Physical Science/P2
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NSC – Memorandum
PHYSICAL SCIENCES P2
MEMORANDUM
COMMON TEST
JUNE 2014
NATIONAL
SENIOR CERTIFICATE
GRADE 12
N.B. This memorandum consists of 7 pages including this page.
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Physical Sciences Grade 12 P2 Eastern Cape June 2014 Memo
Grade 12 Physical Sciences Paper 2 memorandum from the Eastern Cape June 2014, 7 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · Eastern Cape June · 2014. Memorandum, 7 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2014
- Exam period
- Eastern Cape June
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 7
- File size
- 307.1 KB
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Physical Science/P2
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NSC – Memorandum
SECTION A
QUESTION 1
1.1 A (2)
1.2 C (2)
1.3 B (2)
1.4 D (2)
1.5 D (2)
1.6 D (2)
1.7 B (2)
[14]
SECTION B
QUESTION 2
2.1
2.1.1 Carboxylic acid (1)
2.1.2 Butanal (1)
2.1.3
H H
l l
C = C
l l
H H (2)
2.2 Tertiary (1)
2.3 They have the same molecular formula but different functional groups. (2)
[7]
QUESTION 3
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3.1 Hydrocarbons are organic compounds that consist of hydrogen and carbon
only. (1)
3.2
3.2.1 Boiling point increases as the surface area increases OR
Boiling point decreases as the chain become more branched (1)
3.2.2 The more branches the smaller the surface area, the weaker the
intermolecular (Van der Waals) forces. Lesser energy needed to
separate the chains. (the lower the boiling point). (3)
3.3 A. (1)
[6]
QUESTION 4
4.1.1 CH3 CH2 CH2 CH2 -O-H (2)
heat
4.1.2 CH3 CH2 CH2 CH2 -O-H CH3CH2CH=CH2 + H2O.
H2SO4 (conc)
H2O H2SO4 (conc)
Excess heat
CH3CH2CHCH3
O -H (4)
4.2.1 Alcohol (1)
heat
4.2.2 CH3CH=CH2 + H2O H2SO4 (conc) CH3CHCH3 (3)
O -H
heat
4.2.3 CH3CH2CH2Cℓ + NaOH (aq) CH3CH2CH2O-H + HCℓ (3)
[13]
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QUESTION 5
5.1 Exothermic
ΔH = energy of the products – energy of the reactants
ΔH = –197 – 0
ΔH = –197 kJmol–1 (2)
5.2 1580 kJmol–1 (1)
5.3 1777 kJmol–1 (1)
5.4 1537 kJmol–1 (1)
[5]
QUESTION 6
6.1 Zn(s) + 2HCℓ(aq) → ZnCℓ2 (aq) + H2(g) bal (3)
6.2 surface area / state of division (1)
6.3 To make the investigation fair / To have 1 independent variable (1)
6.4 Volume of H2(g) is dependant on the amount of zinc used. (1)
ΔV
6.5 Average reaction rate =
Δt
48 30
=
40 20
= 0,9 cm3.s-1 (3)
6.6 Greater than (1)
6.7 Zn is powder, the more finely divided (greater surface area) the
greater the contact area. This will result in more effective collision
occur simultaneously per unit time. (2)
6.8
6.8.1 Remains the same. (1)
6.8.2 Steeper. (1)
6.9 Copper is a catalyst. Increase the rate of reaction. (2)
[16]
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QUESTION 7
7.1 Homogenous, reactants and products are in the same phase. (2)
7.2 Catalyst (1)
7.3 The equilibrium in graph C is reached faster than in graph A while the
concentration of the products in graph A and C are equal and also the
concentration of the reactants in both graphs A and C are equal. (2)
7.4 Exothermic. (1)
7.5 If the temperature is increased from 5000C to 6000C the concentration of the
products decreases and the reactants concentration increases. The reverse
reaction is thus favoured. According to Le Chateliers’ principle endothermic
reaction is favoured when temperature is increased. Thus the forward
reaction is exothermic. (2)
7.6 H2(g) + I2(g) ⇌ 2HI(g)
n(H2) = 1
2
= 0,5 mol
n(I2) = 127
254
= 0,5 mol
Mole Ratio H2 I2 2HI
Initial mol (0,5 0,5) 0
Change (used/ formed) (0,4 0,4 0,8)
Equilibrium mol (0,1 0,1) 0,8
Equilibrium Concentration 0,1 0,1 0,8
C= n/v 2 2 2
(0,05) (0,05) (0,4)
Kc = [ HI ]2
[ H2 ] . [ I2 ]
Kc = (0,4)2
(0,05) (0,05)
Kc = 64 (8)
[16]
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QUESTION 8
8.1
8.1.1 C (1)
8.1.2 A (1)
8.1.3 E (1)
8.1.4 B (1)
8.2 H2SO4 + 2H2O → 2H3O+ + SO42–
pH = –log [ H3O+]
pH = – log (2 x 0,1)
pH = 0,698 (3)
8.3 Na2CO3 → Na+ + CO32-
CO32- + H2O → HCO3– + OH– (3)
[10]
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QUESTION 9
9.1 Dilute acid contains a small number of moles of acid in proportion to the
volume of water. (1)
9.2 Standard solution is the solution of known concentration. (1)
9.3 c = m
MV
0,2 = m
56 x 0,3
m = 3,36g (4)
9.4 To change the colour at the end point of titration. OR
To indicate/show when equivalent number of moles of acid and base
have reacted. (1)
9.5 Bromothymol blue, because the equivalent point will be approximately
pH = 7 since H2SO4 is a strong acid and KOH is a strong base. (2)
9.6
Ca x Va na
Cb x Vb nb
Ca x 20,15 1
15,04 x 0,2 2
Ca 0,075 mol.dm
3
(4)
[13]
TOTAL SECTION B: 86
GRAND TOTAL: 100
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