# Physical Sciences Grade 12 Paper 2 Gauteng Mock 2020 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng Mock Exam 2020, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2020-afr-eng-p2-gauteng-chemie-memo/adef4c95-801f-48da-8c7f-fecd3bcf4540
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2020
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 12 · Gauteng Mock Exam · 2020. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

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So) GAUTENG PROVINCE

FPCAT ION
2 REPUBLIC OF SOUTH AFRKA

GAUTENG DEPARTMENT OF EDUCATION
GAUTENGSE DEPARTEMENT VAN ONDERWYS
PREPARATORY EXAMINATION
VOORBEREIDENDE EKSAMEN

2020

MARKING GUIDELINES
NASIENRIGLYNE

10842

PHYSICAL SCIENCES: CHEMISTRY
FISIESE WETENSKAPPE: CHEMIE

PAPERI/IVRAESTEL 2

12 pagesibladsye

P.T.O.

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QUESTIONIVRAAG 1
11 Dvv (2)
12 AvY (2)
1.3 AYY (2)
14. Cv (2)
1.5 D¥vv¥ (2)
16 Avv (2)
170 Cvv (2)
18 ByY (2)
19 Avy (2)
110 Dvv (2)
[20]

QUESTION/VRAAG 2
2.1.1 A orlofC ¥ (1)
212 BY (1)
2.1.3 Fv (1)
2.1.4 DY (1)
2.2.1 4,5-dimethyl “hex-2-ene v /4,5-dimetietheks-2-een

Marking criteria 2,.2,1-2.2.3/Nasienriglyne:

Only functional group correct/Slegs funksionele groep korrek: Max/Maks: %B

Whole structure correct/Hele struktuur korrek: % (2)
2.2.2 2,3-dibromo-5-methyl “heptane ~

2,3-dibromo-5-metietheptaan (2)
2.2.3 4-methyl “’pent-2-yne ¥

4-metielpent-2-yn {2)

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EET
H ; c—c i ck he aan: vv
H H H H H HoH (2)
Marking criteria/Nasienriglyne
¢ Only functional group correct/Sfegs funksionele groep korrek:
Max/Maks: Ne
%
* Whole structure correct:/hele struktuur korrek 72
2.3.2 Propanoic acidv ¥
Propanoésuur (1)
2.3.3 Sulphuric acid/H2SO4 ¥
Swawelsuur/ H2SO4 (1)
[14]

QUESTION/VRAAG 3

3.1 The temperature at which the solid and liquid phase are at equilibrium vv”

Die temperatuur waarby die vaste stof en die vioeistof fase in ewewig is. ““ (2)

3.2 C Vv It has a lower vapour pressure ¥

loer:
¥ Cv Dithet ndampdruk ¥, Sonche oun © (2)
3.3.1. Organic compounds with the same molecular formula, but different structural
formulae. ¥v ¥
2 oF zero

Organiese verbindings met dieselfde molekulére formule, maar verskillende
struktuurformules. “ “
2 of nul (2)

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3.3.2

3.4

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Compound B heptane is less branched/has a longer chain/less
compact/less sphericai/has a larger surface area.

Stronger intermolecular forces/Van der Waals forces/dispersion
forces/London forces/Induced dipole- induced dipole forces. “

More energy needed to overcome intermolecular forces”

OR

Compound A/2-methyhexane is more branched/more compact/more
spherical/has shorter chain/has a smaller surface area. “

It has weaker intermolecular forces/Van der Waals forces/dispersion

forces/London forces/Induced dipole- induced dipole forces. V
Less energy needed to overcome intermolecular forces. “

Verbinding B, heptaan, is minder vertak/het jn langer ketting/minder

kompak/minder sferies/ het 'n groter opperviak area.

Sterker intermolekulére kragte/Van der Waals
kragte/dispersiekragte/_ondonkragte/geinduseerde kragte. “
Meer energie benodig om intermolekulére kragte te oorkom._ “

OF

Verbinding A/2-metielheksaan is meer vertak/meer kompak/meer
sferies/het 'n korter kettinglengte/et 'n kleiner opperviak area. “

Dit het 'n swakker intermolekulére krag/Van der Waals

kragte/dispersiekragte/Londonkragte/geinduseerde dipoolkragte. ~“
Minder energie word benodig om die intermolekulére kragte te
oorkom. _“ (3)

Both compounds C and D have hydrogen bonding between
molecules. “Compound C has one site for hydrogen bonding whilst
compound D has two sites for hydrogen bending “

OR

Compound C has less sites for hydrogen bondingAweaker hydrogen
bonding than compound D.

More eneray needed to overcome intermolecular forces in compound
Dfless energy needed to overcome intermolecular forces in
compound E. v

Beide verbindings C en D het waterstofbindings tussen die molekule.
” Verbinding C het een plek vir waterstofbinding terwy! verbinding D

twee plekke het vir waterstofbinding. “
OF

Verbinding C het minder plekke vir waterstofbinding/swakker
waterstofbinding as verbinding D.

Meer energie word benodig om die intermolekulére kragte te oorkom

in verbinding D/minder energie word benodig om die intermolekulére

kragte in verbinding Ete oorkom. ~ (3)
[12]

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QUESTION/VRAAG 4
4.1.1 It contains a double bond/C=C between two carbon atoms in its
hydrocarbon chain/All carbon atoms not bonded to the maximum
number of atoms/four atoms¥
Daar is dubbelbindings/C=C tussen twee koolstofatome in die
koolwaterstof ketting/Alle koolstofatome is nie gebind aan die
maksimum aantal atome/vier atome nie. “ (1)
4.1.2 (a) Addition/Bromination ¥
Addisie/Broominasie (1)
(b) Substitution
Substitusie (1)
4.1.3 Heat/sunlight/ultraviolet light “
Hitte/sonlig/ultraviolet lig (1)
4.1.4 Butanev/butaan (1)
4.1.5 Hydrogen bromide/HBrY
Waterstofbromied/ HBr (1)
4.2.1 Elimination”
Eliminasie (1)
4.2.2 Butv-2-enev
But-2-een (2)
423
v v
i! iy v Hf lew ¥
Heo H+ NaOH ——* H-G— E7o-H + NaCt
HH H HH H (6)
[15]

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QUESTION/VRAAG 5
5.1 Carbon dioxide/CO2 ¥
Koolstofdioksied/CO2 (1)

5.2.1. The decrease in concentration of hydrochloric acid per unit time. “”

Die afname in konsentrasie van waterstofchloried per eenheid tyd.
wv (2)

5.2.2 Concentration (of the acid) ¥

Konsenirasie (van die suur) (1)
5.2.3 Fora fair test/comparison “

Vir n regverdige toets/of vergelyking “ (1)

5.3.1. e Higher acid concentration in experiment 2 means more
particles/molecules per unit volume”
e More particles have kinetic eneray equal to or greater than
activation energy/More particles have enough kinetic energy ¥
e More effective collisions per unit time/frequency of effective
collisions increases
¢ Rate of effective collisions increases. “

Hoér konsentrasie suur in eksperiment 2 beteken meer

deeltjies/molekule per eenheid volume”

Meer deeltjies het kinetiese energie gelyk of groter as die
aktiveringsenergie/Meer deeltjies het genoeg kinetiese energie. v”

Meer effektiewe botsings per eenheid tyd/frekwensie van effektiewe
botsings verhoog.

Tempo van effektiewe botsings verhoog. (4)

5.3.2 n(CaCOs) = mM Marking Criteria

* Subsiituting 100 into formula m/M

e Using ratio 2:1

e Substituting the number of moles into
formula c= nv

= 4/100 = 0,04 mol

n(HCl) : n(CaCOs) = 2: 1

n(HCl) = 2(0,04) ¥ = 0,08mol f e Final answer —
i Nasienriglyne:
¢(HCl) = nv | © Vervang 100 in formule m/M
© Gebruik verhouding 2:1
0.4 = 0,08/VY e Vervang geial mol in formule c=n/V
e Finale antwoord
V=0,2 dm? = 200 cem3v |

(4)

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5.4

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Activated complex
Geaktiveerda kompleks

vom v
Shape

Potential energy
Potensi#le energie

Reaksle kodrdinate
Reaction coordinates

QUESTIONIVRAAG 6

6.1

6.2

6.3
6.3.1

6.3.2

The reaction is reversible. “
Die reaksie is omkeerbaar. “

This reaction is in contact with a catalyst, and therefore is known as the
contact process. ¥

Hierdie reaksie is in kontak met 'n katalisator en daarom staan dit bekend as
die Kontakproses. “

If the temperature is too low the reaction rate will be too slow for the yield to
be enough. Vv

Indien die temperatuur te laag is sal die reaksietempo te stadiq wees vir die
opbrengs om genoeg te wees, ““

A high temperature favours an endothermic reaction ¥
If the temperature is high the reverse reaction will be favoured ”
Less product will be formed which is not favourable.

in Hoé temperatuur bevoordeel die endotermiese reaksie.

Indien die temperatuur hoog is word die teruqwaarise_reaksie bevoordee!, “”
Minder produkte sal vorm en dit is nie voordelig nie. “

(4)
[17]

(1)

()

(2)

(3)

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6.4

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Marking guideline/Nasienriglyne:
e Use the Ratio/Gebruik ratio

© N(SO2)eqewe = n(SO2)initiatbegin — An(SO2)

N(O2)eqewe = x — An(O2)

Divide equilibrium moles by 200 dm3/Deel ewewigsmol deur 200 dm3V

v

°
e Correct Ke expression (formulae in square brackets)“
Korrekte Ke uitdrukking(formules in blokhakies)

* Substitution of values in expression/Vervanging van waardes in

uitdrukking ¥

e Substitution of/Vervanging van 32 g:mol' in m = nM ¥

e Final answer/Finale antwoord: 891,179 “
e Range/Gebied: 891,167 - 891,2

SOz2 O2 SOz
Initial amount(moles)
Aanvangs hoeveelheid 50 x 9
Change in amount(moles) 22 io’
Verandering in hoeveelheid 22 " Ratio
Equilibrium amount(moies)
Hoeveelheid 28 xo 22
Equilibrium concentration(mol-dm) (x11) Divide by
Ewewigkonsentrasie(mol-dm*) 0,14 200 011 200 dm3v’
_ _ [so3]2 ¥
a
isogl* 0 ]
0,11
7,328 = me m=nxM
14) oa) = (278,849)(32)v
x = 27,849 mol = 891,167g ¥
a ne! (7)
Forward ¥ “/ Voorwaarts (2)
[16]

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QUESTION/VRAAG 7

7.1
7.1.4

7.1.2

7.13
7.14

7.156

Acid-base reaction/neutralisation/protolysis
Suur-basis reaksie/ neutralisasie/protoliese ~~

Barium hydroxide
Bariumhidroksied

X-Burette “  Buret
A white precipitate in a yellow solution that gradually turns orange Vv
‘n Wit neerslag in n geel oplossing wat geleidelik oranje word. v~

(a) ¥ Remains constant Barium Hydroxide is a strong base that will
dissociatefionise from the start

Bly konstant. Bariumhidroksied is in sterk basis wat sal
dissosieer/ioniseer van die begin af.

(b) Decrease “ Verminder

(c} Decrease “ Verminder

Marking quidelines/Nasienriglyne
e Any formulae/Enige formule: “

Division by Volume ~/ dee/ deur volume
Use mole ratio “ gebruik molverhouding
Use M = 233 g:mol* inm = nM “
Answer = 1,165 g “

10842/20

(1)

(1)
(1)

(3)

OPTION 1: OPSIE 1 OPTION 2: OPSIE 2
Cae v Ma v c=
CoVp My
cqa(30) 1 O1=
== v ,
1)G60) 1 0,005 mol

Cq = 0,167mol « dm-3
n

c=-

|ast=

S
oS
oT

Vv
n

0,167 = d03
n = 0,005 mol

n(BaSO,) =n(Ba(OH), v
= 0,005 mol
m=nxM
= (0,005)(233)
=1,1659 “

(5)

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7.2
7.2.1 NHa*(aq) + H20(é) > NHaY + Hs0* ¥

7.2.2 CHsCHOO-(aq) + H2O(é) - CHsCOOHY + OHY

QUESTION/IVRAAG 8

8.1 A galvanic/voltaic cellv
Galvaniese sel/voltaiese sel

82 Ags) | Ae?*(aq)(1 mol-dm) V1 Cu2*(aq)(1 mol-dm) | Cu(s) “

Accept Marking criteria: _Nasienriglyne;
i e Oxidation “ Oksidasie
Atl Ags [| Cu2* ICu i ° Saltbridge” soutbrug
e Reduction’ reduksie

10842/20

(2)

(2)
[17]

(1)

(3)

83 AétoCu”
Aéna Cu” (1)
8.4 AL> AS + 3a Vv
Marking quidelines/Nasienrigiyne
1 0
© Af 2AB+3e° Ms AG +3e 2 Ae %
ie}
Ags +3e° Al % Ag +3e AL %
e Ignore if charge omitted on electron./ignoreer indien lading weggelaat
op elektron.
e If charge (+) omitted on Aé* /indien lading (+) weggelaat op Aé*:
Example/Voorbeeld: A¢é > AG +3eV¥
Max./Maks: he _ ;
. — my
i Notes (2)
_ e Accept any other correct formula from the
8.5 E°cat = EP reduction = E°oxidation 7 data sheet./Aanvaar enige ander korrekte
= 0,34 — (-1,66) ¥ formule vanaf gegewensbiad.
=2Vv i (4)

£

j e Any other formula using unconventional
' abbreviations, e.g. E°cay = E°oa- E°ra

j followed by correct substitutions:/Enige
i ander formule wat onkanvensionele

i afkortings gebruik bv. E°sei =.3/4

i e E*ou- E°rmgevolg deur korrekte
'
]
i

vervangings: ¥%,

10

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8.6
8.6.1 Decrease” Verminder (1)

8.6.2 NEGATIVE MARKING FROM 8.6.1/NEGATIEWE NASIEN VANAF 8.6.7

The concentration of Aé*? will increase. According to Le Chatelier the system
needs to decrease the concentration. ¥

The reverse reaction will be favoured. ¥

Therefore the reading will decrease. ¥

Die konsentrasie van AZ sal vermeerder. Volgens Le Chatelier sal die
sisteem die konsentrasie moet verlaag. ~

Die terugwaartse reaksie sal bevoordeel word. “”
Dus sal die lesing verminder. ~ (3)

8.7 Chemical energy to electrical energy”

Chemiese potensiéle energie na elektriese energie. “ (1)
[16]

QUESTION/VRAAG 9
91 Av (1)
92 Cu%+ 2e >CuvY (2)

A 93 Marking quidelines/Nasienriglyne
« Multiply nx2 or mx2 “ = Vermenigvuldig nx2 of mx2

© Maeposite = 31,759
© Miostat anode = (31,75 + 15,8)g9 %
e Answer/antwoord 66,77% ¥

n= >
M ‘Tost at anode

v
Ze, 1,75

(0.25% 2)" se = 2 e100

= 3175 (31,75 + 15,8)

Ce = 66,67% “

0,25 mol Cu formed in 1 800 s so in 3 600s (2x0,25) mol formed
0,25 mol Cu vorm in 7 800 s dus in 3 600 s (2x0,25) moi gevorm (4)

ue Ypurity = —~seeeste4 5100

9.4,1 Electrolyte is a solution/liquid/dissolved substance that conducts electricity
through the movement of ions (2 or 0)

Elektroliete is 1n oplossing/vioeistof/opgeloste stof wat elektrisiteit gelei deur
die beweging van ione (2 of 0) (2)

9.4.2 The rate of oxidation is equal to the rate of reduction. ”

Die tempo waarteen oksidasie plaasvind is gelyk aan die tempo van

reduksie. “ (1)
[10]

11

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QUESTION/VRAAG 10

10.1
10.1.1

10.1.2

10.1.3

10.1.4

10.2

Nitrogen ¥ stikstof
Phosphorus ” fosfor
Potassium ” kalium

Acceptlaanvaar
N:P:K (3)

Potassium/KY
Kalium/ K (1)

Phosphates/P “
Fosfate/P (1)

Haber orfof Ostwald Vv (1)

Eutrophication: The process by which an ecosystem eg. A river or dam,
becomes enriched with inorganic nutrients”, especially phosphorus and
nitrogen, resulting in excessive plant growth ”

Eutrofikasie: Die proses waarby 'n ekosisteem bv. 'n rivier of dam, verryk
word met anorganiese voedingstowwe v, veral fosfor en stikstof, wat 'n
oorgroot toename in plantegroei tot gevoighet. ~“ (2)

Marking quidelines/Nasienriglyne

* m(N) = 2x36 ¥

¢ Double up on Fertilizer BY verdubbel kunsmis B
« Fertilizer A m(N) = 24kg, m(P)=m(K) =6 kg
* Fertilizer B m(N) = 19,09kg, m(P)=m(K) = 5,45 kg
« Answer Fertilizer BY
Fertilizer Al/kunsmis A + Fertilizer Bilkunsmis B
In 100kg 4.1.1(36) In 50 kg 7.2.2(15)
m(N) = £236 + In 100kg 14.4.4(30)
=24kg “ - m(N) = #4230
m(P)=m(K) = 2x36 v = 19,09 kg
=6kg op m(P}=m(k) = Sx30
= 5,45 kg v

Therefore Fertilizer B will be the best “

Kunsmis B sal die beter een wees. (5)
[13]

TOTAL/ITOTAAL: 150

12

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