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NATIONAL
SENIOR CERTIFICATE
NASIONALE
SENIOR SERTIFIKAAT
GRADE/GRAAD 12
PHYSICAL SCIENCES: CHEMISTRY (P2)
FISIESE WETENSKAPPE: CHEMIE (V2)
NOVEMBER 2014
MEMORANDUM
MARKS/PUNTE: 150
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Physical Sciences Grade 12 Paper 2 NSC November 2014 Memo
Grade 12 Physical Sciences Paper 2 memorandum from the NSC November Exam 2014, 20 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · NSC November Exam · 2014. Memorandum, 20 pages. Read online or download the PDF.
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- 2014
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Physical Sciences P2/Fisiese Wetenskappe V2 2 DBE/November 2014
NSC/NSS – Memorandum
QUESTION 1 / VRAAG 1
1.1 C (2)
1.2 B (2)
1.3 D (2)
1.4 D (2)
1.5 A (2)
1.6 B (2)
1.7 B (2)
1.8 A (2)
1.9 D (2)
1.10 C (2)
[20]
QUESTION 2 / VRAAG 2
2.1
2.1.1 B (1)
2.1.2 E (1)
2.1.3 F (1)
2.2
2.2.1 2-bromo-3-chloro-4-methylpentane
2-bromo-3-chloro-4-metielpentaan / 2-broom-3-chloor-4-metielpentaan
Marking criteria / Nasienriglyne:
• Correct stem i.e. pentane. / Korrekte stam d.i. pentaan.
• All substituents correctly identified. / Alle substituente korrek geïdentifiseer.
• Substituents correctly numbered, in alphabetical order, hyphens and commas
correctly used.
Substituente korrek genommer, in alfabetiese volgorde, koppeltekens en
kommas korrek gebruik. (3)
2.2.2 Ethene / Eteen (1)
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Physical Sciences P2/Fisiese Wetenskappe V2 3 DBE/November 2014
NSC/NSS - Memorandum
2.3 H Marking criteria / Nasienriglyne:
2.3.1 • Six saturated C atoms in
H C H longest chain i.e. hexane.
Ses versadigde C-atome in
H H H H H langste ketting d.i. heksaan.
• Three methyl substituents on
second C and fourth C.
H C C C C C C H Drie metielsubstituente op
tweede C en vierde C.
H H H H
H C H H C H
H H
Notes / Aantekeninge:
• If correct structure, but H atoms omitted / Indien korrekte struktuur, maar H-atome
weggelaat: Max / Maks. 1
2
• Condensed or semi-structural formula:
Gekondenseerde of semistruktuurformule: Max./Maks. 1
2
• Molecular formula / Molekulêre formule: 0
(2)
2
2.3.2 H H Marking criteria / Nasienriglyne:
H H O • Whole structure correct / Hele struktuur
H C C C C C H korrek: 2
2
H H H H • Only functional group correct / Slegs
funksionele groep korrek: 1
2
Notes / Aantekeninge:
• If two or more functional groups/Indien twee of meer funksionele groepe: 0
2
• Condensed or semi-structural formula:
Gekondenseerde of semistruktuurformule: Max / Maks 1
2
• Molecular formula / Molekulêre formule: 0
2 (2)
2.4
2.4.1 (Compounds with) the same molecular formula but different functional
goups / different homologous series.
(Verbindings met) dieselfde molekulêre formule, maar verskillende
funksionele groepe / verskillende homoloë reekse. (2)
2.4.2 B&F (1)
[14]
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Physical Sciences P2/Fisiese Wetenskappe V2 4 DBE/November 2014
NSC/NSS - Memorandum
QUESTION 3 / VRAAG 3
3.1 ANY ONE / ENIGE EEN:
• Alkanes have ONLY single bonds.
Alkane het SLEGS enkelbindings.
• Alkanes have single bonds between C atoms.
Alkane het enkelbindings tussen C-atome.
• Alkanes have no double OR triple bonds OR multiple bonds.
• Alkane het geen dubbel- OF trippelbindings OF meervoudige bindings nie.
• Alkanes contain the maximum number of H atoms bonded to C atoms.
Alkane bevat die maksimum getal H-atome gebind aan C-atome. (1)
3.2
3.2.1 ANY ONE / ENIGE EEN:
─C─O─H ─ C ─ OH – OH –O–H
R ─ OH R─O─H
(1)
3.2.2 H Marking criteria / Nasienriglyne:
• - OH group on second C atom of longest
H C H chain.
- OH-groep op tweede C-atoom van langste
H H ketting.
• Tertiary group consisting of four C atoms
H C C C H with methyl group on 2nd C atom.
Tersiêre groep bestaande uit vier C-atome
H H
met metielgroep op 2de C-atoom.
O • If two or more functional groups / Indien
H twee of meer funksionele groepe: 0
2
Notes / Aantekeninge:
• Accept / Aanvaar – OH
• If correct structure and number of bonds, but H atoms omitted / Indien korrekte
struktuur en getal bindings, maar H-atome weggelaat: Max / Maks. 1
2
• Condensed or semi-structural formula / Gekondenseerde of
semistruktuurformule: Max / Maks. 1
2
• Molecular formula / Molekulêre formule: 0
(2)
2
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Physical Sciences P2/Fisiese Wetenskappe V2 5 DBE/November 2014
NSC/NSS - Memorandum
3.3
3.3.1
Criteria for investigative question / Riglyne vir ondersoekende vraag:
The dependent and independent variables are stated.
Die afhanklike en onafhanklike veranderlikes is genoem.
- Ask a question about the relationship between the independent and
dependent variables.
Vra 'n vraag oor die verwantskap tussen die onafhanklike en afhanklike
veranderlikes.
Examples / Voorbeelde:
• How does an increase in chain length / molecular size / molecular
structure / molecular mass / surface area influence boiling point?
Hoe beïnvloed 'n toename in kettinglengte / molekulêre grootte /
molekulêre struktuur / molekulêre massa / reaksieoppervlak die
kookpunt?
• What is the relationship between chain length / molecular size / molecular
structure / molecular mass / surface area and boiling point?
Wat is die verwantskap tussen kettinglengte / molekulêre grootte /
molekulêre struktuur / molekulêre massa / oppervlakte en kookpunt? (2)
3.3.2 • Structure / Struktuur:
The chain length / molecular size / molecular structure / molecular mass /
surface area increases.
Die kettinglengte / molekulêre grootte / molekulêre struktuur / molekulêre
massa / oppervlakte neem toe.
• Intermolecular forces / Intermolekulêre kragte:
Increase in strength of intermolecular forces / induced dipole / London /
dispersion / Van der Waals forces.
Toename in sterkte van intermolekulêre kragte / geïnduseerde
dipoolkragte / London-kragte / dispersiekragte / Van der Waalskragte.
• Energy / Energie:
More energy needed to overcome / break intermolecular forces.
Meer energie benodig om intermolekulêre kragte te oorkom / breek.
OR / OF
• Structure / Struktuur:
From propane to methane the chain length / molecular size / molecular
structure / molecular mass / surface area decreases.
Van propaan na metaan neem die kettinglengte / molekulêre grootte /
molekulêre struktuur / molekulêre massa / oppervlakte af.
• Intermolecular forces / Intermolekulêre kragte:
Decrease in strength of intermolecular forces / induced dipole forces /
London forces / dispersion forces.
Afname in sterkte van intermolekulêre kragte / geïnduseerde dipoolkragte
/ London-kragte / dispersiekragte.
• Energy / Energie:
Less energy needed to overcome / break intermolecular forces.
Minder energie benodig om intermolekulêre kragte te oorkom / breek. (3)
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Physical Sciences P2/Fisiese Wetenskappe V2 6 DBE/November 2014
NSC/NSS - Memorandum
3.4 • Between propane molecules are London forces / dispersion forces /
induced dipole forces.
Tussen propaanmolekule is Londonkragte / dispersiekragte /
geïnduseerde dipoolkragte.
• Between propan-1-ol molecules are London forces / dispersion forces /
induced dipole forces and hydrogen bonds.
Tussen propan-1-ol molekule is Londonkragte / dispersiekragte /
geïnduseerde dipoolkragte en waterstofbindings.
• Hydrogen bonds / Forces between alcohol molecules are stronger or need
more energy than London forces / dispersion forces / induced dipole
forces.
Waterstofbindings / Kragte tussen alkoholmolekule is sterker of benodig
meer energie om oorkom te word as Londonkragte / dispersiekragte /
geïnduseerde dipoolkragte.
OR/OF
Between propane molecules are weak London forces / dispersion forces /
induced dipole forces and between propan-1-ol molecules are strong
hydrogen bonds.
Tussen propaanmolekule is swak Londonkragte / dispersiekragte /
geïnduseerde dipoolkragte en tussen propan-1-olmolekule is sterk (3)
waterstofbindings. [12]
QUESTION 4 / VRAAG 4
4.1
4.1.1 Substitution / chlorination / halogenation
Substitusie / chlorering / halogenering / halogenasie (1)
4.1.2 Substitution / hydrolysis
Substitusie / hidrolise (1)
4.2
4.2.1 Hydrogenation / Hidrogenasie / Hidrogenering (1)
4.2.2 H H H H H
H C C C + H H H C C C H
H H H H H H
Notes / Aantekeninge:
• Ignore/Ignoreer ⇌
• Accept H 2 if condensed. / Aanvaar H 2 as gekondenseerd.
• Any additional reactants and/or products
Enige addisionele reaktanse en / of produkte: Max./Maks. 2
3
• Accept coefficients that are multiples.
Aanvaar koëffisiënte wat veelvoude is.
• Molecular / condensed formulae
Molekulêre-/ gekondenseerde formule: Max./Maks. 2
3 (3)
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Physical Sciences P2/Fisiese Wetenskappe V2 7 DBE/November 2014
NSC/NSS - Memorandum
4.3 Cℓ H H Marking criteria / Nasienriglyne:
• Whole structure correct:/ Hele struktuur korrek: 2
H C C C H 2
• Only ONE Cℓ atom as functional group. / Slegs
H H H EEN Cℓ-atoom as funksionele groep. 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
4.4
4.4.1 Esterification / Condensation
Verestering / Esterifikasie / Kondensasie (1)
4.4.2 (Concentrated) H 2 SO 4 / (Concentrated) sulphuric acid
(Gekonsentreerde) H 2 SO 4 / (Gekonsentreerde) swawelsuur of swaelsuur (1)
4.4.3 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:
• If two or more functional groups/Indien twee of meer funksionele groepe: 0
2
• 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
4.4.4 Propyl ethanoate
Propieletanoaat (2)
4.5 Sulphuric acid / H 2 SO 4 / Phosphoric acid / H 3 PO 4
Swawelsuur / Swaelsuur / H 2 SO 4 / Fosforsuur / H 3 PO 4 (1)
[15]
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Physical Sciences P2/Fisiese Wetenskappe V2 8 DBE/November 2014
NSC/NSS - Memorandum
QUESTION 5 / VRAAG 5
5.1 ONLY ANY ONE OF/ SLEGS ENIGE EEN VAN:
• Change in concentration of products / reactants per (unit) time.
Verandering in konsentrasie van produkte / reaktanse per (eenheids)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 / getal mol/volume / massa van produkte of
reaktanse per (eenheids)tyd.
• Amount / number of moles / volume / mass of products formed or
reactants used per (unit) time.
Hoeveelheid / getal mol / volume / massa van produkte gevorm of
reaktanse gebruik per (eenheids)tyd. (2)
5.2
5.2.1 Temperature / Temperatuur (1)
5.2.2 Rate of reaction / Volume of gas (formed) per (unit) time
Reaksietempo / Volume gas (gevorm) per (eenheids)tyd (1)
5.3 • Larger mass / amount / surface area.
Groter massa / hoeveelheid / reaksieoppervlak.
• More effective collisions per (unit) time. / Frequency of effective collisions
increase./ More particles collide with sufficient kinetic energy & correct
orientation per (unit) time.
Meer effektiewe botsings per (eenheids)tyd. / Frekwensie van effektiewe
botsings verhoog./ Meer deeltjies bots met genoeg kinetiese energie &
korrekte oriëntasie per tyd(seenheid).
IF / INDIEN:
• Larger mass / amount / surface area.
Groter massa / hoeveelheid / reaksieoppervlak.
• More particles collide. / More collisions.
Meer deeltjies bots. / Meer botsings. Max./Maks. 2
3
Notes / Aantekeninge:
IF/INDIEN:
No reference to mass / amount / surface area in answer:
Geen verwysing na massa / hoeveelheid / reaksieoppervlak in antwoord: 0
3 (3)
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Physical Sciences P2/Fisiese Wetenskappe V2 9 DBE/November 2014
NSC/NSS - Memorandum
5.4 Marking criteria / Nasienriglyne:
Compare The reaction in exp. 1 is faster in exp. 1 than in exp. 2 due to the
Exp.1 with higher acid concentration.
Exp. 2: Die reaksie in eksp. 1 is vinniger as dié in eksp. 2 as gevolg van
Vergelyk die hoër suurkonsentrasie.
Eksp. 1 met Therefore the gradient of the graph representing exp. 1 is
Eksp. 2: greater / steeper than that of exp. 2. / Graph of Exp. 1 reaches
constant volume in shorter time than exp. 2.
Dus is die gradiënt van die grafiek wat eksp. 1 voorstel,
groter/steiler as dié vir eksp. 2. / Grafiek van exp. 1 bereik
konstante volume in korter tyd as dié vir eksp. 2.
Compare The reaction in exp. 3 is faster than that in exp. 1 due to the
Exp. 1 with higher temperature.
Exp 3 & 4: Die reaksie in eks. 3 is vinniger as dié in eksp. 1 as gevolg van
Vergelyk die hoër temperatuur.
Eksp. 1 met The reaction in exp. 4 is faster than that in exp. 1 due to the
Eksp. 3 & 4: higher temperature / larger surface area.
Die reaksie in eks. 4 is vinniger as dié in eksp. 1 as gevolg van
die hoër temperatuur / groter reaksieoppervlak.
OR/OF
Graph A represents exp. 4 due to the greater mass of CaCO 3 -
greater yield of CO 2 at a faster rate.
Grafiek A stel eksp. 4 voor as gevolg van die groter massa
CaCO 3 - groter opbrengs CO 2 teen vinniger tempo.
Therefore the gradient of the graphs of exp. 3 & 4 are
greater/steeper than that of exp. 1. / Graphs of Exp. 3 & 4
reaches constant volume in shorter time than exp. 1.
Dus is die gradiënte van die grafieke vir eksp. 3 & 4 is
groter/steiler as dié in eksp. 1./ Grafieke van exp. 3 & 4 bereik
konstante volume in korter tyd as dié vir eksp. 1.
Final answer
Finale C
antwoord (6)
Notes/Aantekeninge
• Compare exp. 1 with exp. 2 / Vergelyk eksp. 1 met eksp. 2:
o Factor & rate / Faktor & tempo.
o Gradient / volume CO 2 per time / gradient / volume CO 2 per tyd.
• Compare exp. 1 with exp. 3 / Vergelyk eksp. 1 met eksp. 3:
o Factor & rate / Faktor & tempo.
• Compare exp. 1 with exp. 4/ Vergelyk eksp. 1 met eksp. 4:
o Factor & rate / Faktor & tempo.
• Compare gradient / volume CO 2 per time of exp 1 with that of exp. 3 & 4
Vergelyk gradient/volume CO 2 per tyd van eksp 1 met die van eksp. 3 & 4
• Final answer / finale antwoord: C
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Physical Sciences P2/Fisiese Wetenskappe V2 10 DBE/November 2014
NSC/NSS - Memorandum
5.5 Marking criteria / Nasienriglyne:
• Divide volume by / Deel volume deur: 25,7
• Use ratio / Gebruik verhouding: n(CO 2 ) = n(CaCO 3 ) = 1:1
m
• Substitute / Vervang 100 in n = .
M
• Subtraction / Aftrekking.
• Final answer / Finale antwoord: 7,00 g to/tot 7,5 g
OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2
V Calculate mass of CO 2 :
n(CO 2 ) = Bereken massa CO 2 :
Vm
V
4,5 n(CO 2 ) =
= Vm
25,7
4,5
= 0,18 mol (0,175 mol) =
25,7
n(CaCO 3 ) = n(CO 2 ) = 0,18 mol = 0,18 mol (0,175 mol)
m
n(CaCO 3 ) = m
M n(CO 2 ) =
M
m
0,18 = m
100 0,18 =
∴ m = 18 g (17,5 g) 44
Ratio/
∴ m(CO 2 ) = 7,92 g (7,7043 g) verhouding
m(CaCO 3 ) not reacted/nie
7,92
gereageer nie): m(CaCO 3 needed / benodig ) = × 100
25 – 18 = 7,00 g (7,49 g) 44
= 18 g (17,5 g)
m(CaCO 3 not reacted/nie gereageer nie):
(Accept range: 7,00 g – 7,5 g) 25 – 18,00 = 7,00 g (7,49 g)
(Aanvaar gebied: 7.00 g – 7,5 g)
(Accept range: 7,00 g – 7,5 g)
(Aanvaar gebied: 7.00 g – 7,5 g) (5)
OPTION 3 / OPSIE 3 OPTION 4 / OPSIE 4
25,7 dm3 : 1 mol 100 g CaCO 3 → 25,7 dm3 CO 2
4,5 dm3 : 0,18 mol x g → 4,5 dm3 CO 2
∴ x = 17,51 g
100 g : 1 mol
x : 0,18 mol Mass not reacted/Massa nie gereageer nie
x = 18 g = 25 – 17,51
m(CaCO 3 not reacted/nie = 7,49 g
gereageer nie): (Accept range: 7,00 g – 7,5 g)
25 – 18 = 7,00 g (Aanvaar gebied: 7,00 g – 7,5 g)
(Accept range: 7,00 g – 7,5 g)
(Aanvaar gebied: 7,00 g – 7,5 g) (5)
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Physical Sciences P2/Fisiese Wetenskappe V2 11 DBE/November 2014
NSC/NSS - Memorandum
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.
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)
6.2 CALCULATIONS USING NUMBER OF MOLES
BEREKENINGE WAT GETAL MOL GEBRUIK
Mark allocation / Puntetoekenning:
• 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.
• Substitution of K C value / Vervanging van K C -waarde.
• Equilibrium concentration of both NO 2 & N 2 O 4 multiplied by 0,08 dm3.
Ewewigskonsentrasie van beide NO 2 & N 2 O 4 vermenigvuldig met 0,08 dm3
• Change in n(N 2 O 4 ) = equilibrium n(N 2 O 4 ) – initial n(N 2 O 4 )
Verandering in n(N 2 O 4 ) = ewewig n(N 2 O 4 ) – aanvanklike n(N 2 O 4 )
• USING ratio / GEBRUIK verhouding: NO 2 : N 2 O 4 = 2 : 1
• Initial n(NO 2 )= equilibrium n(NO 2 ) + change n(NO 2 ).
Aanvanklike n(NO 2 )= ewewig n(NO 2 ) + verandering n(NO 2 ).
• Final answer / Finale antwoord: 1,11 (mol)
Accept range/Aanvaar gebied: 1,11 – 1,12 (mol)
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Physical Sciences P2/Fisiese Wetenskappe V2 12 DBE/November 2014
NSC/NSS - Memorandum
OPTION 1 / OPSIE 1
[N O ]
K c = 2 42 No KC expression, correct substitution / Geen Kc-
[NO 2 ] uitdrukking, korrekte substitusie: Max./Maks. 7
8
[N O ]
171 = 2 42
(0,2) Wrong KC expression /Verkeerde Kc-uitdrukking:
Max./Maks. 5
∴ [N 2 O 4 ] = 171 x (0,2)2 8
= 6,84 mol∙dm-3 (8)
NO 2 N2O4
Initial quantity (mol)
1,11 0
Aanvangshoeveelheid (mol)
Change (mol)
1,094 0,55 ratio
Verandering (mol)
verhouding
Quantity at equilibrium (mol)/
0,016 0,55
Hoeveelheid by ewewig (mol)
Equilibrium concentration (mol∙dm-3) x 0,08
0,2 6,84
Ewewigskonsentrasie (mol∙dm-3)
OPTION 2 / OPSIE 2
[N O ] No KC expression, correct substitution /Geen Kc-
K c = 2 42
[NO 2 ] uitdrukking, korrekte substitusie: Max./Maks. 7
8
[N2 O 4 ]
171 =
(0,2)2 Wrong KC expression /Verkeerde Kc-uitdrukking:
Max./Maks. 5
∴ [N 2 O 4 ] = 171 x (0,2)2 8
= 6,84 mol∙dm-3
Equilibrium moles / Ewewigsmol:
n(N 2 O 4 ) = (6,84)(0,080)
= 0,55 mol
n(NO 2 ) = (0,2)(0,080) x 0,08 dm3
= 0,016 mol
n(N 2 O 4 formed/gevorm) = 0,55 – 0 = 0,55 mol
Ratio / Verhouding:
n(NO 2 reacted / gereageer) = 2n(N 2 O 4 formed/gevorm) = 2(0,55) = 1,094 mol
Initial / Aanvanklike n(NO 2 ) = 0,016 + 1,094 = 1,11 (mol) (8)
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Physical Sciences P2/Fisiese Wetenskappe V2 13 DBE/November 2014
NSC/NSS - Memorandum
OPTION 3 / OPSIE 3
NO 2 N2O4
Initial quantity (mol)
2x + 0,016 0
Aanvangshoeveelheid (mol)
Change (mol) ratio
Verandering (mol) 2x x verhouding
Quantity at equilibrium (mol)/
0,016 x
Hoeveelheid by ewewig (mol)
Equilibrium concentration (mol∙dm-3) x x 0,08 &
Ewewigskonsentrasie (mol∙dm-3) 0,2 0,08 ÷ 0,08
[N2 O 4 ]
Kc = No KC expression, correct substitution/Geen Kc-
[NO 2 ] 2
x uitdrukking, korrekte substitusie: Max./Maks. 7
8
0,08
171 = Wrong KC expression/Verkeerde Kc-uitdrukking:
(0,2)2
Max./Maks. 5
8
∴ x = 0,5472
∴ n(initial/aanvanklik) = 2(0,5472) + 0,016
= 1,11 mol
(8)
OPTION 4 / OPSIE 4
NO 2 N2O4
Initial quantity (mol)
x 0
Aanvangshoeveelheid (mol)
Change (mol) x − 0,016
Verandering (mol) x – 0,016 ratio
2 verhouding
Quantity at equilibrium (mol)/ x − 0,016
Hoeveelheid by ewewig (mol) 0,016
2
Equilibrium concentration (mol∙dm-3) x − 0,016
x 0,08 &
Ewewigskonsentrasie (mol∙dm-3) 0,2 0,16 ÷ 0,08
[N2 O 4 ]
Kc = No KC expression, correct substitution/Geen Kc-
[NO 2 ] 2
uitdrukking, korrekte substitusie: Max./Maks. 7
x − 0,016 8
0,16
171 = Wrong KC expression/Verkeerde Kc-uitdrukking:
(0,2)2
Max./Maks. 5
8
∴ x = 1,11 mol (8)
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Physical Sciences P2/Fisiese Wetenskappe V2 14 DBE/November 2014
NSC/NSS - Memorandum
CALCULATIONS USING CONCENTRATION
BEREKENINGE WAT KONSENTRASIE GEBRUIK
Mark allocation / Puntetoekenning:
• 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.
• Substitution of K C value. / Vervanging van K C -waarde.
• Change in [N 2 O 4 ] = equilibrium [N 2 O 4 ] – initial [N 2 O 4 ].
Verandering in [N 2 O 4 ] = ewewig [N 2 O 4 ] – aanvanklike [N 2 O 4 ].
• USING ratio/GEBRUIK verhouding: NO 2 : N 2 O 4 = 2 : 1
• Initial [NO 2 ] = equilibrium [NO 2 ] + change in [NO 2 ].
Aanvanklike [NO 2 ] = ewewigs [NO 2 ] + verandering in [NO 2 ].
• Equilibrium concentration of [NO 2 ] multiplied by 0,08 dm3.
Ewewigskonsentrasie van [NO 2 ] vermenigvuldig met 0,08 dm3.
• Final answer/Finale antwoord: 1,11 (mol)
Accept range/Aanvaar gebied: 1,11 – 1,12 (mol)
OPTION 5 / OPSIE 5
[N O ] No KC expression, correct substitution/Geen Kc-
K c = 2 42
[NO 2 ] uitdrukking, korrekte substitusie: Max./Maks. 7
8
[N2 O 4 ]
171 =
(0,2)2 Wrong KC expression/Verkeerde Kc-uitdrukking:
Max./Maks. 5
∴ [N 2 O 4 ] = 171 x (0,2)2 8
= 6,84 mol∙dm-3
NO 2 N2O4
-3
Initial concentration (mol∙dm )
13,88 0
Aanvangskonsentrasie (mol∙dm-3)
Change (mol∙dm-3) ratio
-3 13,68 6,84
Verandering (mol∙dm ) verhouding
Equilibrium concentration (mol∙dm-3)
0,2 6,84
Ewewigskonsentrasie (mol∙dm-3)
n(NO 2 ) = cV = (13,88)(0,08) = 1,11 mol (8)
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Physical Sciences P2/Fisiese Wetenskappe V2 15 DBE/November 2014
NSC/NSS - Memorandum
OPTION 6 / OPSIE 6
NO 2 N2O4
-3
Initial concentration (mol∙dm )
x 0
Aanvangskonsentrasie (mol∙dm-3)
Change (mol∙dm-3) x − 0,2 ratio
-3 x – 0,2
Verandering (mol∙dm ) 2 verhouding
Equilibrium concentration (mol∙dm-3) x − 0,2
Ewewigskonsentrasie (mol∙dm-3) 0,2 2
[N2 O 4 ]
Kc = No KC expression, correct substitution/Geen Kc-
[NO 2 ] 2
uitdrukking, korrekte substitusie: Max./Maks. 7
x − 0,2 8
171 = 2 Wrong KC expression/Verkeerde Kc-uitdrukking:
(0,2)2
Max./Maks. 5
8
∴ x = 13,88 mol⋅dm-3
n(NO 2 ) = cV = (13,88)(0,08) = 1,11 mol (8)
6.3
6.3.1 Concentration (of the gases) increases. / Molecules become more condensed
or move closer to each other.
Konsentrasie (van die gasse) verhoog. / Molekule word meer saamgepers of
beweeg nader aan mekaar. (1)
6.3.2 • Increase in pressure favours the reaction that leads to smaller number of
moles / volume of gas.
Toename in druk bevoordeel die reaksie wat tot die kleiner getal mol /
volume gas lei.
• Forward reaction is favoured. / Voorwaartse reaksie word bevoordeel.
• Number of moles/amount of N 2 O 4 / colourless gas increases.
Aantal mol/hoeveelheid N 2 O 4 / kleurlose gas neem toe.
OR / OF
Number of moles/amount of NO 2 / brown gas decreases.
Aantal mol/hoeveelheid NO 2 / bruin gas neem af. (3)
6.4
6.4.1 Darker / Donkerder (1)
6.4.2 Decreases / Verlaag (1)
[16]
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Physical Sciences P2/Fisiese Wetenskappe V2 16 DBE/November 2014
NSC/NSS - Memorandum
QUESTION 7/ VRAAG 7
PENALISE ONCE FOR THE INCORRECT CONVERSION OF UNITS.
PENALISEER EENMALIG VIR VERKEERDE OMSKAKELING VAN EENHEDE.
7.1
7.1.1 Ionises / dissociates completely (in water)
Ioniseer / dissosieer volledig (in water). (1)
7.1.2 NO 3 - / Nitrate ion / Nitraatioon (1)
7.1.3 pH = -log[H 3 O+] / -log[H+] Notes/Aantekeninge:
• If no/incorrect formula/Indien geen/foutiewe
= -log(0,3) formule: Max./Maks: 2
= 0,52 3
• If no substitution step: 2 marks for correct
answer./Indien geen substitusie stap: 2 punte
vir korrekte antwoord. (3)
7.2
7.2.1 n
c=
V
n
2=
0,1
∴ n(HCℓ) = 0,2 mol (3)
7.2.2 Burette / Buret (1)
7.2.3 B
- Titration of strong acid and strong base.
Titrasie van sterk suur en sterk basis.
OR/OF
The endpoint will be approximately at pH = 7 which is in the range of the
indicator.
Die eindpunt sal ongeveer by pH = 7 wees wat in die gebied van die indikator
is. (3)
7.2.4 The number of moles of acid in the flask remains constant.
Die getal mol van die suur in die fles bly konstant. (1)
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Physical Sciences P2/Fisiese Wetenskappe V2 17 DBE/November 2014
NSC/NSS - Memorandum
7.2.5 n
c=
V
n
0,2 =
0,021
n = 4,2 x 10-3 mol
n(HCℓ in excess/in oormaat):
n(HCℓ) = n(NaOH)
= 4,2 x 10-3 mol (3)
7.2.6 POSITIVE MARKING FROM QUESTION 7.2.1 AND 7.2.5.
POSITIEWE NASIEN VAN VRAAG 7.2.1 EN 7.2.5.
Marking criteria / Nasienriglyne:
• n(HCℓ reacted) = initial (from Q7.2.1) – excess (from Q7.2.5).
n(HCℓ reageer) = begin (van Q7.2.1) – oormaat (van Q7.2.5).
• Use mol ratio of acid: base = 2 : 1.
Gebruik molverhouding suur : basis = 2 : 1.
m
• Substitute / Vervang 40 into / in n =
M
m(MgO reacted / reageer )
• × 100 .
4,5
• Final answer / Finale antwoord: 87,11 %
OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2
n(HCℓ reacted/gereageer): n(HCℓ reacted/gereageer):
0,2 – 4,2 x 10-3 = 0,196 mol 0,2 – 4,2 x 10-3 = 0,196 mol
n(MgO reacted/gereageer): m
½n(HCℓ) = ½(0,196) n(HCℓ reacted/gereageer) =
M
= 9,8 x 10-2 mol m
0,196 =
36,5
m
n(MgO reacted/gereageer) = ∴ m(HCℓ reacted/gereageer) = 7,154 g
M
m 40 g MgO .............. 73 g HCℓ
∴ 0,098 =
40 x g MgO ................... 7,154 g
∴ m = 3,92 g ∴ x = 3,92 g
3,92 3,92
% purity/ suiwerheid = × 100 % purity/suiwerheid = × 100
4,5 4,5
= 87,11% = 87,11%
(Accept range: 87 - 87,11 %.) (Accept range: 87 - 87,11 %.)
(Aanvaar gebied: 87 – 87,11 %) (Aanvaar gebied: 87 – 87,11 %) (5)
[21]
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Physical Sciences P2/Fisiese Wetenskappe V2 18 DBE/November 2014
NSC/NSS - Memorandum
QUESTION 8 / VRAAG 8
8.1 • Pressure: 1 atmosphere (atm) / 101,3 kPa / 1,013 x 105 Pa
Druk: 1 atmosfeer (atm) / 101,3 kPa / 1,013 x 105 Pa
• Temperature/Temperatuur: 25 °C / 298 K (2)
8.2 • Platinum is inert / does not react with the H+ ions OR acid.
Platinum is onaktief / reageer nie met die H+-ione OF suur nie.
• Platinum is a conductor (of electricity).
Platinum is 'n geleier (van elektrisiteit). (2)
8.3
8.3.1 Salt bridge / Soutbrug (1)
8.3.2 -0,31 V (1)
8.3.3 2H+ + 2e- → H 2
Marking guidelines / Nasienriglyne:
• 2H+ + 2e- ⇌ H 2 1 H 2 ⇌ 2H+ + 2e- 0
2 2
H 2 ← 2H + 2e
+ - 2 H 2 → 2H + 2e
+ - 0
2 2 (2)
8.4
8.4.1 POSITIVE MARKING FROM QUESTION 8.3.2.
POSITIEWE NASIEN VAN VRAAG 8.3.2.
θ θ θ
E cell = E reduction − E oxidation Notes / Aantekeninge:
Accept any other correct formula from the
2,05 = -0,31 – E Mθ / M2 + data sheet.
E Mθ / M2 + = -2,36 (V) Aanvaar enige ander korrekte formule vanaf
gegewensblad.
M is magnesium/ Mg. Any other formula using unconventional
θ θ θ
abbreviations, e.g. Ecell = EOA − ERA followed
Option 2/ Opsie 2
2+ - by correct substitutions: 4
M→ M + 2e E° = 2,36 (V) 5
X + 2e → X E° = -0,31 (V) Enige ander formule wat onkonvensionele
2+ -
θ θ θ
E° = 2,05 V afkortings gebruik bv. E sel = E OM − E RM
θ
Thus/Dus: E reduction = - 2,36 (V) gevolg deur korrekte vervangings: 4
5
M is magnesium/ Mg.
Notes / Aantekeninge
Give mark for Mg / magnesium ONLY if concluded from -2,36 V.
Ken punt vir Mg / magnesium slegs toe indien afgelei uit -2,36 V (5)
8.4.2 Exothermic / Eksotermies (1)
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Physical Sciences P2/Fisiese Wetenskappe V2 19 DBE/November 2014
NSC/NSS - Memorandum
8.5 The cell reaction reaches equilibrium.
Die selreaksie bereik ewewig.
Notes / Aantekeninge:
Accept: One or more of reactants are used up. / The cell reaction has run to
completion.
Aanvaar: Een of meer van reaktanse word opgebruik. / Die selreaksie het volledig
verloop. (1)
[15]
QUESTION 9 / VRAAG 9
9.1 Electrolytic / Elektrolities (1)
9.2 Q& T Notes / Aantekeninge:
IF more than TWO electrodes, mark first two.
Cu2+ + 2e → Cu
Indien meer as TWEE elektrodes, sien eerste twee na.
Marking guidelines / Nasienriglyne
Cu2+ + 2e ⇌ Cu (1 ) Cu → Cu2+ + 2e- (0 )
2 2
Cu ← Cu2+ + 2e (2 ) Cu ⇌ Cu2+ + 2e- (0 )
2 2 (4)
9.3
9.3.1 Cℓ 2 / chlorine (gas) / chloor(gas) (1)
9.3.2 Cu2+ (ions) / copper(II) ions / CuCℓ 2 / copper(II) chloride
- Cu2+ (ione) / koper(II)-ione / CuCℓ 2 / koper(II)chloried (1)
9.4 Cu is a stronger reducing agent than Cℓ- (ions) and Cu will be oxidised
(to Cu2+).
Cu is 'n sterker reduseermiddel as Cℓ-(-ione) en Cu sal geoksideer word
(na Cu2+).
OR/OF
Cℓ- (ions) is a weaker reducing agent than Cu and Cu will be oxidised
(to Cu2+).
Cℓ-(-ione) is 'n swakker reduseermiddel as Cu en Cu sal geoksideer word
(na Cu2+). (3)
[10]
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Physical Sciences P2/Fisiese Wetenskappe V2 20 DBE/November 2014
NSC/NSS – Memorandum
QUESTION 10 / VRAAG 10
10.1
10.1.1 Nitrogen / N 2 / Stikstof
Hydrogen / H 2 / Waterstof (2)
10.1.2 NH 3 + HNO 3 → NH 4 NO 3 Bal.
Notes / Aantekeninge:
• Reactants Products Balancing:
Reaktanse Produkte Balansering
• Ignore double arrows. / Ignoreer dubbelpyle.
• Marking rule 6.3.10. / Nasienreël 6.3.10. (3)
10.2 Marking criteria / Nasienriglyne:
3
• Use ratio / gebruik verhouding:
9
• x 20 kg
• x 36 / 36 %
• Final answer / Finale antwoord: 2,4 kg
OPTION 1 / OPSIE 1: OPTION 2 / OPSIE 2: OPTION 3 / OPSIE 3:
3 m(nutrients/voedingstowwe): m(N):
% N = (x 36) 36
9 (x 20) = 7,2 kg
= 12 % 100 3 36
× (× 20) (× ) = 2,4 kg
12 9 100
∴ m(N) : (× 20 kg) 3
100 ∴ m(N) = x 7,2
= 2,4 kg 9
= 2,4 kg (4)
[9]
TOTAL/TOTAAL: 150
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