Section A:
1. A
2. C
3. B
4. A
5. A
6. A
7. D
8. D
9. C
10. A
11. C
12. A
13. D
14. B
15. D
16. C
17. C
18. D
19. D
20. B
21. C
22. A
23. B
24. D
25. C
26. C
27. B
28. B
29. C
30. B
Section B (thanks to Thushan!):
SAQ 1a.
Two energy profiles; uncatalysed rising to 335, dropping to 92.4 and tungsten uncatalysed rising to 162, dropping to 92.4.
SAQ 1b.
Tungsten; lower activation energy required means a greater fraction of collisions between ammonia molecules will be successful, and hence there is a higher frequency of successful collisions between ammonia molecules, and thereby a greater rate of decomposition.
SAQ 2a.
A - CH2CH2, C - CH2=CH-CH3, D - CH3COOH, E - CH3CH2OH
SAQ 2b.
B - propan-1-amine, D - ethanoic acid
SAQ 2c.
CH3COOCH2CH2CH3, propyl ethanoate
SEMI STRUCTURAL FORMULA NEEDED HERE.
SAQ 3a.
total energy from ethanol = 1.80/46.0 x 1364 = 53.3 kJ
energy absorbed by water = 100.0 x 4.18 x 15.0 = 6270 J = 6.27 kJ
% lost = 88.3%
SAQ 3b.
- insulate metal container
- insulate spirit burner
- close metal container
Lots of things here.
SAQ 3ci.
CH3(CH2)12 COOCH3 - ester
SAQ 3cii.
Canola oil and biodiesel are mixtures of different compounds with different molar masses, it is hence impossible to calculate the total amount of molecules in either, given the mass of these mixtures.
SAQ 4a.
Two of the following:
- OH (acid) at ~3000
- CH at ~3000
- C=O at ~1700
SAQ 4b.
x = 3, y = 5
Mass spectrum shows two molecular ions at 108 and 110. Take away 35-Cl and the two Os from mass = 108, you get remaining mass 108-35-32=41, which is consistent with a C3H5 group.
SAQ 4ci.
Either of:
- presence of CH3 and a CHCl group
- Cl atom on carbon-2
SAQ 4cii.
CH3CHClCOOH
SAQ 4d.
Cl has 2 isotopes which are present in significant amounts. m/z = 108 is C3H5O235Cl and m/z = 110 C3H5O237Cl.
SAQ 5a.
HO-CH2-COOH
SAQ 5b.
Increase; adding glycolate push equilibrium to left, consuming H3O+, decrease [H3O+].
SAQ 5c.
13 mol L-1
SAQ 5d.
n(glycolate) = 1.3 mol
n(Na2CO3) = 1.3/2 = 0.66 mol (using calc values)
m(Na2CO3) = 70 g
SAQ 5e.
- safety goggles
- gas mask
- ventilation
- protective gloves
Any of these hehe
SAQ 6ai.
K = [HI]^2/[H2][I2]
SAQ 6aii.
Rice table!
Reactant Initial Change Equilibrium
H2 2.00 -1.93 0.07
I2 3.00 -1.93 1.07
HI 0 +3.86 3.86
Concentrations same as amounts since 1 L vessel.
K = 2 x 10^2
SAQ 6bi.
Self explanatory
SAQ 6bii.
Same thing, but quicker (no change in equilibrium position)
SAQ 7a.
non polar: any of glycine, alaine, valine, leucine, isoleucine
acid: glutamic acid or aspartic acid
SAQ 7bi.
cysteine
SAQ 7bii.
serine, asparagine
SAQ 7biii.
lysine, aspartic acid => must say both
SAQ 7biv.
dispersion forces/Van der Waal's forces
SAQ 7c.
- low pH disturbs ionic interactions between COO- and NH3+
- alters tertiary structure therefore shape of active site of trypsin
- trypsin therefore unable to bind to its substrates
SAQ 8a.
oxidation
Not accepting redox, since only the oxidation half equation is shown here.
SAQ 8bi.
m(BaSO4) = 0.474 g
n(BaSO4) = 0.00203 mol
n(SO2) = 0.00203 mol
m(SO2) = 0.130 g
ans = 0.260%
SAQ 8bii.
2.60 x 10^3 ppm
SAQ 8biii.
Ensure all of SO2, even SO2 on the inside of the dried apricot, reacts with the H2O2.
SAQ 8ci.
1:1
SAQ 8cii.
m(I2) = 4.95/500.0 x15.0 = 0.149 g
n(I2) = 5.85 x 10^-4 mol
n(H2O) = 5.85 x 10^-4 mol
m(H2O) = 0.0105 g
SAQ 8ciii.
0.115%
SAQ 9a.
Mg2+ (l) + 2e- -> Mg (l)
2Cl- (l) -> Cl2 (g) + 2e-
SAQ 9b.
Prevent Mg contacting and reacting vigorously with O2.
SAQ 9c.
Zn2+ better oxidant than Mg2+, will be discharged in preference at cathode.
SAQ 9d.
Fe stronger reductant than Cl-, will corrode as anode: Fe (s) --> Fe2+ (l) + 2e-. Even Fe2+ is a better reductant than Cl-, so you can even get Fe2+ (l) --> Fe3+ (l) + e-
At cathode, since you have a mixture of Fe2+, Fe3+ and Mg2+, the former two are better oxidants, so Fe will be discharged at cathode:
Fe3+ + e- --> Fe2+
Fe2+ + 2e- --> Fe
SAQ 10a.
Zn(s) + 2OH- (aq) --> Zn(OH)2 + 2e-
SAQ 10b.
Not sure what VCAA wants. I can think of the following:
- ensure an alkaline environment (for some reason this sounds a little superficial... :/ )
- precipitate out Zn2+ ions to ensure products are solid and don't leak out
SAQ 10c.
Q = It = 0.00236 x 10 x 60 x 60 = 85 C
n(e-) = 85/96500 = 8.8 x 10^-4 mol
n(Zn) = n(e-)/2 = 4.4 x 10^-4 mol
m(Zn) = 65.4 x n(Zn) = 0.029 g
SAQ 10d.
H2 (g) + 2 OH- (aq) --> 2H2O (l) + 2e-
SAQ 11a.
+ Br2 (l) + 2H+ (aq) + 2e- --> 2HBr (aq)
- QH2(aq) --> Q (aq) + 2H+ (aq) + 2e-
SAQ 11b.
Q (aq) + 2 HBr (aq) --> QH2 (aq) + Br2 (l)
SAQ 11ci.
LEFT
SAQ 11cii.
Ensure QH2 and Br2 do not react directly. If they do, heat produced that could have been converted to electrical energy. Reduced efficiency.
SAQ 11d.
2H2O (l) --> 2H2 (g) + O2 (g)
SAQ 11e.
QH2 (aq)+ H2O(l) --> QH- (aq) + H3O+(aq)
K = [H3O+][QH-]/[QH2]
SAQ 11f.
The quinone can be continuously produced at a rate similar to consumption - by constantly growing crops of rhubarb.
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