# Physical Sciences Grade 12 Paper 2 North West Mock 2018 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the North West Mock Exam 2018, 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/nw-2018-paper-2-memorandum/54f8dae1-e861-4113-9ad0-4c36cc44ccc4
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 11

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## About this document

Physical Sciences · Grade 12 · North West Mock Exam · 2018. Memorandum, 11 pages. Read online or download the PDF.

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Education and Sport Development

Department of Education and Sport Development
Departement van Onderwys en Sportontwikkeling
Lefapha la Thuto le Tlhabololo ya Meishameko

NORTH WEST PROVINCE

NATIONAL SENIOR CERTIFICATE
NASIONALE SENIOR SERTIFIKAAT

" PHYSICAL SCIENCES: CHEMISTRY (P2))
: FISIESE WETENSKAPPE: CHEMIE (V2) t
: SEPTEMBER 2018 :
7 MEMORANDUM ii
< 4

MARKS/PUNTE: 150

This memorandum consists of 15 pages./
Hierdie memorandum bestaan uit 15 bladsye.

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Physical Sciences P2/Fisiese Wetenskappe V2 2 NW/September 2018
NSC/NSS — Memorandum

QUESTION 1/VRAAG 7

11 OD (2)
12 C (2)
13° Bs. (2)
14 C (2)
15 OD (2)
16 &G (2)
1.7 OD (2)
18 4A (2)
1980,D (2)
1.10 B (2)
[20]
QUESTION 2/VRAAG 2
2.1. Molecules with the same molecular formulae, but different types of
chains. ¥v ¥
Molekule met dieselfde molekulére formule, maar verskillende tipes
kettings. (2)
2.2
2.2.1. DY and/enF ¥ (2)
2.2.2 2,2 —dimethyl propane /2,2—dimetielpropaan VV (2)
2.3
2.3.1 Saturation Test/versadigingstoets V (1)
2.02 shih INC salogenation/Bromination, v
Htersie/halogenering/bromering (1)
Relclisie
2.3.3 Cv; has a double bond between C-atoms”/unsaturated.
C; het ‘n dubbelband tussen C-atomelonversadigd (2)

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Physical Sciences P2/Fisiese Wetenskappe V2 3 NW/September 2018
NSC/NSS — Memorandum
2.4
H H H H it
Fae et ah ooops 1 ~
H i | i Cc Cc Oo—C H
i
H H 4H H H
Methyl pentanoate /Metielpentanoaat Vv (4)
[14]
QUESTION 3/VRAAG 3
3.1. The pressure exerted by a vapour at equilibrium with its liquid in a closed
system. VW
Die druk uitgeoefen deur ‘n damp wat in ewewig is met sy vioeistoffase,
in ‘n geslote sisteem. (2)
3.2. Butan-1-ol. ¥ (1)
3.3 No side chains _
3.3.1 Use straight chainY primary alcohols. ¥
Gebruik reguitkettingY primére alkohole ¥ (2)
Letseny sy h hags
3.3.2
a) Structure
Chain length/number of carbon atoms in the chain/molecular
size/molecular mass/surface area increases from top to bottom
or from butan-1-ol to hexan-1-ol. %
Struktuur
Kettinglengte/aantal koolstofatome in die ketting/molekulére
grootte/molekulére massa/kontakopperviak neem toe van bo na
onder of van butan-1-ol tot by heksan-1-ol.
b) Intermolecular forces
Intermolecular forces/Van der Waals forces/ London forces/
Dispersion forces increase from top to bottom or
from butan-1-ol to hexan-1-ol. Vv
Intermolekulére kragte
Intermolekulére kragte/ Van der Waalskragte/ Londonkragte/
Dispersiekragte neem toe van bo na onder of van butan-1-ol tot
by heksan-1-ol.
c) Energy
Energy needed to overcome/break intermolecular forces
increases from top to bottom or from butan-1-ol to
hexan-1-ol. ”
Energie
Energie nodig om intermolekulére kragte te oorkom/breek, neem
toe van bo na onder of van butan-1-ol tot by heksan-1-ol. (3)
3.4 Remains the same. Bly dieselfde. v (1)

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Physical Sciences P2/Fisiese Wetenskappe V2 4 NW/September 2018
NSC/NSS — Memorandum

ae

Marking criteria:
e Mole ratio for V(COz) correctly used.
Molverhouding vir V(CO2) korrek gebruik.
° Mole ratio for V(H2O) correctly used.
Molverhouding vir V(H2O) korrek gebruik.
° Mole ratio for V(Oz reacted) correctly used.
Molverhouding vir V(Oz gereageer) korrek gebruik.
e V(O2 excess) = V(Oz initial) - V(O2 change)
V(O2 oormaat) = V(Oz oorspronklik) - V(O2 verandering)
ad Viotai/totaat = 95 cm?

OPTION 1/OPSIE 1

V(COz) = 4V(CaHi0) = V(H20) = 5V(C4Hio) ~~ V(Oz reacted/gereageer) = = V(CiHro)

= (10) ¥ = 5(10) v = > (10) ¥
= 40 = 50 =65 cm?
(Oz excess/oormaat) = 70 — 65Y
ve eee =5cm*
V totavtotaal = 40 + 50 + 5
=95cm> v (5)
OPTION 2 /OPSIE 2
C4Hi0 Oz COz H20
Initial/Aanvangs V(cm*) 10 70 0 )
Change in/Verandering in V(cm*) 10 50 Vv 40Vv 50 v
Final/Finale V(cm?) 0 5v 40 50
V totavtotael =40+50+5=95cm®> Vv [14]
QUESTION 4/VRAAG 4
Af
4.1.1 Addition/Hydrogenation. ¥
Addisie/hidrogenering (1)
4.1.2 Substitution/halogenation/chlorination. ~
Substitusie/halogenering/chlorinering (1)
4.1.3 Elimination/dehydration. “
Eliminasie/dehidrasie (1)
4.2. 2-bromo propane/2-bromopropaan VV (2)

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Physical Sciences P2/Fisiese Wetenskappe V2 5 NW/September 2018
NSC/NSS ~ Memorandum

4.3

4.3.1 Dehydrohalogenation/dehydrobromination. ¥
Dehidrohalogenering/dehidrobromering (1)

4.3.2 Hot” ethanolic strong base.
Warm etanoliese sterk basis (2)
OR/OF
- Concentrated strong base/NaOQH/KOH
Gekonsentreerde sterk basis/NaQH/KOH
OR/OF
- Strong base with no water
Sterk basis met geen water nie
OR/OF
- Strongly heated or hot base
Sterk verhitte of warm basis
OR/OF
- High temperature/heat strongly
Hoé temperatuur/verhit sterk
44

4.4.1 H2O/H2SO4/H3PO4 (1)

H
Be +HCRY >
|

iO Lt

7
!

Hv
hg + H—O ¥
i

(4)

e Accept HC£, H2O0 and OH/ Aanvaar HCf, H20 en OH
e Incorrect balancing — max A

Foutiewe balansering — maks cs
e Molecular/condensed formulae — UA

Molekulére/gekondenseerde formules — 2 ty

e Accept coefficients that are multiples
Aanvaar balanseringsgetalle wat veelvoude is

e Any additional reactant or/and product -¥,

Enige addisionele reactant en/of produk — ¥,

4.5

4.5.1 A reaction in which small molecules join to form very large molecules
by adding on at double bonds. YW
‘n Reaksije waarin klein molekule verbind om baie groot molekule te
vorm deur byvoeging van dubbelbindings. (2)

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Physical Sciences P2/Fisiese Wetenskappe V2 6 NW/September 2018
NSC/NSS — Memorandum

4.5.2 Freezer bags/soft bottles/bendable cables/shopping bags/plastic
crates/plastic pipes/toys. V
Vriessakkies/ sagte bottels/ buigbare kabels/inkopiesakke/plastiek-
kratte/ plastiekpype/ speelgoed (1)
[16]

QUESTION 5/VRAAG 5

5.1 _ ONE/ENIGE EEN:
* Change in concentration of products/reactants per (unit) time. “”
Verandering in konsentrasie van produkte/reaktante per (eenheid) tyd.
* Rate of change of concentration.
Tempo van verandering in konsentrasie.
* Change in amount/number of moles/volume/mass of products or reactants
per (unit) time.
Verandering in hoeveetheid/getal mol/volume/massa van produkte of
reaktante per (eenheid) tyd.
* Amount/number of moles/volume/mass of products or reactants used
per (unit) time.
Hoeveelheid/getal mol/volume/massa van produkte of reaktante gebruik
per (eenheid) tyd. (2)

5.2 Reaction rate/Reaksietempo ¥ (y

5.3 Powder/poeier ¥

Powder has a larger surface area Vand there will be more effective collisions
per unit time. W The reaction rate will increase.
Poeier het ‘n groter kontakopperviakte en meer effektiewe botsings per

tydseenheid vind plaas. Die reaksietempo sal verhoog. (3)
om
5.4 Nw
N= a m\
a i
Spt
15), ico
100 ey |
ZO, CIS Me
= 0,015 mot
lin chal I) tvs
m (CO2) = nM
= (0,015)(44v’)
= 0,66 g¥ (4)
5.5.1 The rate of the reaction will increase /
Die tempo van die reaksie sal verhoog. VV (2)

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Physical Sciences P2/Fisiese Wetenskappe V2 4 NW/September 2018
NSC/NSS — Memorandum

5.5.2 Increasing temperature increases the average kinetic energy (of the
reacting particles). v More effective collisions per unit time/Frequency of
effective collisions increases.v, thus increase the rate of the reaction. v/
Die verhoogde temperatuur verhoog die gemiddeide kinetiese energie (van
die partikels wat reageer). Meer effektiewe botsings per eenheidstyd/
Frekwensie van effektiewe botsings neem toe, dus verhoog die tempo van
die reaksie. (3)

[15]
QUESTION 6/VRAAG 6
6.1 0,5 mol X2Y3 ¥ (1)

6.2 OPTION 1/OPSIE 1

2 mol X + 3 mol Y yields/lewer 1 mol X2.Y3
Thus/dus 1 mol X + 1,5 mol Y yields/fewer 0,5 mol X2Y3_ J”

After/Na 50 s:
4-1 =3molXv
4—1,5 =2,5 mol YY
0,5 mol X2Y¥3

n
c=—:
Vv

3
Vv 5
2 i 2 2

= 1,5 mol.dm® = 1,25 mol.dm® = 0,25 mol.dm®

[xX] =

XPLY}?

0:25
(1,5)°(1,25)°

= 0,057 V (7)

DaYa]

OPTION 2/OPSIE 2

2X(g)| 3¥(g) | X2Ya(g)

Initial moles/Aanvangsmo! 4 4 0

Moles reacted/mo! gereageer 1 1,5 0,5 ¥

Moles at equilibrium/mol by ewewig 3v 2,54 0,5

Equilibrium concentration/ 1,5 1,25 0,25
Ewewigskonsentrasie
“3
(mol.dm™)
Yv (divide by 2/dee/ deur 2)

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Physical Sciences P2/Fisiese Wetenskappe V2 8 NW/September 2018
NSC/NSS — Memorandum

ae,
xPtyrP

_ 0,25
(1,5)°(1,25)°

= 0,057 v

6.3 At70Os/By 70s:

e Increase in temperature increases the number of moles of reactants/
decreases the number of moles of the products. ¥
Verhoging in temperatuur verhoog die aantal mol van reaktante/
verlaag die aantal mol van die produk.

e reverse reaction is favoured ¥
terugwaartse reaksie is bevoordeel

e hence the forward reaction is exothermic.
dus die voorwaartse reaksie is eksotermies

Accept/Aanvaar:

e the reverse reaction is endothermic/ absorbs the heat energy” /
die teruqwaartse reaksie endotermies/ absorbeer die hitte-

energie (3)
6.4 Ke decreases. /Ke verlaagY (1)
[12]

QUESTION 7/VRAAG 7

7.1. Anacid is a substance that produces hydrogen ions (H') hydronium
ions (H30*) when it dissolves in water. VW
‘n Suur is ‘n stof wat waterstofione (H")/ hidroniumione (H;0*) vorm

wanneer dit in water oplos. (2)
tds
7.2.1 Pv (1)
T22.
(a) NH’ +H,O W=NH3+H;0°Y — balv (3)
(b) — Acidic/ suur ¥
Hydronium ions (H30*) are formed in the solution. ¥
Hidroniumione (H;0*) word gevorm in die oplossing (2)
7.3 7.3.1
m
c= — v
MV
ne ee
(40)(0,5)
= 0,2 mol-dm? ¥
OR/OF

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Physical Sciences P2/Fisiese Wetenskappe V2

NW/September 2018

NSC/NSS — Memorandum

m

n=—
M

as
40

= 0,1 mol ¥
n

é=—
V

= Edy
5

= 0,2 mol-dm? ¥

7.3.2 Cae Me y
Covy Ob
c,(25) 1,

(0,2)(12,5) 2

Ca=0,05 moldm®

Ca = 0,05 mol-dm?

‘Ny

OR/OF Cy Vb = Nb

/

(0,2)(12,5 x 10°) =n,

Ny = 2,5 x 10° mol

Na = % (2,5 x 10°)
n

qg=

Vv, /
= 4,25x10° /
25 x 10°
= 0,05 mol-dm®

[H20°] = 2(0,05) ¥
pH = - log [H30*] ¥ ©

pH = -log (0,1) ¥
pH =1v¥

QUESTION 8/VRAAG 8

(7)
[18]

8.1 A solution/liquid/dissolved substance that conducts electricity through

the movement of ions. Vv

'n Oplossing/vioeistof/opgeloste stof wat elektrisiteit gelei deur die

beweging van ione.

(2)

8.2 Temperature at 25°C /298 K/ Temperatuur van 25°C /298 K¥
Concentration of 1 mol-dm/ Konsentrasie van 1 mol-dm’? ¥ (2)

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Physical Sciences P2/Fisiese Wetenskappe V2 10 NW/September 2018
NSC/NSS — Memorandum ‘
8.3. Itis reducedv’, because Y** gains/accepts electrons from metal X “/it is an
oxidising agent/the oxidation number decreases.
Word gereduseer, want Y** neem elektrone op vanaf metaal X/ dit is ‘n
reduseermiddel/ die oksidasiegeta! verminder (2)
84 Y,X%,Zv (1)
8.5 8.5.1 OPTION 1/OPSIE 1 Notes/Aantekeninge
E’con = E®cathoue. - E°aicas Accept any other correct
6 = Fe formula from the data
E’sa = katode!" ‘anode sheet/Aanvaar enige ander
= 90,34 v - (-2,36) ¥ korrekte formule vanaf
x gegewensblad.
= 27V¥ Any other formula using
OPTION 2/ OPSIE 2 unconventional abbreviations,
2+ -_ iy e.g. E® se = EP oa- E®pa followed
{ we ~ Gufs) alee by correct substitutions:/ Enige
¥ \ Mg — Mg" + 2e° = 2,36V¥ ander formule wat
Cu**(aq) + Mg > _Cu(s) + Mg?"(ag) onkonvensionele afkortings
gebruik bv.
EF? se = Eo - E°pu gevolg deur
korrekte vervangings:
Max/Maks: 3/4
(4)
8.5.2 Mg(s)/ Mg** (aq) ¥ // ¥ Cu?*(aq/Cu(s) ¥ (3)
8.5.3 Endothermic/Endotennies. Vv (1)
fxothaumic / Elsotenies [15]
QUESTION 9/VRAAG 9
9.4 Electrical energy is converted to chemical energy. vv
Elektriese energie word omgeskakel na chemiese energie. (2)
92 PY (1)
93 Qv (1)
94 Cus) — Cu*(aq)+2e VY (2)
9.5 The rate of oxidation (at the anode) equals to the rate of reduction

(at the cathode). VW
Die tempo van oksidasie (by die anode) is gelyk aan die tempo van reduksie
(by die katode) (2)

OR/OF

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Physical Sciences P2/Fisiese Welenskappe V2 11 NW/September 2018
NSC/NSS - Memorandum

The rate at which copper (Cu) is oxidized equals to the rate at which
copper ions(Cu”") are reduced.

Die tempo waarteen koper (Cu) geoksideer word, is gelyk aan die tempo
waarteen koperione (Cu**) gereduseer word.

96 PY
2Clr - Ch+2e vv (3)
9.7 Decreases/ Verminder ¥ ‘ (1)
[12]
QUESTION 10/VRAAG 10
10.1. Contact (process)/Kontak (proses)¥ (1)
10.2
10.2.1 2SO,(g) +O.” = 280.(g) VY (3)
10.2.2 Vanadium pentoxide or (V2O0s) VW (2)
Vanadiumpentoksied (V20s).
10.3
10.3.1 Gmmanic./ cenmontciky Vv (1)
10.3.2 2NHa(g) + HeSO.(aq) Y > (NHa)2SO4(aq) VY (3)
10.4
M (NHa)2SOg = 2(14) +4(2) + 32 + 4(16)
= (3gq-mol 'v
28
%*N=>V x 100 ¥
132
= Ualyv (4)
[14]

TOTALITOTAAL: 150

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