Physical Sciences Grade 12 Paper 2 NSC November 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) NOVEMBER 2019 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 2 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 C  (2) 1.3 B  (2) 1.4 D  (2) 1.5 C  (2) 1.6 B  (2) 1.7 B  (2) 1.8 A  (2) 1.9 A  (2) 1.10 C  (2) [20] QUESTION 2/VRAAG 2 2.1 2.1.1 C n H 2n - 2  (1) 2.1.2 H Marking criteria/Nasienriglyne  H C H • Functional group correct.  Funksionele groep korrek. H H • 2 methyl substituents.   2 metielsubstituente. H C C C C C H  • Whole structure correct:/Hele struktuur korrek: 3 H H 3 H C H H (3) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 3 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2 2.2.1 Compounds with the same molecular formula,  but different positions of the side chain/substituents/functional groups on the parent chain. Verbindings met dieselfde molekulêre formule, maar verskillende posisies van die syketting/substituente/funksionele groepe op die stamketting. (2) 2.2.2 Pentan-3-one/3-pentanone  Pentan-3-oon/3-pentanoon Marking criteria/Nasienriglyne • Functional group and correct position i.e. 3 /Funksionele groep en korrekte posisie nl. 3.  • Whole name correct/Hele naam korrek.  Accept for ONE mark/Aanvaar vir EEN punt Pentanone with the 3 in incorrect place, e.g. penta-3-none. Pentanoon met die 3 in foutiewe plek, bv. penta-3-noon. (2) 2.2.3 Marking criteria/Nasienriglyne H H H H O • Whole structure correct:/Hele  struktuur korrek: 2 H C C C C C H 2  • Only functional group correct/Slegs H H H H funksionele groep korrek Max: 1 2 OR: Any correct structure of an aldehyde with five carbon atoms. OF:Enige korrekte struktuur van ‘n aldehied met vyf koolstofatome. H H H O  H H H O  H C C C C H  OR/OF H C C C C H  H H H H H C H H C H H H H H C H H O  OR/OF H C C C H  H H C H H (2) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 4 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3 2.3.1 Tertiary (alcohol)/Tersiêre (alkohol)  The C atom bonded to the functional group/hydroxyl (group)/-OH is bonded to three other C atoms. /The C-atom bonded to the hydroxyl (group) has no hydrogen atoms.  Die C-atoom gebind aan die funksionele groep/hidroksiel(groep)/-OH is gebind aan drie ander C-atome./ Die C-atoom gebind aan die hidroksiel (groep) het geen waterstofatome nie. (2) 2.3.2 2-methylbutan-2-ol/2-methyl-2-butanol/2-metielbutan-2-ol/2-metiel-2-butanol Marking criteria/Nasienriglyne • 2-methyl/2-metiel  • Butan-2-ol/2-butanol  • Any error e.g. hyphens omitted and/or incorrect sequence: Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 1 (2) 2 2.3.3 2-methylbut-2- ene/2-methyl-2-butene/2-metielbut-2-een/2-metiel-2-buteen Marking criteria/Nasienriglyne • 2-methyl/2-metiel  • But-2-ene/2-butene/But-2-een/2-buteen  • Any error e.g. hyphens omitted and/or incorrect sequence: Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 1 (2) 2 [16] QUESTION 3/VRAAG 3 3.1 Marking guidelines/Nasienriglyne The underlined key phrases must be used in the CORRECT CONTEXT (pressure/boiling). /Die onderstreepte frases moet gebruik word in die KORREKTE KONTEKS (druk/kook). The temperature  at which the vapour pressure of a substance equals atmospheric/external pressure.  Die temperatuur waar die dampdruk van 'n stof gelyk is aan atmosferiese/ eksterne druk. (2) 3.2 (Q, R and S) have same molecular mass/formulae/number of carbon and hydrogen atoms/are (chain) isomers.  (Q, R en S) het dieselfde molekulêre massa/formule/aantal koolstof en waterstofatome/ is (ketting)isomere. OR/OF The compounds are all alkanes /same homologous series and have the same number of carbon atoms. Die verbindings is almal alkane /dieselfde homoloë reeks en het die dieselfde aantal koolstofatome. (1) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 5 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne Marking guidelines/Nasienriglyne • 55 (°C)  • Compare all three compounds or Q and S in terms of branches/chain lengths / surface area.  Vergelyk al drie verbindings of Q en S in terme van vertakkings/kettinglengte/ oppervlakarea. • Compare strengths of all three or Q and S’s IMF’s / Vergelyk sterkte van al drie of Q en S se IMK'e. • Compare energy of all three /Vergelyk energie van al drie.  3.3 55 (°C)  Compare compound R with compounds Q and S: • Compound R is less branched/compact/spherical/surface area than compound Q and more branched/compact/spherical/surface area than compound S.  OR Q is the most branched/compact /spherical/surface area and S is least branced/compact/spherical/surface area. • Intermolecular forces in compound R are stronger than in compound Q and weaker than in compound S.  • More energy needed to overcome intermolecular forces in compound R than in compound Q and less energy needed to overcome (break) intermolecular forces in compound R than in compound S.  OR • Compound R has a longer chain length than compound Q and a shorter chain length than compound S.  OR S has the longest chain length and Q the shortest. • Intermolecular forces increase with increase in chain length.  • More energy needed to overcome intermolecular forces as chain length increases.  Vergelyk verbinding R met verbindings Q en S:. • Verbinding R is minder vertak/kompak/sferieseoppervlak as verbinding Q en meer vertak as verbinding S. OF Q is die meeste vertak/kompak en S is die minste vertak/kompak/series/oppervlak. • Intermolekulêre kragte in verbinding R is sterker as in verbinding Q en swakker as in verbinding S. • Meer energie word benodig om intermolekulêre kragte in verbinding R te oorkom as in verbinding Q, en minder energie word benodig om intermolekulêre kragte in verbinding R te oorkom / breek as in verbinding S. OF • Verbinding R het 'n langer kettinglengte as verbinding Q en 'n korter kettinglengte as S. OF S het die langste ketting en Q die kortste. • Intermolekulêre kragte neem toe met toename in kettinglengte. • Meer energie word benodig om intermolekulêre kragte te oorkom wanneer kettinglengte toeneem. (4) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 6 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4 3.4.1 P  (2) 3.4.2 Marking guidelines/Nasienriglyne • Name type of IMFs in P/pentanal.  Noem tipe IMK'e in P/pentanaal. • Name type of IMFs in/Noem tipe IMK'e in T/pentan-1-ol. • Compare strength of IMFs. /Vergelyk sterkte van IMK'e.  OR/OF Compare energy needed to overcome IMFs./Vergelyk energie benodig om IMK'e te oorkom. • In P/ pentanal/aldehydes: dipole-dipole forces (in addition to London forces/dispersion forces/induced dipole forces). • In T/pentan-1-ol: Hydrogen bonding. (in addition to London forces/dispersion forces/induced dipole forces). • Intermolecular forces in P/pentanal are weaker than in T/pentan-1-ol OR dipole-dipole forces are weaker than hydrogen bonds OR intermolecular forces in T/pentan-1-ol are stronger than in P/pentanal. OR More energy needed to overcome/break intermolecular forces in T. • In P/pentanaal/aldehiede: dipool-dipoolkragte (tesame met Londonkragte/ dispersiekragte/geïnduseerde dipoolkragte). • In T/pentan-1-ol: Waterstofbinding. (tesame met Londonkragte/ dispersiekragte/geïnduseerde dipoolkragte). • Intermolekulêre kragte in P swakker as in T/pentan-1-ol OF intermolekulêre kragte in T/pentan-1-ol sterker as in P/pentanaal OF dipool-dipoolkragte is swakker as waterstofbindings. OF Meer energie benodig om intermolekulêre kragte te oorkom/breek in T. (3) [12] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 7 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 Haloalkane/alkyl halide  Haloalkaan/alkielhalied (1) 4.2 4.2.1 Elimination/dehydrohalogenation  Eliminasie/dehidrohalogenering (1) 4.2.2 Substitution/hydrolysis  Substitusie/hidrolise (1) 4.2.3 Esterification/condensation  Esterifikasie/kondensasie/verestering (1) 4.3 4.3.1 • (Mild) heat/Heating/(matige) hitte/ verhitting  • Dilute (strong base)/Verdunde (sterk basis)/(NaOH/KOH/LiOH)  OR/OR Add water/H 2 O/Voeg water/H 2 O by (2) 4.3.2 Propan-1-ol/1-propanol  Marking criteria/Nasienriglyne: • Correct stem and functional group i.e. propanol/Korrekte stam en funksionele groep, d.i. propanol.  • Whole name correct:/Hele naam korrek: propan-1-ol  (2) 4.4 H H H Marking criteria/Nasienriglyne • Whole structure correct:/Hele struktuur H C C C H korrek: 2 2 H O H • Only functional group correct/Slegs H  funksionele groep korrek: 1 2 Notes/Aantekeninge • Accept –OH as condensed. /Aanvaar –OH as gekondenseerd. • Condensed or semi-structural formula: Gekondenseerde of semi-struktuurformule: Max./Maks. 1 2 • Molecular formula/Molekulêre formule: 0 2 • If functional group is incorrect/Indien funksionele groep verkeerd is: 0 2 • If more than one functional group: Indien meer as een funksionele groep: 0 2 (2) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 8 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 4.5 POSITIVE MARKING FROM Q4.3.2 ONLY IF THE COMPOUND IN Q4.3.2 IS AN ALCOHOL. /POSITIEWE NASIEN VANAF V4.3.2 SLEGS INDIEN DIE VERBINDING IN Q4.3.2 'N ALKOHOL IS. 4.5.1  Marking criteria/Nasienriglyne H H H O H • Whole structure correct:/Hele struktuur H C C C O C C H  korrek: 2 2 • Only functional group correct/Slegs H H H H funksionele groep korrek: 1 2 Notes/Aantekeninge • Condensed or semi-structural formula: Gekondenseerde of semistruktuurformule: Max./Maks. 1 2 • Molecular formula/Molekulêre formule: 0 2 • If functional group is incorrect/Indien funksionele groep verkeerd is: 0 (2) 2 4.5.2 (Concentrated) sulphuric acid/(Gekonsentreerde) swawelsuur/H 2 SO 4  (1) [13] QUESTION 5/VRAAG 5 5.1 Exothermic/Eksotermies  - H < 0/Energy is released/Energie word vrygestel  (2) 5.2 ∆m OR/OF rate/tempo = − ∆t ∆m rate/tempo = − 0,25 − 2  ∆t = − 30  − 1,75  = − = 0,06 (g·s-1)  30  (0,0583 g·s-1) = 0,06 (g·s-1)  (0,0583 g·s-1) (3) Notes/Aantekeninge -1 Accept negative answer i.e./Aanvaar negatiewe antwoord d.i. - 0,06 g·s . Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 9 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3 Marking guidelines • Calculate/Bereken: m(CaCO 3 ) reacted/reageer or / of V(CO 2 ) produced/gevorm.  • Substitute/Vervang: 100 g∙mol-1.  • USE mol ratio/GEBRUIK molverhouding: n(CO 2 ) : n(CaCO 3 ) = 1 : 1  • Use of/ /Gebruik van 22,4 dm-3∙mol-1. • Final answer/Finale antwoord: 0,18 dm3 (0,1792 dm3)  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 40 For 2 g antacid/teensuurtablet: m(CaCO 3 ) = x2 100 g CaCO 3 ......22,4 dm3  CO 2 100 = 0,8 g 2 g CaCO 3 .....0,448 dm3  m n(CaCO 3 ) reacted = 100% CO 2 .......... 0,448 dm3  M 40% CO 2 ............. 0,18 dm3  0,8 = OPTION 3/OPSIE 3 100  = 8 x 10-3 mol 100% CaCO 3 .........2 g 40% .......................0,8 g  n(CO 2 ) = n(CaCO 3 )  100 g ……. 1 mol = 8 x 10-3 mol 0,8 g .......... 8 x 10-3 mol  V(CO 2 ) = 8 x 10-3 x 22,4  1 mol ...............22,4 dm3  = 0,18 dm3  8 x 10-3 mol ......0,18 dm3  (5) 5.4 ANY ONE/ENIGE EEN: • Concentration (of acid)/Konsenterasie (van suur)  • Size/mass of tablet/Identical tablet /Type of tablet. Grootte/massa van tablet/Identiese tablet./Tipe tablet. • State of division / Surface area / Toestand van verdeeldheid / reaksieoppervlak. (1) 5.5 Criteria for conclusion/Riglyne vir gevolgtrekking: Dependent [(reaction) rate/time] and independent (temperature) variables correctly identified.  Afhanklike [(reaksie)tempo/tyd] en onafhanklike (temperatuur) veranderlikes korrek geïdentifiseer. Relationship between the independent and dependent variables correctly stated./Verwantskap tussen die afhanklike en onafhanklike veranderlikes  korrek genoem. Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 10 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne Examples/Voorbeelde: 1 • Reaction rate ( ) increases with increase in temperature. time 1 Reaksietempo ( ) neem toe met toename in temperatuur. time 1 • Reaction rate ( ) decreases with decrease in temperature. time 1 Reaksietempo ( ) neem af met afname in temperatuur. time • Time taken for reaction decreases when temperature increases. Tyd vir die reaksie neem af wanneer temperatuur toeneem. • Time taken for reaction increases when temperature decreases. Tyd vir die reaksie neem toe as temperatuur afneem. IF/INDIEN Reaction rate is DIRECTLY proportional to temperature: Max. 1 2 Reaksietempo is DIREK eweredig aan temperatuur: Maks. 1 2 (2) 5.6 • Increase in temperature increases the average kinetic energy/molecules move faster. /Toename in temperatuur verhoog die gemiddelde kinetiese energie/molekule beweeg vinniger.  • More molecules have enough/sufficient kinetic energy/More molecules have E k > E a.  Meer molekule het genoeg/voldoende kinetiese energie/Meer molekule het E k > E a. • More effective collisions per unit time/second. /Frequency of effective collisions increases.  Meer effektiewe botsings per eenheidtyd/sekonde./Frekwensie van effektiewe botsings neem toe. (3) 5.7 Marking guidelines/Nasienriglyne • For each value of temperature, the CURVE Y must be above the given CURVE. / Vir elke waarde van temperatuur, moet kurwe Y bo die gegewe kurwe wees.  • CURVE Y must have an increasing rate with an increase in temperature. / KURWE Y moet ‘n toenemende tempo het soos die temperatuur toeneem.  1 Y (s-1) time / tyd Temperature/Temperatuur (°C) (2) [18] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 11 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 (The stage in a chemical reaction when the) rate of forward reaction equals the rate of reverse reaction.  (Die stadium in 'n chemiese reaksie wanneer die) tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. (2 or/of 0) OR/OF (The stage in a chemical reaction when the) concentrations of reactants and products remain constant. (Die stadium in 'n chemiese reaksie wanneer die) konsentrasies van reaktanse en produkte konstant bly. (2 or/of 0) (2) 6.2 CALCULATIONS USING NUMBER OF MOLES BEREKENINGE WAT AANTAL MOL GEBRUIK 6.2.1 Marking guidelines/Nasienriglyne • Substitute/Vervang: 44 g∙mol-1.  • Equilibrium concentration of CO 2 multiply by 3 dm3 Ewewigskonsentrasie van CO 2 vermenigvuldig met 3 dm3  AND/EN n(CO) eq divide by /deel deur 3 dm3 • Use mole ratio/Gebruik molverhouding: 1:2 / n(CO) = 2n(CO 2 ).  • n(CO 2 ) change = n(CO 2 ) initial - n(CO 2 ) final  n(CO) eq/ewe = n(CO) initial/begin + ∆n(CO) • Correct K c expression (formulae in square brackets).  Korrekte K c -uitdrukking (formules in vierkanthakies). • Substitution of concentrations into K c expression.  Vervanging van konsentrasies in K c -uitdrukking. • Final answer/Finale antwoord: 12,24 (range/gebied: 11,85 – 12,66)  Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 12 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 1/OPSIE 1 m n(CO 2 ) = M 60,8 = 44  = 1,382 mol CO 2 CO Initial quantity (mol) 1,382 0 Aanvangshoeveelheid (mol) Change (mol)  1,22 Use ratio/Gebruik 2,44 Verandering (mol) verhouding  Quantity at equilibrium (mol)/ 0,162 2,44 Divide/multiply by Hoeveelheid by ewewig (mol) -3 Equilibrium concentration (mol∙dm ) 3/Deel/ vermenig- 0,054 0,813 vuldig met 3  Ewewigskonsentrasie (mol∙dm-3) [CO] 2 Kc =  No Kc expression, correct substitution/Geen Kc- [CO 2 ] 6 uitdrukking, korrekte substitusie: Max./Maks. (0,813 ) 2 7 =  0,054 Wrong Kc expression/Verkeerde Kc- uitdrukking: = 12,24  Max./Maks. 4 7 OPTION 2/OPSIE 2 m n(CO 2 ) = M 60,8 = 44  = 1,382 mol n(CO 2 ) change = n(CO 2 ) initial/begin - n(CO 2 ) final/finaal = 1,382 – 0,162 = 1,22 mol n(CO) change = 2(CO 2 )  = 2(1,22)  = 2,44 mol n(CO) eq = n(CO) change = 2,44 mol n c(CO) = V 2.44 = 3 = 0,813 mol∙dm-3 [CO] 2 Kc =  [CO 2 ] (0,813)2 =  0,054 = 12,24  (Accept range/Aanvaar gebied: 11,85 – 12,66). Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 13 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne CALCULATIONS USING CONCENTRATION BEREKENINGE WAT KONSENTRASIE GEBRUIK Marking guidelines/Nasienriglyne • Substitute 44 g∙mol-1.  • Initial n(CO 2 )divide by 3 dm3.  Aanvanklike n(CO 2 ) gedeel deur 3 dm3. • USE ratio/GEBRUIK verhouding: c(CO 2 ) : c(CO) = 1 : 2  • ∆c(CO 2 ) = c(CO 2 ) initial/begin - c(CO 2 ) eq/ewe .  c(CO) eq/ewe = c(CO) initial/begin + ∆c(CO). • Correct K c expression (formulae in square brackets).  Korrekte K c uitdrukking (formules in vierkanthakies). • Substitution of concentrations into K c expression.  Vervanging van konsentrasies in K c -uitdrukking. • Final answer/Finale antwoord: 12,15 (range/gebied: 11,85 – 12,66)  OPTION 3/OPSIE 3 m n(CO 2 ) = M 60,8 = 44  = 1,382 mol CO 2 CO Initial concentration (mol∙dm-3) Divide by /Deel 0,4607 0 Aanvanklike konsentrasie (mol∙dm-3)  deur 3 dm3  Change (mol∙dm-3) ratio  0,4067 0,813 Verandering (mol∙dm-3) verhouding Equilibrium concentration (mol∙dm-3) 0,054 0,813 Ewewigskonsentrasie (mol∙dm-3) [CO] 2 No Kc expression, correct substitution/Geen Kc- Kc =  [CO 2 ] uitdrukking, korrekte substitusie: Max./Maks. 6 7 (0,813) 2 =  Wrong Kc expression/Verkeerde Kc-uitdrukking: 0,054 = 12,15  Max./Maks. 4 7 (7) 6.2.2 POSITIVE MARKING FROM Q6.2.1/POSITIEWE NASIEN VANAF V6.2.1 n(C) reacted/reageer = Marking guidelines n(CO 2 ) reacted/reageer • USE mol ratio/ GEBRUIK molverhouding: = 1,22 mol  n(C) = n(CO 2 ).  -1 • Substitute/Vervang: 12 g∙mol .  m(C) = nM • Final answer/Finale antwoord: 14,64 g.   = 1,22(12) = 14,64 g  (3) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 14 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 6.3 6.3.1 Remains the same/Bly dieselfde  (1) 6.3.2 Decreases/Afneem  - • (When pressure is increased) the reaction that leads to the smaller amount/number of moles/volume of gas is favoured.  (Wanneer die druk verhoog word,) word die reaksie wat tot die kleiner hoeveelheid/aantal mol/volume gas lei, bevoordeel. • The reverse reaction is favoured. / More CO 2 is formed.  Die terugwaartse reaksie word bevoordeel./ meer CO 2 word gevorm. (3) 6.4 6.4.1 Endothermic/Endotermies  - • When the temperature increases the mol/percentage CO(g)/product increases/forward reaction is favoured./Wanneer die temperatuur toeneem, neem die mol/persentasie CO(g)/produk toe/voorwaartse reaksie word bevoordeel.  • An increase in temperature favours the endothermic reaction/Toename in temperatuur bevoordeel die endotermiese reaksie.  (3) 6.4.2 POSITIVE MARKING FROM Q6.2.1./POSITIEWE NASIEN VANAF V6.2.1. Marking guidelines/Nasienriglyne • Calculate total volume/mol of gas at equilibrium/Bereken totale volume/mol gas by ewewig: 0,162 + 2,44 = 2,606 dm3 /mol  OR/OF Calculate the total concentration at equilibrium/Bereken die totale konsentrasie by ewewig: 0,054 + 0,813 = 0,867 mol∙dm-3 • Calculate percentage of ANY one gas/Bereken persentasie van ENIGE een gas (CO 2 or/of CO).  • Final answer/Finale antwoord: T = 827 °C  OPTION 1/OPSIE 1 V total eq = 0,162 + 2,44  = 2,606 dm3 0,162 % CO 2 = x 100  OR/OF 2,606 2,44 % CO = x 100  = 6,225 % 2,606 = 93,63 % OPTION 2/OPSIE 2 c total eq = 0,054 + 0,813 = 0,867 mol∙dm-3 0,054 0,813 % CO 2 = x 100  OR/OF % CO = x 100  0,867 0,867 = 6,228 % = 93,77 % ∴ T = 827 °C  (3) [22] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 15 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 - Strong (acid)/Sterk (suur)  Large/Groot K a value/waarde/ K a > 1 / (HBr) ionises completely/ioniseer volledig  (2) 7.2 H2O  - Br  (2) 7.3 7.3.1 Marking guidelines/Nasienriglyne n c × Va n a • Formula/Formule: c = /n = cV / a =  V c b × Vb nb • Substitution of/Vervanging van: (0,5)(0,0165)/(0,5)(16,5)  • Use mol ratio/Gebruik molverhouding: 1:1/n(HBr) = n(NaOH)  • Substitute/Vervang: V = 0,09 dm3 /90 cm3  • Formula/Formule: pH = -log[H 3 O+]  • Substitute [H 3 O+] in pH formula.  • Final answer/Finale antwoord: pH = 1,04 (range/gebied: 1,036 – 1,05)  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 n(NaOH) reacted/reageer = cV  c a Va na = 0,5(0,0165)  =  c b Vb nb = 0,00825 mol c a (90)  1 =  n(HBr) excess/oormaat = n(NaOH) = 0,00825 mol  (0,5)(16,5) 1 n c a = 0,092 mol·dm-3 c(H 3 O+) = V 0,00825 pH = -log[H 3 O+]  = = -log(0,092)  0,09  = 1,04  = 0,092 mol·dm-3 pH = -log[H 3 O+]  = -log(0,092)  = 1,04  (7) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 16 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 7.3.2 Marking guidelines/Nasienriglyne • Calculate/Bereken n(HBr) initial/aanvanklik : substitute/vervang (0,45)(0,09) in n = cV  • Subtraction/Aftrekking: n(HBr) reacted/reageer = n(HBr) initial/aanvanklik – n(HBr) reacted with/reageer met NaOH .  OR/OF: c(HBr) reacted/reageer = c(HBr) initial/aanvanklik – c(H 3 O+) excess/oormaat • Use mol ratio/Gebruik molverhouding: n(Zn(OH) 2 ) : n(HBr) = 1 : 2  • Substitution of/Vervanging van: 99 g∙mol-1  • Final answer/Finale antwoord: 1,5964 g (range/gebied: 1,58 – 1,68)  POSITIVE MARKING FROM Q7.3.1/POSITIEWE NASIEN VANAF V7.3.1 OPTION 1/OPSIE 1 n(HBr) initial/begin = cV = (0,45)(0,09)  = 0,0405 mol n(HBr reacted with/reageer met Zn(OH) 2 ) = 0,0405 – 0,00825  = 0,03224 mol n(Zn(OH) 2 ) = ½n(HBr) = ½(0,03224) = 0,016125 mol m(Zn(OH) 2 ) = nM = (0,016125)(99)  = 1,596 g  OPTION 2/OPSIE 2 c(HBr) = 0,45 – 0,092  = 0,358 mol∙dm-3 n(HBr reacted/reageer) = cV = 0,358 x 0,09  = 0,0322 mol n(Zn(OH) 2 ) = ½n(HBr) = ½(0,0322) = 0,01611 mol m(Zn(OH) 2 ) = nM = 0,01611 x 99  = 1,595 g  (1,60 g) (6) [17] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 17 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Chemical to electrical/Chemies na elektries  (1) 8.2 Provides path for movement of ions./ Completes the circuit./Ensures electrical neutrality in the cell./Restore charge balance.  Verskaf pad vir beweging van ione./Voltooi die stroombaan./Verseker elektriese neutraliteit in die sel./Herstel balans van lading. (1) 8.3 OPTION 1/OPTION 1 θ E cell θ = E cathode θ − E anode  Notes/Aantekeninge   θ • Accept any other correct formula from the data 1,49 = 1,36 − E anode sheet. /Aanvaar enige ander korrekte formule θ E anode = 1,36 − 1,49 vanaf gegewensblad. • Any other formula using unconventional = − 0,13 ( V )  abbreviations, e.g. E° cell = E° OA - E° RA X is Pb/Lead/Lood  followed by correct substitutions:/Enige ander formule wat onkonvensionele afkortings gebruik, bv. E° sel = E° OM - E° RM gevolg deur korrekte vervangings : 4 5 OPTION 2/OPSIE 2 Cℓ 2 + 2e- → 2Cℓ- E θ = 1,36 V   X → X2+ + 2e- E θ = 0,13 V  Cℓ 2 + X → X2+ + 2Cℓ- E θ = 1,49 V  X is Pb/Lead/Lood  (5) POSITIVE MARKING FROM Q8.3/POSITIEWE NASIEN VANAF V8.3 8.4 X/Pb/Lead/Lood  (1) 8.5 8.5.1 Reaction reached equilibrium./(In each half cell) the rate of oxidation is equal to rate of reduction./Rate of the forward reaction is equal to the rate of the reverse reaction.  Reaksie bereik ewewig./(In elke halfsel) die tempo van oksidasie is gelyk aan tempo van reduksie./Tempo van die voorwaartse reaksie is gelyk aan die tempo van die terugwaartse reaksie. (1) 8.5.2 Increases/Toeneem  (1) 8.5.3 • [Cℓ-] decreases/neem af.  • Forward reaction is favoured./Voorwaartse reaksie word bevoordeel.  (2) [12] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 18 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Marking guidelines/Nasienriglyne If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte frases in die korrekte konteks uitgelaat is, trek 1 punt af. The chemical process in which electrical energy is converted to chemical energy.  Die chemiese proses waarin elektriese energie omgeskakel word na chemiese energie. OR/OF The use of electrical energy to produce a chemical change. Die gebruik van elektriese energie om 'n chemiese verandering teweeg te bring. OR/OF The process during which an electrical current passes through a solution/molten ionic compound. Die proses waar 'n elektriese stroom deur 'n oplossing/gesmelte ioniese verbinding gestuur word. (2) 9.2 9.2.1 2H 2 O(ℓ) + 2e-  H 2 (g) + 2OH-(aq)  Ignore phases/Ignoreer fases Marking guidelines/Nasienriglyne • H 2 (g) + 2OH-(aq) ← 2H 2 O(ℓ) + 2e- ( 2 ) 2H 2 O(ℓ) + 2e- ⇌ H 2 (g) + 2OH-(aq) 2 ( 1 ) 2 H 2 (g) + 2OH-(aq) ⇌ 2H 2 O(ℓ) + 2e- ( 0 ) 2H 2 O(ℓ) + 2e- ← H 2 (g) + 2OH-(aq) 2 ( 0 ) 2 • Ignore if charge omitted on electron./Ignoreer indien lading weggelaat op elektron. • If charge (-) omitted on OH- /Indien lading (-) weggelaat op OH-: Example/Voorbeeld: 2H 2 O(ℓ) + 2e-  H 2 (g) + 2OH(aq)  Max./Maks: 1 2 (2) 9.2.2 Water/ H 2 O  - (1) 9.3 H2O is a stronger oxidising agent  than Na+  and will be reduced  (to H2). H2O is 'n sterker oksideermiddel as Na+ en sal gereduseer word (na H2). OR/OF - + Na is a weaker oxidizing agent than H 2 O and therefore H 2 O will be reduced  (to H 2 ) Na+ is ‘n swakker oksideermiddel as H 2 O en daarom sal H 2 O gereduseer word (na H 2 ) OR/OF The half-reaction that produces H2(g) has a more positive reduction potential (-0,83 V)  than the half-reaction that produces Na (-2,71 V).  Therefore water/H2O will be reduced  to H2./Na+ will not be reduced to Na. Die halfreaksie wat H2(g) vorm, het 'n meer positiewe reduksiepotensiaal (-0,83 V) as die halfreaksie wat Na vorm (-2,71 V). Daarom word water/H2O na H2 gereduseer./Na+ sal nie gereduseer word na (3) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 19 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne Na nie. [8] Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 20 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 10.1.1 Hydrogen/Waterstof/H 2  (1) 10.1.2 Nitrogen monoxide/Stikstofmonoksied/NO  (1) 10.1.3 Nitric acid/Salpetersuur/HNO 3  (1) 10.2 10.2.1 (Catalytic) oxidation/Redox/(Katalitiese) oksidasie/Redoks  (1) 10.2.2 NH 3 + HNO 3  → NH 4 NO 3  Bal  Notes/Aantekeninge • Reactants  Products  Balancing  Reaktanse Produkte Balansering • Ignore double arrows (⇌) and phases./Ignoreer dubbelpyle (⇌)en fases. • Marking rule 6.3.10./Nasienreël 6.3.10. (3) 10.3 10.3.1 (Total) percentage of nutrients/fertiliser/N,P,K.  (Totale) persentasie nutriente/ kunsmis/N,P, K. (1) Copyright reserved/Kopiereg voorbehou Please turn the page/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 21 DBE/November 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 10.3.2 Marking guidelines/Nasienriglyne • Calculate mass fertiliser in A./Bereken massa kunsmis in A  • Calculate mass fertiliser in B./ Bereken massa kunsmis in B  • Calculate mass P in A and B ./Bereken massa P in A en B • Final answer/Finale antwoord: B has more phosphorous than/het meer fosfor as A.  OPTION 1/OPSIE 1 OPTION 3/OPSIE 3 Mass fertiliser in A: Mass phosphorous in A/ Massa kunsmis in A: Massa fosfor in A: 21 3 m= × 50  = 10,5 kg %P = × 21= 7,88% 100 8 Mass fertiliser in B: 7,88 /Massa kunsmis in B: m(P) = × 50  = 3,94 kg 100 27  m= × 40  = 10,8 kg Mass(P) in B 100 Massa (P) in B: Mass phosphorous in A/ 3 Massa fosfor in A: %(P) = × 27 = 10,13% 8 3 10,13 × 10,5 = 3,94 kg × 40  = 4,05 kg 8 m= 100 Mass phosphorous in B/  Massa fosfor in B: Fertiliser B has more phosphorous 3 than fertiliser A. /Kunsmis B het meer × 10,8 = 4,05 kg fosfor as kunsmis A  8 Fertiliser B has more phosphorous OPTION 4/OPSIE 4 than fertiliser A.  Mass fertiliser in A: Massa kunsmis in A: OPTION 2/OPSIE 2 21 m= × 50  = 10,5 kg Mass phosphorous in A/ 100 Massa fosfor in A: Mass fertiliser in B: 3 21 /Massa kunsmis in B: m= × × 50  = 3,94 kg 8 100 27 m= × 40  = 10,8 kg Mass(P) in B  100 Massa (P) in B: For the same NPK ratio  3 27 the bag with more fertiliser will have m= × × 40  = 4,05 kg 8 100 more phosphorous ∴ bag B Vir dieselfde NPK verhouding, die sake Fertiliser B has more phosphorous met meer kunsmis sal meer fosfor het than fertiliser A. /Kunsmis B het ∴ sak B meer fosfor as kunsmis A. (4) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou