie aanvaar: hidrobromasie (1) 4.1.2 H H H H H H H H H- C C C C H + HBr H C C C C H     H H H H Br H (4) Notes: Reactants Product Marking Rule 6.3.10. 4.1.3 2-bromobutane / 2-broombutaan/2-bromobutaan Notes ∙ 2-bromo  ∙ butane  (2) 4.2.1 Substituition  /Hydrolysis / substitusie/ hidroliese (1) 4.2.2 Notes: H H H H ∙ Functional group correct / Funksionele groep korrek H C C C C H H H OH H  ∙ Whole structure correct /  Hele struktuur korrek (2) 4.2.3 Butan-1-ol / 1- butanol  (2) Marking Rules: 4.3 H H H H Functional group correct / korrekte H C C C C H funksionele groep  Whole structure correct  / hele H H  struktuur reg (2) 4.4 4.4.1 Alkene  / alkene (1) 4.4.2 Potassim bromide/KBr / kaliumbromied Water /H 2 O  (2) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com 5 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION/VRAAG 5 5.1 Carbon dioxide/CO 2 / koolstofdioksied (2) 5.2.1 The decrease in concentration of hydrochloric acid per unit time. / die afname in soutsuur konsentrasie per eenheidstyd (2) (2 or 0) 5.2.2 Concentration(of the acid) / konsentrasie (1) 5.2.3 For a fair test/comparison / vir ‘n regverdige toets (1) 5.2.4 Higher acid concentration in experiment 2 means: • More particles/molecules per unit volume • More particles have kinetic energy equal to or greater than activation energy/More particles have enough kintetic energy  • More effective collisions per unit time/frequency of effective collisions increases/Rate of effective collisions increases.  (3) Hoër konsentrasie suur vir eksperiment beteken: • Meer deeltjies/ eenheidsvolume • Meer deeltjies het genoeg E K gelyk of groter as aktiverings energie/ genoeg E K • Meer effektiewe botsings/ eenheidstyd m 5.2.5 n(CaCO 3 ) = 𝑀𝑀 4 = 100  Marking Guidelines: 𝑚𝑚 = 0,04 mol ∙ Substitution into n = 𝑀𝑀 1 n(HCℓ):n(CaCO 3 ) = 2 : 1 ∙ Ratio n(HCℓ) = n(CaCO3) 2 ∴ n(HCℓ) = 2(0,04) = 0,08 mol ∙ Substitution in c = n/V ∙ Final answer 𝑛𝑛 c(HCℓ) = 𝑉𝑉 0,08 0,4 = 𝑉𝑉  ∴ V = 0,2 dm3 = 200 cm3 (4) 5.3 POTENTIAL ENERGY VERSUS REACTION COORDINATE GRAPH Activated Complex / geaktiveerde kompleks Ea Potential energy/ Potensiële energie  Shape / ∆H  vorm (4) Reaction Coordinate / reaksie ko-ordinate [17] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION/VRAAG 6 6.1 3 to 4 minutes (Any value from 3 min to 4 min can be accepted) / 3 - 4 (1) minute 6.2.1 Greater than/ groter as (1) 6.2.2 Equal to / gelyk aan (1) 6.3 CALCULATION USING NUMBER OF MOLES / bereken met aantal mol Marking Criteria: • Correct Kc expression/ korrekte Kc uitdrukking • Substituting the Kc value in the expression/ vervanging van Kc waarde • Substituting concentration value of PCℓ5(g) in Kc/ vervanging van [PCℓ5]] in Kc • Equilibrium concentration PCℓ3 and Cℓ2 = 0,91 mol·dm-3/ [ewewig] PCℓ3 = Cℓ2 = 0,91 • Using n = cV to find the n of PCℓ5 (multiplying c by 2)/ x 2 om n van PCℓ5 te kry • Ratio of n(PCℓ5): n(PCℓ3): n(Cℓ2) = 1:1:1/ verhouding [PCℓ5]: [PCℓ3]: [Cℓ2] = 1:1:1 • Calculating n (PCℓ5) initial = 2,12 mol/ berekening van aanvanklik PCℓ5 = 2,12 Assume that the equilibrium concentration of PCℓ 3 (g) is x [PCl3][Cl2] Kc = [PCl5]  ( x )( x ) 5.55 = 0,15  ∴x =0.91 mol·dm-3  PCℓ 5 PCℓ 3 Cℓ 2 Initial Quantity (mol) 2,12  0 0 Change in quantity (mol) -1,82 +1,82 +1,82 R Equilibrium quantity (mol) 0,30  1,82 1,82 X2 Concentration at equilibrium 0,15 0,91 0,91 (mol·dm-3) V = 2 dm3 𝑚𝑚 n = 𝑀𝑀 m 2,12 = 102  m = 216,24 g  Copyright reserved Please turn over

Downloaded from hlayiso.com 7 Physical Science P2 Memo NCS Limpopo DoE September 2016 CALCULATION USING CONCENTRATION/ berekening met [ ] • Correct Kc expression/ korrekte Kc uitdrukking • Substituting the Kc value in the expression/ vervanging van Kc waarde • Substituting concentration value of PCℓ5(g) in Kc / vervanging van [PCℓ5]] in Kc • Equilibrium concentration for PCℓ3 and Cℓ2 = 0,91 mol·dm-3 / [ewewig] PCℓ3 = Cℓ2 = 0,91 • Ratio of [PCℓ5]: [PCℓ3]: [Cℓ2] = 1:1:1 / verhouding [PCℓ5]: [PCℓ3]: [Cℓ2] = 1:1:1 • Initial concentration of PCℓ5(g) / aanvanklike [PCℓ5] • Substituting 2 in c = n/V / deel deur 2 vir [ ] • Substituting 102 in n = m/M / vervanging van 102 in n = m/M • Final answer / finale antwoord Assume that the equilibrium concentration of PCℓ 3 (g) is x [PCl3][Cl2] Kc = [PCl5]  ( x )( x ) 5.55 = 0,15  ∴x =0.91 mol·dm-3  PCℓ 5 PCℓ 3 Cℓ 2 Initial concentration (mol·dm-3 ) 1,06  0 0 aanvanklik Change in concentration (mol·dm-3 ) -0,91 +0,91 +0,91 Ratio/ Verandering verhouding Equilibrium concentration (mol·dm-3 ) 0,15 +0,91 +0,91 Ewewig 𝑛𝑛 OR/ of 𝑐𝑐 = 𝑉𝑉 m = cMV n 1.06 = 2  =(1,06)(102)(2) n = 2,12 mol =216,24 g m = n M = (2,12)(31+5 x 31,5)  = 443,02 g  (9) Copyright reserved Please turn over

Downloaded from hlayiso.com 8 Physical Science P2 Memo NCS Limpopo DoE September 2016 6.4 Endothermic.  endotermies Decrease in temperature increases [PCℓ 5 ] and decreases [Cℓ 2 ] and [PCℓ 3 ], which implies that the reverse reaction is favoured/equilibrium position shifts to the left. According to Le Chatelier ‘s Principle, the decrease in temperature favours the exothermic reaction. Therefore, the forward reaction is endothermic. / Afname in temperatuur verhoog [PCℓ 5 ] en verlaag [Cℓ 2 ] en [PCℓ 3 ], dus word terugwaartse reaksie bevoordeel/ ewewig skuif na links Volgens Le Chatelier se beginsel sal ‘n afname in temperatuur die eksotermiese reaksie bevoordeel (d.w.s terugwaartse reaksie is endotermies ) Dus is die voorwaartse reaksie endotermies (3) [15] QUESTION/VRAAG 7 7.1 7.1.1 Concentrated acid- contains a large amount (number of moles) of acid in proportion to the volume of water. Dilute acid – contains a small amount (number of moles) of acid in proportion to the volume of water./ (2) Gekonsentreede suur: bevat ‘n groot hoeveelheid /mol suur in verhouding to die volume water Verdunde suur bevat ‘n klein hoeveelheid / aantal mol suur in verhouding tot die volume water 7.1.2 It ionises completely in water to form a high concentration of H 3 O+ ions. // dit ioniseer volledig in water en vorm ‘n hoë konsentrasie H 3 O+ ione (2) Note: 2 or 0 7.1.3 pH = -log[H 3 O+] = - log(0,20) = 0,7 (3) 7.2.1 Basic / basies (1) 7.2.2 CO 3 (aq) + H 2 O (ℓ)  HCO 3 (aq) + OH (aq) 2- - - (reactants  products ) Excess OH- (aq) ions are produced/it is a salt of strong base and weak acid (and the resultant solution is basic.)  ‘n Oormaat OH- ione word geproduseer/ dit is die sout van ‘n sterk basis en ‘n swak suur (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 9 Physical Science P2 Memo NCS Limpopo DoE September 2016 Marking Criteria / nasienkriteria: 7.3.1 V n(CO2) = • Substitution into/vervanging in Vm 𝑉𝑉 4,48 n = =  𝑉𝑉𝑚𝑚 22,4 • Using mole ratio/ gebruik molverhouding = 0,2 mol • Multiplying by/ vermenigvuldig met 106 • Percentage purity / pesentasie n(Na 2 CO 3 ): n(CO 2 ) = 1:1 suiwerheid ∴ n(Na 2 CO 3 ) = 0,2 mol  m(Na 2 CO 3 ) = n·M = (0,2)(106)  = 21,2 g mass of pure Na2 CO3 ∴ % purity = × 100% mass of impure Na2 CO3 21,2 = × 100  25 (5) = 84,8% 7.3.2 Mass of impurity/ massa onsuiwerheid = 25 -21,2 = 3,8 g OR % impurity/ onsuiwerheid = 100-84,8 = 15,2% 15,2 x 25 = 3,8 g 100 (2) [18] QUESTION/VRAAG 8 8.1.1 Reduction is a gain in electron(s) (by a chemical substance)// die (2) bykry van elektrone Note: 2 or 0 8.1.2 Aluminium is a stronger reducing agent than Sn/ Ag/ Hg and therefore it is easily oxidised and a current flows, simulating the pain sensors./ (3) Al is ‘n sterker reduseermiddel as sn/Ag/Hg en word daarom maklike geoksideer. ‘n Stroom vloei wat pyn veroorsaak 8.2.1 2Cr2+(aq) + Cu2+(aq) 2Cr3+(aq) + Cu(s) (reactants, products, balancing) (3) 8.2.2 Eθcell = Eθcathode − Eθanode  =+0,34-(-0,41) = 0,75 V  OR Eθcell = Eθcathode − Eθanode  = +0,34+0,41 (4) = 0,75 V 8.2.3 Increases  / verhoog Increase in temperature increases the rate of electron transfer as well as the movement of ions in the electrolyte. (This increases the rate of flow of charge in the cell.)/ (2) Copyright reserved Please turn over

Downloaded from hlayiso.com 10 Physical Science P2 Memo NCS Limpopo DoE September 2016 ‘n verhoging in temperatuur verhoog die tempo van electron oordrag en die beweging van ione in die elektroliet 8.2.4 As the cell operates, [cation]/[Cr 3+ ]increases in the anode half cell and [cation]/[Cu 2+]decreases in the cathode half cell. Anions from the salt bridge move into the anode half-cell to balance out excess cations. Cations from the salt bridge move into the cathode half-cell to balance out shortage of cations./ (3) 3+ Soos die sel werk, verhoog [katione] / [Cr ] in die anode/Cr halfsel en [katione] / [Cu 2+] verlaag in die katode/Cu halfsel. Anione / negatiewe ione va die soutbrug beweeg na die anode/ Cr halfsel om die oormaat katione/ Cr 3+ uit te balanseer Katione / negatiewe ione van die soutbrug beweeg na die katode/ Cu halfsel om die tekort van katione/ Cu2+ te balanseer [17] QUESTION/VRAAG 9 9.1 Bauxite / Bauksiet (2) 9.2 Carbon / Koolstof (1) 9.3 The anode reacts with the oxygen produced to form CO 2. // Die (1) anode reageer met die suurstof wat geproduseer word en vorm CO 2. 9.4 Cryolite / Krioliet (1) 9.5 Aℓ + 3e- Aℓ (2) 9.6 1890 C~ 0,176 g Aℓ 96500 C ~ x g Aℓ (96500 )(0,176 ) ∴x =  (1890 ) = 8,98 g 1 mol of metal ~ 27 g Since 8,98 g of metal are discharged by 1 mol of electrons, 27g 27 g of metal are discharged by  =3 moles of electrons. 8,98g If 1 mole of metal needs 3 moles of electrons, the charge on the cations is +3/ 1 mol metaal ~ 27 g Omdat 8,98 g metaal ontlaai word deur 1 mol elektrone, sal 27g 27 g metaal ontlaai word deur  =3 mol elektrone 8,98g Indien 1 mol metal 3 mol elektrone benodig, is die lading van die katioon +3 (3) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com 11 Physical Science P2 Memo NCS Limpopo DoE September 2016 QUESTION/VRAAG 10 10.1 10.1.1 Increase in population requires an increase in the production of food. OR Removing plant material from the soil depletes the nutrients so more fertilizer must be added to the soil. / Toename in bevolking vereis ‘n toename in voedselproduksie OF (1) verwydering van plantmateriaal vanuit die grond veroorsaak uitputting van voedingstowwe so meer kunsmis moet tot die grond toegevoeg word 10.1.2 Allows nutrients to be absorbed through roots with water./ laat toe (1) dat voedingstowwe deur die wortels opgeneem word 10.1.3 Run off/ leaching results in the cost of lost nutrients/ eutrophication/ (1) eutrifikasie/ afloop /loging veroorsaak verhoogte koste as gevolg van verlore voedingstowwe 10.2 10.2.1 Contact process/ kontakproses (1) 10.2.2 Oleum/ pyrosulphuric acid/ fuming sulphuric acid/oleum/rokende (1) swael 10.2.3 H 2 S 2 O 7 + H 2 O  2H 2 SO 4 (reactants, products, balancing) Note: Marking rule 6.3.10 (3) 10.2.4 Ammonium sulpahte /(NH 4 ) 2 SO 4 / ammoniumsulfaat (1) 10.2.5 (a) For stems and leaves development / vir ontwikkeling van (1) blare en stamme (1) (b) Fractional distillation of liquid air / fraksionele distillasie van vloeibare lug [11] GRAND TOTAL : 150 Copyright reserved Please turn over