# Physical Sciences Grade 12 Paper 2 Gauteng Mock 2015 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng Mock Exam 2015, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gauteng-2015-sept-chemistry-p-2-memo/40a5fd4c-59ed-46c5-8d4b-65a713e21821
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 12 · Gauteng Mock Exam · 2015. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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### Page 1

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TENG PROVINCE

GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATIONS

2015

MEMORANDUM

SUBJECT: PHYSICAL SCIENCES P2 (10842)

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10842/15
GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATIONS
PHYSICAL SCIENCES
(Second Paper)
MEMORANDUM
QUESTION 1
14 Cv (2)
1.2 BYY (2)
1.3 Dvv (2)
1.4 Cvv (2)
1.5 ByY (2)
1.6 AVY (2)
1.7 Avy (2)
1.8 AVY (2)
1.9 BYY (2)
1.10 Cry (2)

[20]

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1O842/15
QUESTION 2
2.7.4 Cr oD)
24.2 AY (1)
224 Bulanal (i)
2.2.2 2-methyl™ butan -2-o0l 2)
OR 2-methyl-2-butanol
2.3.1 Waiter /H.0 “ qd)
2.3.2 |
« Correct functional group “
e Methyl group on the second
carbon
e Whole siructure ”
if
HOG
Notes:
a Condensed or semi-structural formula: vA
Aa)
e Molecular formula: %
(3)

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2.3.3

2.4.1

2.4.2

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10842/15

Addition / Hydration” (1)

Process by which molecules of two monomers with different functional (2)

groups~ react with the elimination/loss of water molecules. ¥

Fy (1)
[13]

QUESTION 3

3.1.1

Organic compounds having the same molecular formulaY with the same

functional group situated at different positions. Yor same side chain at
different positions
OR

Organic compounds with the same molecular formula, ¥ but different

positions of the side chain / functional groups. “

an
H—¢—C—C=C— H v
H H
(1)

But -1-yne ¥ /1-butyne ¥

|
H H (1)

But-2-yne v /2-butyne

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3.2.1

3.2.2
3.3.4

3.3.2

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108425
ee ee
HoH H HOH
H
H ia H-G—H
|
HoH H
ReactantsY” Products (3)
Condensed/semi-structural formulae: Max: ly
Molecular formula; Of,
Any additional reactant or products: Max: 24,
Propene~ and methylpropane “ (2)

The temperature Vat which the vapour pressure of a liquid is equal to the

external atmospheric pressure “ OR

The temperature Wat which a liquid changes phase into the gaseous
phase. 1/2 (2)

Propan -1-.ol has only one hydrogen bond “between its molecules.

Thus, propan — 1 — ol has weaker intermolecular forces between its

molecules. Vv Less energy is needed to overcome these intermolecular
forces, “thus propan — 1 ~ ol has a lower boiling point. OR
Propanoie acid has two hydrogen bonds between its molecules.

The intermolecular forces between the propanoic acid molecules are thus

stronger” ». more energy is needed to overcome these forces, “

thus propanoic acid has a higher boiling point. (3)
[16]

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10842/15
QUESTION 4
4.41 Q: Substitutionv /halogenation/chlorination (1)
4.1.2 R: Elimination/ dehydrohalogenation/ dehydrobrominationY (1)
4.2.1 e Functional group
HH oH e Whole structure
| | Vv correct
H- r — H
H H
Accept -OH
condensed in
v structural formula (2)
Condensed/semi — structural formulae Max: l/,
Molecular formula of,
4.2.2 Propan — 2 — ol ¥ (1)
2-propanol
4.3.1 Hydrogen bromide v (1)
v
4.3.2 H H H H
\ lov | |
a ep—e—H + H-Brv¥ —>y-Cc C—HY
Hob W | |
H H H
e Functional group¥
e Whole structure correct”
Accept HBr
Ignore: =
e Condensed/semi-structural formulae Max: 3/4
¢ Molecular formula o%,
Any additional reactant or products: | Everything correct, wrong balancing
Max: 3/4 Max: 3/4
(4)
[10]

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108425

QUESTION 5
S11 Surace area/siate of division/reaction surface“ @
5.1.2 There are more particles available for reaction.“ /more coll

There are a greater number of elfective collisions” per un me
“, reaction rate increases Q)

isions,

5.1.3 Mass of 16% of original 2 g=0,3 gv
Number of moles of CaCQs 0,3 g:
A
Po we

M

= 03
100”

=0,003 mol. ¥ for both equations
NicacOg: ANC?
1:2
0,003: 0,006“

ft
Cret FT

0,2 = 9,006
Vv

V =0,03 dm® ¥ (5)
5.2.1 Concentration” (1)
5.2.2 Equalio “ ()
5.2.3 The same number of moles/mass/ number of particles of CaCO; ¥ (produce

same number of moles of carbon dioxide). CaCOsis the limiting agent/Acid is in
excess. (1)

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10842/15
5.3.1 (Experiment) 6” (1)
5.3.2 3 Criteria for graph Mark
3 Correct shape for both graphs v
3 Graph of experiment 5 reaches 2 higher maximum/peak
E value than experiment 6. v
Ss Graph of experiment 6 shows more molecules of higher
5 kinetic energy v (3)
g
= xp 6
fy
el
€
a3
2
kinetic energy
[if no labels on axes max:2/3 |
5.4.1 exothermic’ (1)
5.4.2 The energy/potential energy of the products is lower than that of the reactants v
OR (1)
More energy released than absorbedY
OR [18]

Energy is released

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10842/15

QUESTION 6

6.1 CALCULATIONS USING NUMBER OF MOLES:

Mark allocation:

4 23
USING formula rx 30110

6,02.x10"

@

e Change in n(NO) = Initial mol — equilibrium mol”

e Using ratio NO(g) : Oo(g) : NO2(g) = 2: 1: 2% to determine change in
n(©2) and change in n(NQ2)

e Equilibrium [NO] = equilibriurn n(NO) /0,05
Equilibrium [Q2| = equilibrium n(Q,) /0,05 v
Equilibrium [NOz] = equilibrium n(NO») /0,05
e® Correct K, expression (formulae in [ })~
*  Subsiliution of concentrations and K, value into K, expression”
2 [0] at equilibrium”
@ n(Oz) at equilibrium (x 0,05) v

® (Oz) initial = n at equilibrium + n change”

GPTION 1
e 23
n(NOQ) = Sotto" = 5x 10° mol
6,02 x 10
NO) O2(g) NO2(g)
Initial moles 1 0,5006”” Qo ;
Change in moles 0,995v 0,4975 0,995 ¥ mole ratio
Equilibrium roles 0,005 0,0031% 0,995 ¥ divide by 0,05
Equilibrium concentration | 0,1 0,0615 19,9
[No,F
* [NoF To, |
644x105 = (19.9)
(0,"0,]
10, ] = 0,0615 mol: dn? [9]

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10842/15
6.2 Large K. means a high yield of products compared to reactantsY (1)
6.3 Darker” (1)
6.4 Temperature is decreased and the reaction will adjust to resist the change. The
exothermicY reaction is favoured thus the forward reaction’ is favoured and the
contents become darker.” (8)
6.5 No effect v a)
[15]
QUESTION 7
7A Brensted-Lowry base accepts a protony ¥ (2)
7.2.1 pH =-log[H30*]~
= -log(0,12)”
= 0,92” (3)
7.22 Cty (1)
7.3.1 it can act as an acid or a base/ it can accept or donate an H* ionY (1)
7.3.2 HSO¢ + HO = H30* + so,
v¥ reactants ¥ products ¥ balancing
ignore > (3)
7.4 Drops of water will change the concentration of the acid/ make it more dilutew
(1)
7.5.1 Option 1: Option 2:
m m
n=— =
M “uv
= 8,0 8.0
56Y =O
56 x 0,25
=0,14 mol V y
n =0,57 mol-dm?v
ory
= 0,14
0,25¥
= 0,57 mol-dm’s ¥ (3)

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~
Aa

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Cava Me %

CyoVb Ay (dilute acid)
G40) VW = Av
(0,57)(25) 2

OR
Gam 1x0,57 x 25% Caz 1x0,57 x 0,025”
2x40 v OR 2x 0,04 ¥v
=0,178 moldm® “ = 0,178 mol-drn®
753 Concentrated acid
th (moles concentrated) = nz (moles diluie)
v
cx Vj = Ce X Vo
c; = 0,178 x (490+10) ¥ OR c= 0,178 x0,5
10 v 0,01 v
= 8,9 moldm =“ = 8,9 mol-dm?/
QUESTION 8
8.1 Electrical energy to chemical energy”
8.2 Cathode
At cathode, Ag’ / silver ions gain electrons” and are, reduced to Ag/ silver
¥ metal
8.3.1 Ag/silver “ a
Ag’te=Ag (34)
83.2 Ag'+e — Agy¥ tee (0/
Ag > Agi+e (%5)
Ag—Ag've (2)
Ag=Ag™+e (94)
8.4.1 Electrode X becomes eroded/ smaller/ thinner.”
8.4.2 A (silver) layer forms on the medal.”
8.5 The rate at which Ag*/ silver ions are reduced at the cathode is equal to rate

at which Ag metal/ silver is oxidised at the anode.v “

10

10842N5

(4)

(4)
[22]

(1)
(1)

(2)
[14]

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10842/15
QUESTION 9
9.1 Galvanicw cell (1)
9.2 There will no reading/ The reading will be zero/ OVY (1)
9.3 Temperature “ and the initial concentrations of the electrolytes Vv

9.4

9.5.1

9.5.2

9.6.1

9.6.2

9.7

But cannot give 25 °C and 1 mol-dm® because the investigation is not carried
out under standard conditions because the emf values are not for standard

conditions. (2)
The different type of metal/the different haif-cells“” (1)
Voltmeter’s terminals” have been connected incorrectly”

OR

Incorrect connection “ (+ to anode and — to cathode) ~ (2)
Aluminium is a stronger reducing agent than zinc” and zinc is a stronger

reducing agent than copper.” (2)
Aluminium/ Atv (4)
Zine/ Zn ¥ (1)

2AL (s) +3 Zn*(aq)¥—> = 2Af**(aq) + 3Zn(s)_ balancing ¥

ReactantsY Products” BalancingY
Ignore =

Marking rule 3.9:
If only a reactant(s) followed by an arrow, or only a product(s) preceded

by an arrow, is/are written, marks may be awarded for the reactant(s) or
product(s). If only a reactant(s) or only a product(s) are written, without
an arrow, no marks are awarded for the reactant(s) or product(s).

Examples: Na + 3H2 v-> 2NH3” bal, ”

No + He >Y yA
— NH3”% %
No + Ha %
NHs %
4)

11

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1OB42N5
QUESTION 10
10.1.1 lror oxide/iron/aluminium oxide“ )
10.1.2 more collisions per unit time
greater surlace aree/greater number of exposed particles. “ (2)
10.1.3 Increase the number of moles of Ne and He (reactants) favours the forward
reaction”
OR
Increase pressure ~ and thus favours the forward reaction“ (2)
10.2.1 Calcium hydroxide increases the pH/alkalinity of the soil/ reduce the acidity of
the soil.”
Ammonium sulphate is a fertiliser. “ (2)
10.2.2 Ammonia gas is formed which will be released and reduce the
concentration/number of moles/ of fertiliser.“ (1)
10.3 NO) =e 2” 230 mov
M 24
MHNOs) = n(NO) = 30 mol” (3)
m( HNO) = nM = (30)(63) = 1890 g ¥ C1]
TOTAL: 150

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