ATAR Notes: Forum
Archived Discussion => Mathematics/Science/Technology => 2014 => Exam Discussion => Victoria => Chemistry => Topic started by: acinod on November 11, 2014, 12:17:23 pm
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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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Omfg yes I guessed right WEW
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pretty sure 30 is B lol
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Q16. Isn't the answer B? Can't a graph be extrapolated down accurately but no up?
Josh
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is 15 C or D?
and 21 is C I think haha
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Isn't 11A?
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I believe 15 is C, 21 is C and 30 is B. Could very possibly be wrong myself.
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Isn't 11A?
It's C because you have to balance the charges too :/ I missed that too. Forgot it was a redox equation.
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is 15 C or D?
and 21 is C I think haha
Yeah I had 15 as C and 21 as C also. But I don't rate myself on Chemistry, especially section A.
I had 11 as B as well, and 30 was B too
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It's C because you have to balance the charges too :/ I missed that too. Forgot it was a redox equation.
Crap.i thought there were electrons there. Silly me
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Thanks for your corrections guys and I've actually been secretly editing the answers for the past 10 minutes.
15 is C, 21 is C, 30 is B
I believe 16 is C though since I don't think you can extrapolate (you can but its not the MOST accurate)
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It's C because you have to balance the charges too :/ I missed that too. Forgot it was a redox equation.
Crap.i thought there were electrons there. Silly me
FFS, knew I was doing something wrong.
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FFS, knew I was doing something wrong.
I thought that 6+4 =12, don't worry.
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I believe 16 is C though since I don't think you can extrapolate (you can but its not the MOST accurate)
I had this debate in my head for so long. I was reading the word MOST accurate over and over again, ended up picking A, 0-5.
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I believe 16 is C though since I don't think you can extrapolate (you can but its not the MOST accurate)
I have read that accuracy is maintained when you extrapolate down.
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Thanks for your corrections guys and I've actually been secretly editing the answers for the past 10 minutes.
15 is C, 21 is C, 30 is B
I believe 16 is C though since I don't think you can extrapolate (you can but its not the MOST accurate)
I think that anything below the data points that you have got can be extrapolated as a linear region. However above the highest data value it cannot be used as the graph cannot be necessarily be assumed to be linear. That is how I have interpreted it anyway. Will see how Thushan and VCAA see it :)
Thanks for the solutions though!
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I have read that accuracy is maintained when you extrapolate down.
I think you are correct in that we can extrapolate down: http://en.wikipedia.org/wiki/Standard_addition
So 16 is B (for now I guess)
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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
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)
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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One question I did not understand was 23. B is the energy that methane would give out at SLC. However there is the component to the question where H2O is involved. In other VCAA exams and in chapter questions - when water is in a molecule, the energy release is less as energy is consumed evaporating the water. (Brown coal on the 2013 exam for example where wet coal releases less energy than dry coal). With this reasoning, I would have thought 23 was A as it was slightly less that answer B. Don't know what others thought?
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Can 3ci not also be a carboxylic acid?
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Can 3ci not also be a carboxylic acid?
I did that too, but I think biodiesel molecule means ester
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What are everyone's thoughts for the A+ cut off this year??
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What are everyone's thoughts for the A+ cut off this year??
I am in no way an expert at all but I thought it was harder than last years which was 85% So I would guess personally 80-83%
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One question I did not understand was 23. B is the energy that methane would give out at SLC. However there is the component to the question where H2O is involved. In other VCAA exams and in chapter questions - when water is in a molecule, the energy release is less as energy is consumed evaporating the water. (Brown coal on the 2013 exam for example where wet coal releases less energy than dry coal). With this reasoning, I would have thought 23 was A as it was slightly less that answer B. Don't know what others thought?
You're right but the wording of the question is why it's B. It asks for 'The amount of energy released by the complete combustion of methane extracted from a 1kg sample of methane hydrate...'. The methane extraction energy cost is ignored because all it's asking is about the energy from the methane after it was extracted.
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Agh, wasn't consistent with my sig figs. Really hope I'm not caught out on the question they decide to scrutinise. :/
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For Question 7biii I wrote lysine,and aspartic acid. Would that be fine?
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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
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
8a. Wow really, surely theyll also accept redox reaction?
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For 11D doesnt adding the two h20 equations give you 2h20->h2+O2+2Hpositive?
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are there no electrons in 10d's equation? :-\
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8a. Wow really, surely theyll also accept redox reaction?
Idk man cause the oxidation reaction is given?
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Isn't 8bii 1300ppm?
For the question of Br and Q , is it ok if my equations do not have H in them?
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@apreslapulie - Caught me napping there. Nah, there are electrons. I fixed it.
@Chalkhous - Nah, you should get 2H2O -> 2H2 + O2. And whoops, 7biii - you're right, you should write both lysine and aspartic acid.
@Edward Elric - I don't know :/ I wouldn't since they only showed oxidation half equation. Redox is short for reduction-oxidation, meaning both reduction and oxidation have to occur.
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Isn't 8bii 1300ppm?
For the question of Br and Q , is it ok if my equations do not have H in them?
RKTR - I might have made a mistake, how did you get 1300 ppm? :/
As for not having the H, not sure. Question said Br2 converted to HBr. But not having the H+ is acceptable with me, but not sure abt VCAA.
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RKTR - I might have made a mistake, how did you get 1300 ppm? :/
As for not having the H, not sure. Question said Br2 converted to HBr. But not having the H+ is acceptable with me, but not sure abt VCAA.
0.130g / 100mL =1300mg/ L =1300ppm
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Ah - it asked for the concentration of the SO2 in the apricot sample, which is 0.130 g in 50.00 g of apricot. :(
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0.130g / 100mL =1300mg/ L =1300ppm
Yeah I did that at first too, then realised that's the concentration in the water.
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0.130g / 100mL =1300mg/ L =1300ppm
I got 2.6 x 10^4 ppm
I did 0.130g in 5g
Therfore, 0.026g in 1g
Then, 0.026 x 10^6 ppm = 2.6 x 10^4
Is this right?
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Ah - it asked for the concentration of the SO2 in the apricot sample, which is 0.130 g in 50.00 g of apricot. :(
Ohno T.T
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I think you are correct in that we can extrapolate down: http://en.wikipedia.org/wiki/Standard_addition
So 16 is B (for now I guess)
I also put C because my tutor also kept telling me never go outside the max and min of standard solutions. the question does state 'most' so if u look at it is there a region that is even more accurate than 1-4?
^thats pretty much my thinking process when i did the q
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I got 2.6 x 10^4 ppm
I did 0.130g in 5g
Therfore, 0.026g in 1g
Then, 0.026 x 10^6 ppm = 2.6 x 10^4
Is this right?
Working is fine, but it's 50 g of apricot, not 5g :/
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Ah - it asked for the concentration of the SO2 in the apricot sample, which is 0.130 g in 50.00 g of apricot. :(
Used 5g instead of 50g :-[
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Working is fine, but it's 50 g of apricot, not 5g :/
Just realised :/
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I just had a look at MCQ 16, my opinion:
I'm going with C for the following reason:
- you can extrapolate down, yes, but it wont be as accurate (not 100%, maybe 99%) as between 1-4 ppm; you cannot be fully certain of its accuracy
- in 2011 VCAA there was a similar question and the position was that you can't extrapolate "beyond the range of the graph" (i.e between the points)
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I just had a look at MCQ 16, my opinion:
I'm going with C for the following reason:
- you can extrapolate down, yes, but it wont be as accurate (not 100%, maybe 99%) as between 1-4 ppm; you cannot be fully certain of its accuracy
- in 2011 VCAA there was a similar question and the position was that you can't extrapolate "beyond the range of the graph" (i.e between the points)
I feel like people are also getting a bit confused, because most of the time there is a "blank" or standard of 0 concentration put in. Thus, you can interpolate between 0 and 1. However, they only put 1-4 this time. idk
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wouldnt 10a. not have the OH bit in there and just be Zn(s)->Zn2+(s)+2e-, since if it had the OH- bit in there, what would the reduction equation be? Since there is no other products apart from Zn(OH)2 what would the O2 and H2O become?
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wouldnt 10a. not have the OH bit in there and just be Zn(s)->Zn2+(s)+2e-, since if it had the OH- bit in there, what would the reduction equation be? Since there is no other products apart from Zn(OH)2 what would the O2 and H2O become?
I did the same as you
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and isnt the role of potassium hydroxide for 10b to be the electrolyte?
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and isnt the role of potassium hydroxide for 10b to be the electrolyte?
I wrote to provide OH- ions for the reaction since the question assumes acidic electrolyte but idk
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Oh yeah, that's an entirely valid answer.
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I do t get why 30 is b, if you have more stuff left over (un reacted mass) that would make it seem like you have less cuo initially present.......
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I also don't really get the difference between Q30 Multi answers B and D. I think they would both have the same (correct) effect? o.o
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In last s.a did everyone say change in heat would be same since heat is dependent on how much reactant actually reacts which isn't affected by catalyst?
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I do t get why 30 is b, if you have more stuff left over (un reacted mass) that would make it seem like you have less cuo initially present.......
For 30 did you choose C?
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For 30 did you choose C?
Yeah but I think I'm wrong since I kind of guessed between c and d
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I went with B, but I can't see why A isn't true either. If CuO was contaminated with carbon, heating it will do nothing (there's no oxygen in the tube) to the carbon, so it's similar to having unreacted CuO.
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SAQ 10a.
Zn(s) + 2OH- (aq) --> Zn(OH)2 + 2e-
Hmm.. I sorta interpreted it as
anode: Zn(s) -> Zn2+(aq) + 2e-
cathode: H20(l) + O2(g) + 4e- -> 4OH-(aq)
And the fact that Zn2+ and OH- precipitated was a separate thing so I used that anode equation.... Might've lost a mark for that :/
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I went with B, but I can't see why A isn't true either. If CuO was contaminated with carbon, heating it will do nothing (there's no oxygen in the tube) to the carbon, so it's similar to having unreacted CuO.
Wouldn't the oxygen produced from CuO -> Cu + O2 be able to react with the carbon contamination?
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What will the A+ cutoff be?
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Shouldn't both the reactions in 11a produce a solid product since the cell is rechargeable?
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I just had a look at MCQ 16, my opinion:
I'm going with C for the following reason:
- you can extrapolate down, yes, but it wont be as accurate (not 100%, maybe 99%) as between 1-4 ppm; you cannot be fully certain of its accuracy
- in 2011 VCAA there was a similar question and the position was that you can't extrapolate "beyond the range of the graph" (i.e between the points)
I'm gonna go with you here Thushan.
Found this from the textbook solutions:
'Techniques such as AA, GLC, HPLC, UV, etc. do not directly produce measures of concentration. Standards must be used. More than one standard should be used and the unknown sample should lie between these standards. This is because a zero standard may be contaminated with trace amounts of the chemical being tested . Calibration graphs are often non-linear and multiple standards increase the chance of detecting incorrectly prepared solutions'
So MCQ 16 is likely to be C
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I did the same as you
Yeah i did this too
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For question 7b) I listed the actual side chains instead of naming the amino acid. Will I still get marks?
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Wouldn't the oxygen produced from CuO -> Cu + O2 be able to react with the carbon contamination?
Smart. That could be true, yes.
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For question 7b) I listed the actual side chains instead of naming the amino acid. Will I still get marks?
Probz not :(
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No answer to question 12 yet?
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Does anyone know what the exact questions for 4a was or even have a screenshot/photo? Because I don't remember seeing any questions before the x/y one... woops.
EDIT: Don't worry, I found the scanned copy. I definitely didn't see the question, but I'm hoping I might have done it anyway?
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does anyone know where a copy of the Chemistry exam can be located? Thanks
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for question 10a and 11a, can the OH- and H+ be left out since they are spectators to the reaction?
or are they not spectators
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does anyone know where a copy of the Chemistry exam can be located? Thanks
2014 Exam paper
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did anyone realise VCAA didn't put + or - for Q1 short answers for the delta H?
I forgot to say delta H= +92.4 kj mol-1
I just put delta H=92.4 on the graph, do you think they will take a mark off for that?
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Probz not :(
But some parts specifically asked for side chains.....
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But some parts specifically asked for side chains.....
yea i remember there were 2 that asked for side chains so i wrote the side chains for those
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hmm i didnt notice that two required "side chains" only... i guess it depends on how VCAA determine the marking scheme - but i'd assume that writing the name of the acid in those cases (instead of the side chain) would be wrong :(
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Guys i'm really confused about question 6 mc answer. I thought it was a weak acid added to a weak base, because the PH levels out at 2..and doesn't decrease even after you add excess acid. If it was HCL wouldn't it decrease to around 1?
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Could you say KOH was a catalyst, speeds up the reactions for Question 10.
I know it is an electrolyte, I don't even know why I didn't write that in the first place but I couldn't be bothered explaining the exchange of OH- between the anode and cathode. I'll probs lose a mark for saying its a catalyst LOLL but nothing in the question suggests it can't be.
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Guys i'm really confused about question 6 mc answer. I thought it was a weak acid added to a weak base, because the PH levels out at 2..and doesn't decrease even after you add excess acid. If it was HCL wouldn't it decrease to around 1?
Omg same too, I thought it was weak acid and weak base, because theres no clear equivalence point. Then I looked at the solutions and saw that it started pH11, probably a strong base then?
But then I just googled and a weak base has pH between 7-11 :/ So I don't know but I picked C
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Omg same too, I thought it was weak acid and weak base, because theres no clear equivalence point. Then I looked at the solutions and saw that it started pH11, probably a strong base then?
But then I just googled and a weak base has pH between 7-11 :/ So I don't know but I picked C
No it's definitely a weak base. the Ph is around 10.3 if you look at the question again. If it was a strong acid the line should continue to decrease..
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But some parts specifically asked for side chains.....
Yeah that's right, they wouldn't have specified "amino acid side chains" if they only wanted the names of the amino acids themselves. So the answer would be the "amino acid side chains"
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I think this is a little nitpicky and between you and me, I think when the writers wrote the questions they wouldn't have picked up on the fact students would notice the difference (I don't think they put that 'amino acid side chain' deliberately).
Personally, I'd pay both naming the amino acids and writing the side chains themselves.
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Guys i'm really confused about question 6 mc answer. I thought it was a weak acid added to a weak base, because the PH levels out at 2..and doesn't decrease even after you add excess acid. If it was HCL wouldn't it decrease to around 1?
The key thing is is that its "volume added" not "moles added". Now what does that mean, well if the concentration of HCL is 0.01 M, then, no matter how much volume you add, the pH will not go below 2.
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The key thing is is that its "volume added" not "moles added". Now what does that mean, well if the concentration of HCL is 0.01 M, then, no matter how much volume you add, the pH will not go below 2.
wait so what do you think the answer is
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With the 6 MC, if you look at the approximate end point of the titration, the pH is below 7, at approx 5, and this implies a strong acid and a weak base, I think.
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With the 6 MC, if you look at the approximate end point of the titration, the pH is below 7, at approx 5, and this implies a strong acid and a weak base, I think.
yeah thats what I think too, there is definitely an end point present, not like a weak base, weak acid titration where you can't really see it
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Does anyone know if a little bit of red pen is picked up by the scanner?
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Could you say KOH was a catalyst, speeds up the reactions for Question 10.
I know it is an electrolyte, I don't even know why I didn't write that in the first place but I couldn't be bothered explaining the exchange of OH- between the anode and cathode. I'll probs lose a mark for saying its a catalyst LOLL but nothing in the question suggests it can't be.
I think I said something about it being a catalyst last minute too...I was running out of time and it was the only thing I was able to come up with lol
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I think I said something about it being a catalyst last minute too...I was running out of time and it was the only thing I was able to come up with lol
I doubt catalyst will be accepted
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I wonder how many people fell into the trap in the MC chromatography question where the origin was actually at 2cm rather than 0cm.
Also I'm getting kinda nervous I'm not gonna get the full mark for some worded questions, because when I compare what I wrote to thushan's, mine sounded significantly much more general and idk if i used enough chemical metalanguage to earn the full mark :-[
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With the 6 MC, if you look at the approximate end point of the titration, the pH is below 7, at approx 5, and this implies a strong acid and a weak base, I think.
Definitely strong acid-weak base titration.
pH at equivalence point is 5 and there's a pretty decently sharp endpoint. Weak acid-weak base basically have no sharp endpoint; it's practically a near-horizontal line (yes horizontal, not vertical) near the equivalence point.
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I wonder how many people fell into the trap in the MC chromatography question where the solvent front was actually at 2cm rather than 0cm.
omg I know right!! I was about to then checked again. So many cheeky MCs. another one was the first question which asked for reverse reaction. cheeky cheeky
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Could you say KOH was a catalyst, speeds up the reactions for Question 10.
I know it is an electrolyte, I don't even know why I didn't write that in the first place but I couldn't be bothered explaining the exchange of OH- between the anode and cathode. I'll probs lose a mark for saying its a catalyst LOLL but nothing in the question suggests it can't be.
LOL -> exact same answer and thought process as me aha
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I doubt catalyst will be accepted
much sad haha 1 mark oh well, I wish I would've been bothered to write "it is an electrolyte" but you probably have to explain what an electrolyte does, that's why I was like maybe catalyst will cut it haha
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I'll look through these solutions after my final exam tomorrow.
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For the very last SA question, I wrote insulate beaker, use a stirrer and using something like wool to cover up the bits of the holes in the lid where the stirrer and thermometer go (with elaboration of why for these all of course) and that catalysts don't change molar enthalpy (ranting about how it only affects rate too).
Is this enough for 5 marks?
Well that is not what I wrote but I think it's completely right. I wrote how the volume and conc. of H2O2 should be the same and the catalyst amount should be the same just to show the only variable is the type of catalyst. And same as you for molar enthalpy, it only lowers activation energy.
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I wrote how the volume and conc. of H2O2 should be the same and the catalyst amount should be the same just to show the only variable is the type of catalyst. And same as you for molar enthalpy, it only lowers activation energy.
Oh my god I literally had the same answer as you, in the same order haha
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For the Quinone question I wrote Br2 + 2e- -> 2Br- which does not include the H+
And I wrote QH2 as aqueous but Q as liquid.
Do I lose any marks for those? Thank you
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For the Quinone question I wrote Br2 + 2e- -> 2Br- which does not include the H+
And I wrote QH2 as aqueous but Q as liquid.
Do I lose any marks for those? Thank you
I wrote q as liquid too bro - but I don't see why we would lose marks because quinone isn't on study design so we weren't supposed to know if it was soluble in water.
I think I lost around 6 marks in this paper I hope I can still hit a 46/47 but idk....ahha bloody dumb mistakes
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For the Quinone question I wrote Br2 + 2e- -> 2Br- which does not include the H+
And I wrote QH2 as aqueous but Q as liquid.
Do I lose any marks for those? Thank you
Not including the H+...I don't know, it could go either way with VCAA, it's technically correct since HBr (aq) is just H+ and Br-, but the question explicitly stated HBr was formed.
As for the QH2...don't think you'll get the mark for writing (l) since the question stated "aqueous solutions of QH2" or something along those lines. :(
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Not including the H+...I don't know, it could go either way with VCAA, it's technically correct since HBr (aq) is just H+ and Br-, but the question explicitly stated HBr was formed.
As for the QH2...don't think you'll get the mark for writing (l) since the question stated "aqueous solutions of QH2" or something along those lines. :(
I would be REALLY cheesed off if HBr(aq) was the only answer accepted because come on. That thing is a stronger acid than hydrochloric acid.
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Not including the H+...I don't know, it could go either way with VCAA, it's technically correct since HBr (aq) is just H+ and Br-, but the question explicitly stated HBr was formed.
As for the QH2...don't think you'll get the mark for writing (l) since the question stated "aqueous solutions of QH2" or something along those lines. :(
Shit, do they take a mark of for every equation you had the wrong state in or just the first one.....how did I miss the aqueous wow
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When I did Chemistry in 2011 you only lost maximum 1 mark for incorrect states from memory.
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I would be REALLY cheesed off if HBr(aq) was the only answer accepted because come on. That thing is a stronger acid than hydrochloric acid.
Nothing on fluoroantimonic acid though :P
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So 8 marks are gone :(
I am desperate to see whether I lose another mark for writing grounding apricots to powder to increase the rate of reaction. Assume my MC matches yours (16C and has not counted Q12 yet), I may lose up to 9 marks already :(
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Oh yeah and I don't think I would lose another mark for writing ions since in the Q2 of 2013 exam they asked to write a balanced equation from (NH4)2C2O4 aqueous to CaC2O4.H2O solid and they accepted a range of answers. But not sure with the "overall equation" in this year's question though.
But I lose a mark for wrong state anyway
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So 8 marks are gone :(
I am desperate to see whether I lose another mark for writing grounding apricots to powder to increase the rate of reaction. Assume my MC matches yours (16C and has not counted Q12 yet), I may lose up to 9 marks already :(
depends on how lenient VCAA is but idk, seems rather superficial in the context of the question (why does it matter there's a greater reaction rate if not all the SO2 has been reacted?)
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Oh my god I literally had the same answer as you, in the same order haha
yeah cause that's the orser it was in the table they gave us :P
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So 8 marks are gone :(
I am desperate to see whether I lose another mark for writing grounding apricots to powder to increase the rate of reaction. Assume my MC matches yours (16C and has not counted Q12 yet), I may lose up to 9 marks already :(
Sigh lost like 7-10 marks too depending how harsh they are. What does 110-113 even give you? Raw score wise
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Sigh lost like 7-10 marks too depending how harsh they are. What does 110-113 even give you? Raw score wise
that's pretty good. i'd say deffs 45+ maybe around 47
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that's pretty good. i'd say deffs 45+ maybe around 47
Thanks I really wanted 45+ :) hope it turns out okay.
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Thanks I really wanted 45+ :) hope it turns out okay.
lznxl (soz may have spelt it wrong)
Lost 6 marks and got a 50 :)
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for question 10a and 11a, can the OH- and H+ be left out since they are spectators to the reaction?
or are they not spectators
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Hi, I believe question 1a section b is wrong. They gave us the delta H for 2mol NH3 reacting, yet the graph below only wanted the enthalpy for 1mol NH3. Am I correct?
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I wonder how many people fell into the trap in the MC chromatography question where the solvent front was actually at 2cm rather than 0cm.
I did
I didn't even look at the origin
why would they even label it that way, it makes no sense lol
I also failed at Q1 as well, I noticed it said reverse on my first read through in reading time but I did the short answer stuff first so I forgot :(
so yeah a few people hahaha those questions were just silly
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Ya Habibi - I see where you are coming from, but it's just a labelling thing; I don't think they specifically mean (1 NH3).
Having said that, if you explicitly wrote your products with the coefficients halved on the products side in the diagram, I'd give full marks.
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interested in this too! on the one hand they are technically spectating, but on the other hand, the ions are explicitly mentioned in the question. my instinct would be to pay both as a result.
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I did
I didn't even look at the origin
why would they even label it that way, it makes no sense lol
I also failed at Q1 as well, I noticed it said reverse on my first read through in reading time but I did the short answer stuff first so I forgot :(
so yeah a few people hahaha those questions were just silly
Yep, I got question 1 MC wrong, too. I had done so many practice exams that it became second nature to do the forward reaction. fml
And thushan, what if they meant to do 1mol NH3? Then at least 85% of the state would get it wrong. It could have been a sneaky thing they tried.
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It could have been a sneaky thing they tried.
So many sneaky things VCAA did/may have done, I'm going to kill somebody
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lznxl (soz may have spelt it wrong)
Lost 6 marks and got a 50 :)
Was this last year?
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Was this last year?
yes
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Hi, I believe question 1a section b is wrong. They gave us the delta H for 2mol NH3 reacting, yet the graph below only wanted the enthalpy for 1mol NH3. Am I correct?
With the ammonia activation energy graph question (Section B, Q1) at first I was suspicious about the mol ratio as well, and thought about halving. First of all, I'm pretty sure if we halved the enthalpy it would've actually been less that the activation energy, even with the catalyst, so that probably wouldn't have worked out. Secondly, I'm pretty sure both the y-axis units and the units given in the enthalpy were both KJ/mol - hence there's no need to halve the equation - even the activation energies were given in those units. Sneaky from VCAA though.
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Strike that first point - activation energies would've theoretically halved too, but hopefully the second explanation helps you get my drift. :)
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Strike that first point - activation energies would've theoretically halved too, but hopefully the second explanation helps you get my drift. :)
I disagree. I was taught that the unit refers to the coefficient of the subject reactant.
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Would 1-propyl ethanoate be accepted? Because technically couldn't there be a 2-propyl ethanoate formed from propan - 2 - ol. Also, would you consider C13H29COOCH3 a semi structural formula?
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Would 1-propyl ethanoate be accepted? Because technically couldn't there be a 2-propyl ethanoate formed from propan - 2 - ol. Also, would you consider C13H29COOCH3 a semi structural formula?
I'm not even sure that's possible. You're asking for an alkyl group of C13H29, which implies a C13H30 alkane, but thirteen carbons can only bond to at most 28 hydrogens.
You really should write CH3(CH2)12COOCH3 if you want to be more precise with your semi-structural formula.
I think it's implied with the ester that propyl ethanoate means 1-propyl ethanoate. Someone might want to check that though; VCAA doesn't seem to be picky with it.
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I'm not even sure that's possible. You're asking for an alkyl group of C13H29, which implies a C13H30 alkane, but thirteen carbons can only bond to at most 28 hydrogens.
You really should write CH3(CH2)12COOCH3 if you want to be more precise with your semi-structural formula.
I think it's implied with the ester that propyl ethanoate means 1-propyl ethanoate. Someone might want to check that though; VCAA doesn't seem to be picky with it.
Haha, I meant C13H27COOCH3 - so the right number, just probably not correct because the first half is not in groups??
So if I did 1-propyl ethanoate do you think I'd get that wrong?