Physical Sciences Grade 12 P2 Western Cape Mock 2014 Memo

Downloaded from hlayiso.com CAPE WINELANDS EDUCATION DISTRICT PHYSICAL SCIENCES P2 CHEMISTRY MEMORANDUM SEPTEMBER 2014 MARKS: 150 TIME: 3 hours

Downloaded from hlayiso.com QUESTION 1 1.1 B

(2) 1.2 D

(2) 1.3 C

(2) 1.4 A

(2) 1.5 A

(2) 1.6 A

(2) 1.7 B

(2) 1.8 D

(2) 1.9 D

(2) 1.10 B

(2)[20] QUESTION 2 2.1.1 CnH2n+2
(1) H H H H H | 2.1.2 | | | | H H-C-H H
H–C—C—C–C-H
OR | | | | | | | H–C— C—C-H H H-C-H H H | | | I H H-C-H H H I H 2-methylbutane
2,2-dimethylpropane
(2) H O H | ║ | H–C— C—C-H 2.1.3 | |
functional group
whole molecule (2) H H 2.2 The stronger the intermolecular forces, the lower the vapour pressure.
correct variables
correct relationship When IMFs are stronger, it requires more energy to break the IMFs to allow molecules to evaporate,
thus fewer molecules evaporate. (3) 2.3 (as you go down the table), chain length/molecular size increases,
thus strength of Van der Waals IMFs increases
thus vapour pressure decreases (2) 2.4.1 1-butanol has an –OH group
which forms stronger Hydrogen bonding IMFs
than the Van der Waals IMFs in pentane (2) 2.4.2 Secondary alcohols are more spheric in shape / have a smaller surface area for bonding / molecules cannot get as close to each other
as the more linear primary alcohols, thus IMFs are weaker,
thus easier to evaporate. (2) 2.5 Ethanoic acid
(no positive marking → ) (1) 2.6 Ethanoic acid has a lower vapour pressure
than 1-propanol thus stronger IMFs
thus more energy will be required to break IMFs.
(3)[18] 2

Downloaded from hlayiso.com QUESTION 3 3.1 A compound that has one or more double (or triple) bonds
between C atoms.
(2) 3.2
H H Br H
H H H H H H H
| | l | l l | | | l l H – C -- C -- C – C – C – C --H H – C -- C ═ C – C – C – C --H + Br2 | l | | l l | | | l l H Br H H H H H H H H H (3) 3.3 Light/heat
(1) 3.4.1 Addition (hydrohalogenation)
(1) 3.4.2 Catalyst of strong acid / H2SO4
in (excess water) (1) 3.4.3 (Dilute solution) of strong base (NaOH or KOH)
( X conc base) , mild heat
(2) 3.4.4 but-2-ene
H H H (2) | | |
H–C—C═C–C-H | | | H H H 3.5.1 ethene
(1) 3.5.2 addition
(1)[14] QUESTION 4 4.1 the change in concentration of reactants or products per unit time.
(1) 4.2.1 concentration
(1) 4.2.2 reaction rate
(except time) (1) 4.3.1 5. Repeat the experiment but add 2 ml of HCl and 3ml of distilled water 6. Repeat the experiment but add 1 ml of HCl and 4ml of distilled water
HCl volume is changed
the conc of HCl is changed (water added to keep volume the same)
the steps indicate a decrease in conc of HCl. (3) 4.3.2 volume of sodium thiosulphate needs to be kept constant / temp constant / judgement of when cross disappears needs to be constant (same learner) / swirling needs to be the same / stopwatch needs to be started at same time in process (any one of the above is correct)
(1) 4.4 Sulphur / S.
(1) 4.5 Experiment 1.
(1)[9] 3

Downloaded from hlayiso.com QUESTION 5 5.1 The rate of the forward reaction equals the rate of the reverse reaction

(2) 5.2 Closed
(1) 5.3.1 turns more orange

(1) 5.3.2 increases the H+ concentration
which favours the forward reaction
(2) 5.4 Kc = Cr2O72-

[CrO42-]2 [H+]2 (2) 5.5 No (1) 5.6 Yellow (1) + 5.7 The presence of hydroxide will decrease the conc of H
as they react to form water (or increase amount of H2O); according to Le Chatelier’s Principle, this will cause the rxn that increases the conc of H+ to be favoured
, thus reverse rxn / equilibrium shifts to the left.
(3)[13] QUESTION 6 6.1 Endothermic

formulas – if used (1) m 35,9 n 0,62 6.2 nNaCl = = = 0,62mol _(0,6138mol ) c= = = 6,2mol ⋅ dm −3 M 58,5
V 0,1
+ −

SUBSTITUTION
K c = [Na ][Cl ] = (6,2)(6,2) = 38,44 (37,66 if no rounding) (6) 6.3 Increase temp.
An increase in temp will favour the endothermic rxn
which is the forward rxn;
this will increase the concentration of products, thus KC.
(4) 6.4 Less soluble.
(1)[12] QUESTION 7 7.1 NaOH → Na+(aq) + OH-(aq) [OH-] = 0,11
(or: conc OH- = conc NaOH) Kw = 1x10-14 = [H3O+][OH-]
1x10-14 = [H3O+](0,11) [H3O+] = 9,09x10-14
( +→) pH = -log[H3O+]
= - log (9,09x10-14) (substitution) = 13,04
(Except calculation of pOH and subtracting from 14). (5) 7.2 nNaOH = c x V = 0,11 x 0,0285
= 0,0031 mol
(0,003135 if no rounding off) (2) 7.3 nacid : nbase = 1:1
nacid = 0,0031 macid in 25cm3 = nxM = 0,0031 x 60 = 0,186g
(+→) macid in 100 cm3 = 0,186 x 4
= 0,744g
0,744 % ethanoic acid = × 100 = 9,9%
(9,9 – 10 depending on rounding off). (5)[12] 7,5 4

Downloaded from hlayiso.com QUESTION 8 8.1 Contact
(1) 8.2 H2S2O7 / oleum/pyrosulphuric acid / disulphuric acid / fuming sulphuric acid
(1) 8.3 V2O5 / Venadium pentoxide
(1) 8.4 An acid that ionises completely.
(1) 8.5 H2SO4 + 2 H20 → SO42- (aq) + 2 H3O+ (aq) (
LHS
RHS
bal) (3) 8.6 H3O+
(1) 8.7 ampholite (1) 8.8 diprotic (1) 8.9.1 oxidising agent (1) 8.9.2 Copper was oxidised to Cu-ions
thus lost electrons
(2) [13] QUESTION 9 9.1 Chemical energy to electrical energy.
(1) 9.2 Temperature of 25 OC / 298 K
Concentrations of 1 mol•dm-3.
NOT Pressure of 1 atm or 101,3 kPa. (2) 9.3 Mg (magnesium)
(1) 9.4 Mg (s) → Mg2+ + 2e (
LHS
RHS
bal) (3)
formulas – if used m 0,96 2 9.5 nMg = = = 0,04mol
nFe = n Mg × = 0,027
M 24 3
mFe = n × M = 0,027 × 56 = 1,51g
(1,49 if no rounding off earlier) (5) 9.6 Eθcell = Eθreduction - Eθoxidation
(or other equivalent eqn) = - 0,04
- (- 2,37)
= 2,33 V.
YES,
bulb will light up (5) 9.7 Increase
(1) 9.8 Decrease
(neg →); an increase in [Mg2+] will inhibit the forward reaction / will favour the reaction that decreases [Mg2+]
and thus voltmeter reading will decrease (2)[20] 5

Downloaded from hlayiso.com QUESTION 10 10.1 bauxite
(1) 10.2 Electrolytic
(1) 10.3 Cathode
(1) 10.4 Negative
(1) 10.5 2 O2-
→ O2(g) + 2e-
(2) 10.6 DC.
(neg→) To produce Al (product), the reaction must not be reversed
(which is what would happen if power supply was AC). (2)[8] QUESTION 11 11.1 Haber process
(1) 11.2 Reaction rate would be too slow.
(1) 11.3 NO (1) 11.4 4 NH3 + 5 O2 → 4 NO + 6 H2O (
LHS
RHS ) (2) 11.5 NH4NO3
(1) 11.6 Can cause “blue-baby” syndrome, stomach cancer; reproductive problems, birth defects, endocrine, immune and nervous system malfunctions
(1) 5 11.7 18,75 = x 30
x= 8.
thus K = 1
or N:P:K = 5:2:1 (4)[11]
x 6