# Physical Sciences Grade 12 P2 2017 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the KwaZulu Natal November 2017, 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2017-chemistry-p2-memo-kzn/2375687f-8817-4c6a-aa00-959ad5fcdabd
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2017
- Exam period: KwaZulu Natal November
- Paper: 2
- Curriculum: CAPS
- Pages: 9

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## About this document

Physical Sciences · Grade 12 · KwaZulu Natal November · 2017. Memorandum, 9 pages. Read online or download the PDF.

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## Document text

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### Page 1

Education

KwaZulu-Natal Department of Education
REPUBLIC OF SOUTH AFRICA
* PHYSICAL SCIENCES P2 (CHEMISTRY) *
: PREPARATORY EXAMINATION ‘
: MEMORANDUM :
" SEPTEMBER 2017 :
rrr a
NATIONAL SENIOR
CERTIFICATE
GRADE 12 — |
| TIME: 3 hours

150

| MARKS:

This memorandum consists of 9 pages.

### Page 2

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P2 Preparatory Examination September 2017
NSC - Grade 12

QUESTION 1: MULTIPLE CHOICE

11 Dy (2)
12 Byy (2)
13. AvY OR C (2)
14. Dvv (2)
15 Cvv (2)
16 Byy (2)
17 AVY (2)
18 Dyvv (2)
19 Avv (2)
1.10 Cvv (2)
[20]
QUESTION 2
2.1 2-bromo-5-chloro-2,5-dimethyl” heptane “ (2)
2.2 C,Ho, ¥ (1)
23° oH HCIH HHH (3)
Potopoto toa Y correct structural formula
H C-C-C-C-C-C-C-H 4
od roto rod Y Clin correct position
HoH H HHH Y ethyl group in correct position
H-G—H
H —¢ -H
H
2.4 Triple bondv/ Accept Alkynes (1)
25 Oo H (2)
UN 1 vv
C-O-C-H
i !
4 4 Accept any correct formula except the molecular
2.6 Ketones ¥ (1)

[10]

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### Page 3

pnysical sciences DOWNloaded-fssom hlayiso.com

P2 Preparatory Examination September 2017
NSC - Grade 12

QUESTION 3
3.1. CY OR Ethanoic acid (1)

C 3.2 Ethanoic acid has the strongest intermolecular forces “
More energy is required to overcome the intermolecular forces “

least amount will evaporate “ (3)
33 BY (1)

C@ Ethanal has the weakest intermolecular forces ¥ (3)

Least energy is required to overcome the intermolecular forces v
Ethanal has the lowest boiling point and therefore the highest vapour

pressurev
[8]
QUESTION 4
4.1 H,SO,/ H;PO,/Heat” (1)
4.2 Pt/Pd/Niv (1)
43 CsH, + 50,———* 3CO, + 4H,0 V reactants/left hand side (3)
If any structural formulae used 2. ” products/right hand side
3 balancing
44 HHH H HH (3)

' b= 4 Hc 4 ' é é 4  reactants/left hand side
H= ¢ ae + c rot lv products/right hand side
H H cl H structural formulae
lf condensed formulae used 2

3
4.5 Sulphuric acid v (1)
46 HoH OH OW WOH ” correct functional group 2)
H~G~G-G-O-C-C~C~G-H whole structure correct
H HH H H H
4.7. Propyl butanoate “YY Any correct name given for 4.6 (2)
[13]
QUESTION 5
51 A molecule that consists of a large number of atoms. Vv” (2)
5.2 Polymerisation v (1)
53 HH H (2)
H H
[5]

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### Page 4

physical Scenes 9 DOWNIoaded, from hlayiso.com

P2 Preparatory Examination September 2017
NSC - Grade 12

QUESTION 6
6.1 How does the rate of a reaction change with time? vv (2)
Any given investigative question
6.2. Time’ Mark based on given investigative question. (1)
6.3 (2)
[Time (s) 0 1 2 3 4 5 6 7 8 9
Mass of flask and its contents (g) | 178 |176,2|174, 1|172,7|172,3|172,2/172,1|172,1|172,1|172,1
Mass of CO, (g) produced (g) 0 | 1,8 | 3,9 | 5,3 | 5,7 | 5,8 | 5,9 | 5,9 | 59 | 59] w
64 Graph of mass of CO, produced vs time ~ Correct shape
im TPP Tae ~Y plotting of all points
— Clee] [Appropriate scale and labels on both
ete 1 laxes
. ~ Correct heading
8 But plotting 5 or less points subtract one
g ark.
2
6)
oO 1 2 324 5 6 7 8 9
Time (s)
6.5 The reaction has reached completion. v (1)
OR
One of the reactants is finished. ~
6.6 m y
n=
58,
44
= 0,13 mol
1 mol occupies 24,46 dm?®
0,13 mol will occupy 3,18 dm? v
Accept range 3.18 — 3.28 @)
6.7 Graph N has a steeper gradient vand finishes at the same point as the (2)
original graphw
6.8 Higher temperature, molecules have greater kinetic energy “ (3)
More effective collisions per unit time
Reaction rate increases “
[19]

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### Page 5

Physical Sciences Downloaded-:20 m h layiso . com

P2 Preparatory Examination September 2017
NSC - Grade 12

QUESTION 7
7.1 CALCULATIONS USING NUMBER OF MOLES
BEREKENINGE WAT GETAL MOL GEBRUIK

Mark allocation/Puntefoekenning:
e Calculating number of moles of H2, Cl, and HCI. ¥

Molar ratio. “

Number of moles at Equilibrium. “
Dividing by 0,5 to get the concentration. v
K, expression. ¥

e Substitution in the K, expression “

e Calculating concentration of Cl, ¥

¢ — Substitution in n= cV.v

e Substitution in m=nM “

e _ Final answer”

7.1 H+ ~~) Ch = 2HCI (10)
Initial mass 40g 355g Og Calc no. of
Initial n (5 5 0) | “mol
React/Prod Cx Xx 2x_) | ~ Ratio
In at equilibrium] (5 - x 5-x ax Ji v
‘Conc. at 5-x 5-x 2x ae
equilibrium 05 05. 05 v dividing by 0,5
x, = omy INo K, expressi t substituti 9
GJEb1 lo K, expression, correct substitution max
2
(> Wrong KC expression max
60 = 05 y 10
ven 5x 5-x
CYC)
0,5 0,5
x = 3,97
Cone.Cl, at equilibrium = rad
n=C-V
_ 5-397 95
0,5
= 1,03 mol
m=n-M
= 1,03-71”
= 73,139 %

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### Page 6

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P2 Preparatory Examination September 2017
NSC - Grade 12

7.2 When the equilibrium in a closed system is disturbed, the system will re-
instate a new equilibrium by favouring the reaction that will oppose the

disturbance. vv (2)
7.3

7.3.1 Decrease “ (1)

7.3.2 Increase “ (1)

74 The temperature of the system was increased, according to Le Chatelier’s
principle, the system reacts by favouring the endothermic reaction.“ Hence,
the reverse reaction is favoured.”

(2)
[16]

QUESTION 8
8.1 8.1.1. Strong acid “ (1)

8.1.2 7 (5)
f rl ined v CV, = ©, V3
Vol. of conc.HCl require (10\(x) = (0,35)(0,9"

n=CV v x = 0,0125dm?
= 0,25 x 0,5” OR = 12,5em°¥
=0,0125 v

n
“V7
0,0125

V
= 0,0125 “dm?
= 12,50m°

1. H
8.1.3 HCI tho _, H,0° + Cr

1

c
vo
0

Vv

0,25 mol.dm* 0,25 mol.dm®?

pH = -log [H,07} v

= -log (0,25)
= 0,60 ,

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### Page 7

Physical Sciences Downloaded..f£rom hlayiso.com

P2 Preparatory Examination September 2017
NSC - Grade 12
8.2 8.2.1 Option 1 n(HCl) = cW= (0,25) (0,5¥
= 0,125mol

n(NaOH) = cV = (0,2)(0,14¥
= 0,028mol

n(HCl) reacted with (NaOH) = 0,028mol”
n(HCl) reactedwith (CaCO,) = 0,125-0,028%
= 0,097 mol

n(CaCO,) = >* 0,09™

= 0,0485mol
mass of CaCO, = 0M ,
= 0,0485x100

= 4,85¢%
Option 2
Volume of HCI reacted with NaOH
om, n
aaa
CM, OM,
(0,25)V 4
a: ce y
(0,2)(140 ~ 14
V. = 12cm?

a

Volume of HCI reacted with CaCO,
Vici = 500 - 112

v
= 388 cm?
= 0,388 dm?
No. of mol of HCI reacted with CaCO 3
n
Cy
n
0,25 = 0388
n = 0,097 mol
No. of mol of CaCO reacted with HCI
Net * "eaco, = 2:4
0,097
"caco, a
= 0,0485 mol
Mass of mol of CaCO; reacted with HCI
me=n-M
= 0,0485 -100%
= 4,859 v (8)

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### Page 8

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P2 Preparatory Examination September 2017
NSC - Grade 12

8.2.2 (3)
% purity = mass of CaCO, 100
mass of sample
=42 100 %
5
=97% ¥
[20]
QUESTION 9
9.1 9.1.1 cathode v (1)
9.1.2 P to silver medal “ (1)
9.1.3. Electrode P is made of silver. Rate of oxidation is equal to the rate of
reduction.v” (2)
91.4 Agv (1)

9.2 9.2.1 Temp =25°CVv

Concentration of the electrolyte solution = 1 mol.dm?v (2)
9.2.2 Chemical energy to electrical energy “ (1)
9.2.3 Maintain electrical neutrality between the two half cells “ (2)
Completes the circuit ~
9.2.4 6 _ 8 . £e
Eco = cathode E node
2,70" = 0,34°- E°
anode
Eo = -2,36V%
anode
The anode must be magnesium v (5)
+ marking
9.2.5 Mg > Mg” +2evv OR (2)

9.2.6 (3)

vov v
Mg/Mg""//Cu**/Cu. OR YIY?*#1Cu2*/Cu
9.2.7 Remain the same” (1)

9.2.8 Ag’ will form a precipitate with the Cl’. This decreases the concentration
of the Cl in solution, but not the concentration of Cu2*v “
The reading on the voltmeter can only be affected by a change of
concentrations of either Cu2* or Mg2*/ Y2+.” (3)

QUESTION 10
10.1 10.1.1 Haber process

(1)
10.1.2 Fractional distillation of air V (1)
10.1.3 Ostwald process ¥ (1)
10.1.4 Platinum v (1)
10.1.5 HOY OR (H20 +02) (1)

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### Page 9

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Physical Sciences

pP2 Preparatory Examination September 2017
NSC - Grade 12
10.1.6 Enables plants to grow better and faster “ (2)
Increases food production “
10.1.7 Ammonium nitrate ” (1)

10.2 10.2.1 The ratio of N:P:K or primary nutrients in the bag “
Amount of fertilizer in the bag V (2)

10.2.2 or — S x22 = 9,68% v

wp = 19x22 = 4,40% Vv
50
YN = Lox 22 = 7.92% v (3)
OR
22% VV

10.2.3 Potassium”

Shortage of potassium causes poor quality flowers and fruit (colour
and taste) and make leaves have brown or yellow edges. ~

(2)
[15]
TOTAL MARKS: [150]

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