Physical Sciences Grade 12 P2 NSC June 2012 Memo (Afrikaans)

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 1 DBE/2012 SCE/SSE — Memorandum basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE EXAMINATION SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCE P2/NATUUR- EN SKEIKUNDE V2 CHEMISTRY/CHEMIE HIGHER GRADE/HOER GRAAD 2012 DEPARTMENT OF BASIC MEMORANDUM EDUCATION 2oi2 -05- 25 MARKSI/PUNTE: 200 BAG X 110 PRRETORIA 0001 \ PUBLIC EXAMINATIONS This memorandum consists of 11 pages. Hierdie memorandum bestaan uit 11 bladsye. ede Vos | . A EY Nilho (. t wil als vol. Sppaanné Ge a MQ 5 Copyright reserved/Kopiereg voorbehou wt a . ara Lats ee : \4-08 WIT Fes |rae

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 GENERAL COMMENT: Do not penalise for rounding ALGEMENE OPMERKING: Moenie vir afronding penaliseer nie QUESTION/VRAAG 1 1.4 1.2 ve 1.4 1.5 1.6 | A Cc D B Cc D B B A B D B c B Cc VV VVVY VANS VVVV VVVV VV VV VVVV VV VVVS VV S VV VV VV VV SSE 2 DEPARTMENT OF BASIC EDUCATION 1 L i 2012 -05- 25 PRIVATE BAG X 110 PRETORIA 0001 PUBLIC EXAMINATIONS (15 x4) DBE/2012 [60]

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 3 DBE/2012 QUESTION/VRAAG 2 an 2.1.1 hydrogen (bonding) / waterstof(binding)v (2) 2.1.2 van der Waals or/of dipole-dipole/dipool-dipoolY ¥ (2) 2.1.3. HBr has larger molecular mass/more electrons/heavier molecule, “ resulting in stronger intermolecular forces/Van der Waals forces ¥ (2) OR HCé has smaller molecular mass/less electrons/lighter molecule, Vv resulting in weaker intermolecular forces/Van der Waals forces V HBr het ‘n groter molekulére massa/meer elektrone/swaarder molekuul, wat lei tot sterker intermolekulére kragte/Van der Waals kragte OF HC¢ het ‘n kleiner molekulére massa/minder elektrone/ligter molekuul, wat lei tot swakker intermolekulére kragte/Van der Waals kragte =i Pi _ Po 22.1 paTvv kes OROF TT @) OR/OF The pressure (p) is directly proportional to the temperature (T) Die druk (p) is direk eweredig aan die temperatuur (T) ORIOF If p increases, T increases as well As p toeneem, neem T ook toe. “DEPARTMENT OF BASIC | 309 Pi _P2 v ORIOF PVs _ P2Vo EDUCATION T, Ts T, Te v _{ 100 156° ie iE ocel cal PRIVATE BAG X 110 PRETORIA 0001 T) = 447 KY PUBLIC ExaminATiONS | _ pV 7 nRTv if 100 kPa and 1 dm® were used, max ( % ) “C100 000 0,00TY= nx 8)v 6.31 x29) Indien 100 kPa en 1 dm® gebruik is, maks ( 5% ) oi o 0,04 mol -m(SO2) = nMY No positive marking/Geen positiewe nasien = 0,04 x@av = 2,56 gv (2,58 g) (6) 2.2.4 Learner B used a smaller mass/amount/less moles of SO.v”V (2) Leerder B het ‘n kleiner massa/hoeveelheid/minder mol SO2 gebruik. un INOT volume/ NIE volume nie _ | We [20} )

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 4 DBE/2012 SSE QUESTION/VRAAG 3 _ 3.1.1 changes from purplev’ to white” or colourless or brown (2) verander van pers na wit” of kleurloos of bruin If only a colour is given, no mark. DEPARTMENT OF BASIC “Change from purple” or “change to white” (%) EDUCATION As slegs een kleur gegee is, geen punte. 2012 -05- 25 “verander van pers” of “verander na wit” ( Me) PRIVATE BAG X 110 | PRETORIA 0001 , eu " Or reducing agent/reduseermiddelv A ABLIC EXAMINATIONS 3.1.3. Itreduces Mn*’v v (to Mn*?) OR it reduces MnOj (to Mn’) OR SO; is oxidized (to soz) OR sulphur loses electrons OR KMn0O, is reduced OR SOz is oxidised (2) Dit reduseer Mn*’ “ (na Mn") OF dit reduseer MnO; (na Mn*?) OF SO; word geoksideer (tot SO7-) OF swawel verloor elektrone OF KMnO. word gereduseer OF SOz word geoksideer 3.1.4 SO. + H20 > H2SO3 LHSY RHSY Balv (3) ORIOF SO, + H20 -— H*+ HSO3_ | Ignore/ignoreer-= >. 3.2.1 Ch + H2O -» HCf + HOC? (HCO) LHS“ RHSY BAL“ (3) Accept/Aanvaar HCLO 3.2.2 Smaller than/Kleiner asV (1) 3.2.3 (HOCL/C2) is a bleaching agentY “/(HOC#CE,) is ‘n bleikmiddel ~ (2) NOT oxidising agent/ NIE oksideermiddel nie 3.3 3.3.1. dilute/verduny (2) 3.3.2 NOZ + 4H* + 3e > NO + 2H,0VV¥ (2) NO; 4H*+3e = NO+2H,0 (14) NO; + 4H’ + 3e"< NO+2H20 (9%) NO +2H20 <- NO3 + 4H" + 3e° (14) NO +2H,O = NO3 + 4H" + 3e" (94) " So . v (2) ae [20] wg 3.3.3 Cu?*/Cu(H20)—"*Vvv copper ion/ koper-ioon ( 4)

C Downloaded from hlayiso.com WGA mec Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 § . ' at DBE/201 2 *E BAG ae UESTION/VRAAG 4 ORIAD : a “UN ATION ° 4.1. 2NH4C2(s) + Ca(OH)2(s) > 2NH3(g) + CaCf,(s) + 2H20(2) LHS” RHS7 Bal” (3) Accept/Aanvaar 3H: + No — 2NH3 4.2 downward displacement of airy V /afwaartse verplasing van lug (2) 4.3 ammonia is less dense/lighter (than air)Y /ammoniak is minder dig/ligter (as lug) ammonia is soluble in water /ammoniak is oplosbaar in water (2) No negative marking from 4.2 to 4.3 / Geen negatiewe nasien van 4.2 na 4.3 44 NH3 + HC? — NH,Ct (NHZCfl) LHS” RHS¥Balv (3) = | _ 10 Ignore/ignoreer = NotNie NHS + Cf [10] QUESTION/VRAAG 5 5.1. decreases activation energy/verminder aktiveringsenergie VV (2) OR/OF provides an alternative path of lower activation energy/ verskaf alternatiewe roete van laer aktiveringsenergie OR/OF Valid explanation using energy graph/ Geldige verduideliking deur gebruik van energie-grafiek. 5.2.1 reaction is complete/reactants used up/no more products formed/ mass of H2O (formed)/ mass of H20 is 202 g/ mass (or amount) of H2O (or solution) remains constantv ¥ (2) reaksie voltooi/reaktante op gebruik/geen produkte word meer gevorm nie/ massa H2O (gevorm)/ massa H20 is 202 g/ massa (of hoeveelheid) H2O (of oplossing) bly konstant NOT equilibrium / NIE ewewig nie 5.2.2 produces a gas (which escapes)” V/produseer ‘n gas (wat ontsnap) OR/OF It is an open system/It is not a closed system Dit is ‘n oop sisteem/dit is nie ‘n geslote sisteem nie (2) >, |. —- (2) 5.2.4 steepest gradient/slope/highest concentration of H2O2/largest loss of mass (per time period)/largest mass of gas formed (per time period)vv (2) steilste gradiént/helling/ hoogste H2O2-konsentrasie/grootste massa-verlies (per tydinterval)/ grootste massa gas gevorm (per tydinterval) tl uw

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 6 DBE/2012 SSE oo “BESARTMENT OF BAs: EDUCATION 210 B tA bee dd Leer a io) a eee S see below for mark allocation) _ ision onder vir puntetoewysing) ‘ a nan \ais Re oe pe----o bown-=s feoveedonnnne foveeeee 7 ' A ‘ Experiment1 | : Experiment 3 \ ' ' ‘ i ; ‘ ‘ (6 2.6) Vviv (see below for mark allocation/sien gnder vir puntetoewysing) ‘ ; 200 0 10 20 30 40 50 60 70 Time (s) For 5.2.5 and 5.2.6 e Starting at the same point as experiment 1v ¢ For experiment 2 joins the horizontal line later/For experiment 3 joins the horizontal line sooner” e shapeY (3 + 3) Ignore labels on axis [16] Vir 5.2.5 en 5.2.6 e Begin by dieselfde punt as eksperiment 1“ e Vir eksperiment 2 sluit later by horisontale lyn aan/ Vir eksperiment 3 sluit gouer by horisontale lyn aan “ e = vorm” 7 !gnoreer byskrifte op asse [16] If graph for Experiment 1 was not drawn: 5.2.5 Award maximum 1 mark for shape of Experiment 2 5.2.6 Compare Exp 3 to Exp 2 and award max 3. lf separate graphs were drawn: Award marks only if values indicated makes a clear comparison possible. Indien grafiek vir Eksperiment 1 nie geteken is nie: 5.2.5 Ken maksimum 7 punt vir vorm van Eks 2 5.2.6 Vergelyk Eks 3 met Eks 2 en ken maksimum 3 punte toe Indien afsonderlike grafieke geteken is: Ken slegs punte toe indien waardes aangetoon, duidelike vergelyking moontlik maak. R -

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 if DBE/2012 SSE QUESTION/VRAAG 6 6.1 SUBSTANCE QUANTITY OPTION/STOFHOEVEELHEID OPSIE e n(SOs)equilibrium = 6 molY e n(SOz)ewewig = 6 mol” e use of ratio” e gebruik van verhouding “ e subtracting’ © aftrekw” e ¢ (dividing by 2) ¥ « ¢ (deel deur 2) “ e« Kc expression’ e = Ke-uitdrukking~” e substituting” * substituering’” e« answer = 9v e antwoord = 9” substance/stof 2802 O2 2803 n(initial in mol) 8 5 0 n(aanvanklik in mol) n(change in mol) -6 3 +6 Y (rati n(verandering in mol) (ratio) n(equilibrium in mol) C2 | 2 Berean 6Y (n(S0;)eq) | n(ewewig in mol) c(equilibrium in mol.dm) i ot _7 c(ewewig in mol.dm) cry cry cry | a G = 2 oy 2 2 fs <a | = 3 >| v2) - [80s] _ v= 3) v =9V (7) [SO,}°[0.] (1)7(1) “BEBARTMENT OF BASIC | EDUCATION | “aga -05- 25 H PRIVATE BAG X 110 1 PRETORIA 0001 PUBLIC EXAMINATIONS

Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 8 © (initial) = 4 and 2,5 mol.dm? v c(SOs3)equilibrium = 3 mol.dm? v Downloaded from hlayiso.com DBE/2012 SSE CONCENTRATION OPTION/KONSENTRASIE OPSIE * ¢(begin) = 4 en 2,5 mol.dm>v (dividing by 2) (deel deur 2) c(SO3)ewewig = 3 mol.dm? ~ use of ratio” gebruik van verhouding ” ing’ e subtracting © aftrek” ion’ Gy © Ke-uitdrukking” Pag * substituting * substituering” =9v 2 ee ¢ antwoord = 9” bs = a substance/stof 2802 Oz 2803 te uy 2 c(initial in moi.dm*=) n . 7 = mA 3 Ce. c= — fal ike c(aanvanklik in. mol.dm™~) Vv V =o. _ 5 ) 2 a5) (change in moi.dm™) 3 ABV +3 (ratio) c(verandering in mol.dn>) c(equilibrium in mol.dm®°) 1 4 3v c(ewewig in mol.dm®°) c(SOs3)eq (equilibrium in mol.dm™®) 1 3 ¥ (subtract) c(ewewig in mol.dm®) \ = (sos? y. G7 7 ov (7) [SO,}"[02] (4)7(1) exothermic” /eksotermies (1) 6.2 i For Ke to decrease, concentration of products decreases/concentration of reactants increases” Reverse reaction is favoured” According to Le Chateliers principle when the temperature of an system at equilibrium is increased the endothermic reaction is favouredY (3) Vir Ke af te neem, moet konsentrasie van produkte afneem/konsentrasie van reaktanse toeneem Terugwaartse reaksie bevoordeel~” Volgens Le Chateliers se beginsel wanneer die temperatuur van ‘n ewewig- i sisteem toeneem word die endotermiese reaksie bevoordeelw”

~ Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 9 DBE/2012 SSE QUESTION/IVRAAG 7 7.1.1 No/Neev Op, .2 It does not undergo complete ionization vv OR concentration of F is much less (lower) than 0,10 mol.dm® Dit ioniseer nie volledig nie “~~ OF konsentrasie van F is is baie kleiner as 0,10 mol.dm? 7.1.3 Increases/Neem toe’ Vv 7.1.4 [F]increased ¥ OR Acommon ion F is added to the solution. This favours the reverse reaction. v~ OR F + H,O — HE + OHY .. {H*] decreases and OH + H+ > HO v (3) .. Increase in [F] decreases [H*] ¥ [F] het toegeneemv OF ‘n Gemeenskaplike ioon F is is by die oplossing gevoeg. Dit bevoordeel die terugwaartse reaksie. “| OF =F + H20 -» HF + OH ‘[H'] neem af“ en OH + H+ > HO .. Toename in [F] verminder [H"] ~ 7.2.1 n(OH) = VW e n(OH) = cvv = 0,25 x 08% |, cubstituting/substitueer (0,25 x 0,8)” = 0,2 mol e calculating/bereken n(OH" ik Y N(OH )useaigebruik” = 0,2-0,12” : ( )useaigebrute = 0,08 mol e substituting/ substitueer 0,12 in aboveY Iculation/in bg. vgl. n(HCf) : n(NaOH) = 1:1¥ ae 20: 1d * n(HCt) : n(NaOh) e141 “ n(HC8) = 0,08 mol ¥ e n(HCf) = 0,08 mol (ar n(OH)usea calculated 2. n(HC2) = cV (of N(OH) gebruix bereken 0,08 =x x O5Vv e substituting/substitueer 0,5v x = 0.16 mol.dmv 2 _answer/antwoord = 0,16 mol.dmv (8) 722 fon} = 2v = _8%2 y = 0,092 moLam*v 3) a V (0,5+0,8) OR/OF 7.2.3 [H*JJOH] = 1x 10%v pOH = -logl[OH]~ . (H'] x 0,092 = 1 x 10% = -log(0,092) v . ae = -13, “(AY a | = 104 eee pH = 14~pOHY = -log(1,09 x 10") = 14~1,04 2 = 12,96v Ua) + + a 23] Accept [H30*] for [H*] Aanvaar [H30°] ipv [H"] (/ ee [ WW § es se

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 10 DBE/2012 SSE QUESTION/VRAAG 8 8.1.1 Cuv (1) 8.1.2 Fe is a stronger reducing agent vthan Ag” and reduces Cu** to Cu” (3) Fe is ‘n sterker reduseermiddel “as Ag “en reduseer Cu** tot Cu” ORI/OF: Fe is a stronger reducing agent “than Cuv and reduces Cu?* to Cu” Fe is ‘n sterker reduseermiddel vas Cu” en reduseer Cu** tot Cu” 8.1.3 Ag is a weaker reducing agent than Cu vand therefore cannot reduce” Cu** to CuY ® “BASIC Ag is n swakker reduseermiddel Cu “en kan daarom a nie Cu** tot Cu reduseer nie” 8.2.1 salt bridge/soutbrugv 8.2.2 maintains electrical neutrality/completes the circuityv VY behou elektriese neutraliteit/voltooi die stroombaan provides a path for ion flow/ voorsien ‘n pad vir ioonvioei 8.2.3 Pb”*v v Accept lead ion/Aanvaar lood-ioon (2) 8.2.4 DUE TO AN ERROR IN THE DATA SHEET, ONLY THREE MARKS ARE GIVEN IN THIS SUB-QUESTION. AS GEVOLG VAN ‘N FOUT IN DIE DATABLAD, WORD SLEGS DRIE PUNTE IN HIERDIE SUBVRAAG TOEGEKEN. O — Eo Oo i} = 8 8 v OR & col = E oxidising agent ~ E reducing agent E cell/sel = E cathode/katode ~ E anode’ 6 - a) 6 OF E sel ~ oksideermiddel ~ iS reduseermiddel —,r”r.——_._UCdCidisis 8 = 1. E8snode = -1,66V X is aluminium (3) DUE TO THE ERROR, THE FOLLOWING IS ACCEPTED: E% cer = E° oxidising agent /E“teducing agent” OR E%en = E° cathode /E’anodeY 1,53¥ = -0,13V / E® reducing agent 1 1,538Y = -0,138V 1 Eeoncde AS GEVOLG VAN DIE FOUT, WORD DIE VOLGENDE AANVAAR: E%cen = E* oksideermiddel /E° reauseermidtel ¥ OF E% sey = E°katode /E’anodo” 1,53V = -0,13V / E® seduseenidee 1 1,53V = -0,13Y 1 E® node 8.2.5 2Al + 3Pb% -> 2A + 3Pb LHS RHSVBalv (3) 2X +. 3Pb7* - 2X + 3Pb LHSV RHSVBalv OR __ Using the element obtained in 8.2.4 (pos marking) OF Gebruik die element in 8.2.4 verkry (pos nasien) 8.2.6 decreases / neem afvv . (2) 8.2.7 0VYY Accept/Aanvaar 0 (2)

Downloaded from hlayiso.com Physical Science/HG/P2/Natuur- en Skeikunde/HG/V2 11 DBE/2012 SSE QUESTION/VRAAG 9 9.1.1 DY = [CCR] (1) 9.1.2 CY [CHC8s] (1) 9.1.3 EY [CH3;COOH] (1) 9.1.4 BY = [C4Hio] (1) 9.2 2C4Hin + 1302 > 8CO2 +10H20 LHSY RHSv Balv (3) 9.3.1. methyl ethanoate / metieletanoaatv ¥ (2) eG hig U H-C—C—C—O-H 7 ORIOF H-C—C—O—C—H¥¥ moo HOH propanoic acid/propanoésuurv ethyl methanoate/etieimetanoéaalv (3) Condensed structural formula / Gekondenseerde struktuurformule: max (%) Only bonds (sticks), no H / Slegs bindings, geen H: max (%) 9.3.3. methanol” v and ethanoic acid (acetic acid)v Vv (4) metanol en etanoésuur (asynsuur [16] TOTALITOTAAL: 198 DEPARTMENT OF BASIC EDUCATION 2012 -05- 25 PRIVATE BAG X 110 PRETORIA 6001 PUBLIC EXAMINATIONS