Can anyone do the answers to the exam?

[Christafa]

Yeah, SA was far too easy though, I mean "What are the complementary bases?" come on, are you serious?

But yeah, I think Q4 did the job, stuffed a few people up. But that was basically the only one.

It was so easy that I originally stuffed up the "how many hydrogen bonds" question. It's deceptive, and I think a lot of people will have just written "5".

[Anaath]

Hmm I think I also got this wrong too as I failed to add but regarding question 10.
Can you just assume that 1mol of the CO2 produced in the first equation would still be the same volume as the CO2 produced in the last equation?
My logic was that if you started with 1mol C6H12O6:
1C6H12O6 ---> 2CH3CH2OH + 2CO2
If all the CH3CH2OH was then converted
2CH3CH2OH + 2O2 --> 2CH3COOH + 2H2O
4CH3COOH + 2CaCO3 ---> 2Ca(CH3COO)2 + 2CO2 + 2H2O
2Ca(CH3COO2) + 2H2O ---> 4CH4 + 2CO2 + 2CaCO3

Therefore methane : carbon dioxide
                         4: 2 + 2 + 2
                         4: 6
                         2: 3 D?
I stuffed up counting all the CO2 however, wouldn't that be more appropriate than simply just adding up the mols produced of each?
I haven't heard of anyone else have the same logic though so if i'm wrong, why doesn't that work? 

[4231]

i also got 2:3, im fairly sure that is the correct answer, not 1:2

[Water]

Annath , logic doesn't work like that,

The mol of the first equation is taken to 2nd equation. Doesn't mean that its 2CH3CH2OH though.

[jane1234]

Hmm I think I also got this wrong too as I failed to add but regarding question 10.
Can you just assume that 1mol of the CO2 produced in the first equation would still be the same volume as the CO2 produced in the last equation?
My logic was that if you started with 1mol C6H12O6:
1C6H12O6 ---> 2CH3CH2OH + 2CO2
If all the CH3CH2OH was then converted
2CH3CH2OH + 2O2 --> 2CH3COOH + 2H2O
4CH3COOH + 2CaCO3 ---> 2Ca(CH3COO)2 + 2CO2 + 2H2O
2Ca(CH3COO2) + 2H2O ---> 4CH4 + 2CO2 + 2CaCO3

Therefore methane : carbon dioxide
                         4: 2 + 2 + 2
                         4: 6
                         2: 3 D?
I stuffed up counting all the CO2 however, wouldn't that be more appropriate than simply just adding up the mols produced of each?
I haven't heard of anyone else have the same logic though so if i'm wrong, why doesn't that work?  

Doubling the second line doesn't do anything because you come out with 2CH3COOH, and you need the 2CH3COOH for the next stage, which is already there. So no more CO2 is produced with doubling the second equation.

[Milkshake]

Step 1: glucose broken down into 2 ethanols    -> 2Co2 here
Step 2: Each ethanol makes one ethanoic acid, therefore we make 2 ethanoic acids
Step 3: 2 ethanoic acids come in and make one calcium thingo, aswell as one CO2
Step 4: One calcium thingo comes in and makes 2 methane, and 1 CO2
Therefore, 2 methanes, and 4 CO2 total

[mil0006]

what did people say for the mc question about the weak and strong acid. I said it couldn't be determined using a titration which is probably wrong???

[Milkshake]

Same number of mole. Equivalence point is where you mix them in mole ratios, and since they're both monoprotic, then its the same

[BoredSatan]

Same number of mole. Equivalence point is where you mix them in mole ratios, and since they're both monoprotic, then its the same

FML

[miettargh]

what did people say for the mc question about the weak and strong acid. I said it couldn't be determined using a titration which is probably wrong???

Same amount required for each titration. I'm about 95% sure this is the answer as it was on a practice exam I did during SWOTVAC.

[dakes112]

can somebody please printout the exam, do it, scan it back onto the computer and then post it on here. it would save so much time and confusion. I would do it but my chemistry skills and knowledge isn't up to the standard of some people on here.

[DisaFear]

Same number of mole. Equivalence point is where you mix them in mole ratios, and since they're both monoprotic, then its the same

Pays to do past exams  :crazy2:

[Anaath]

Hmm I think I also got this wrong too as I failed to add but regarding question 10.
Can you just assume that 1mol of the CO2 produced in the first equation would still be the same volume as the CO2 produced in the last equation?
My logic was that if you started with 1mol C6H12O6:
1C6H12O6 ---> 2CH3CH2OH + 2CO2
If all the CH3CH2OH was then converted
2CH3CH2OH + 2O2 --> 2CH3COOH + 2H2O
4CH3COOH + 2CaCO3 ---> 2Ca(CH3COO)2 + 2CO2 + 2H2O
2Ca(CH3COO2) + 2H2O ---> 4CH4 + 2CO2 + 2CaCO3

Therefore methane : carbon dioxide
                         4: 2 + 2 + 2
                         4: 6
                         2: 3 D?
I stuffed up counting all the CO2 however, wouldn't that be more appropriate than simply just adding up the mols produced of each?
I haven't heard of anyone else have the same logic though so if i'm wrong, why doesn't that work? 

Doubling the second line doesn't do anything because you come out with 2CH3COOH, and you need the 2CH3COOH for the next stage, which is already there. So no more CO2 is produced with doubling the second equation.

Fair enough then. Yeah I had my doubts about it but I was thinking 1:2 seems way to easy to be the answer. Ahh overthinking the answers ftl.

[BoredSatan]

What was the answer to the petrol question?

[DisaFear]

What was the answer to the petrol question?

I put GC, as did the other few people I asked