VCE Chemistry Question Thread

[steenasonson21]

Also the answer for this one is butyl methanoate, can't be methyl butanoate as it says there there is one singlet in a proton NMR, and methyl butanoate would not result in a singlet

But why not? The 'methyl' part of the molecule is next to an oxygen right, so it has no neighboring H atoms = singlet?

[steenasonson21]

- With efficiency questions, if it says its at e.g. 40% efficiency, if it asks for the mass required to produce x amount of energy, once the mass has been calculated, simply divide by 40 then times by 100 to get the actual mass required, due to the system only being 40% efficient, that means more mass would be required then at 100% efficiency.
- With dilution questions they usually follow the process of
- original sample  (e.g. 10ml) is diluted to (e.g.100ml) , sometimes they say they take a sample of 20 mls from the 100 but this doesn't matter cause the new concentration is still the same. Usually you will use molar concentration calculations to find the concentration of the 100ml sample, this may be from a titration or even graph. Then simply always remember to use the equation C1V1=C2V2, sub the found conc in for C2 and 10ml volume for C1 and 100ml volume for C2, and the concentration of the original undiluted sample can be found

Hope this helps

Dividing by 40 or by 0.4? then multiplying by 100 is to make it into a percentage???

[markmorstentine]

Dividing by 40 or by 0.4? then multiplying by 100 is to make it into a percentage???

U can divide by .4 or what I do cause its visually sets it out for me is /40 *100
e.g. 5/.4 = 12.5
        5/40*100 = 12.5
Also for the other one I think they were not specific enough in the IR splitting, and both are correct based on what they gave.

[steenasonson21]

Is H20 considered a greenhouse gas? and why?

[colline]

Is H20 considered a greenhouse gas?

Yes, it is.

[steenasonson21]

Can someone explain why the answer isn't A?
I did option A and got 22 x 89.0 - (21 x 18.0) = 1580

[Moonblossom]

I'm confused about inert gases. When do they affect what reaction is favoured and when do they have no effect at all???

My understanding is that, when added, pressure is increased and so, the side of the reaction with fewer particles is favoured. However, they do not affect concentration of reactants or products.

[markmorstentine]

I'm confused about inert gases. When do they affect what reaction is favoured and when do they have no effect at all???

My understanding is that, when added, pressure is increased and so, the side of the reaction with fewer particles is favoured. However, they do not affect concentration of reactants or products.

Inert gases have no effect on equlilibrium. Whilst more collisions would occur if adding an inert gas, there would also be a greater proportion of unsuccessful collisions so they are seen as having a neutral effect, where the equalibrium system is not altered as well as the r.o.r

[Coolgalbornin03Lo]

When pH is altered is it just the side chains or also R-groups of Amino acids containing NH/COOH which are affected?

I thought it was both but is that correct?

[ArtyDreams]

When pH is altered is it just the side chains or also R-groups of Amino acids containing NH/COOH which are affected?

I thought it was both but is that correct?

The 'R' groups are definitely affected!! - So you have to keep that in mind when drawing the structures when it asks for a drawing at a high ph etc.
Its like in denaturation, a high pH affects the structure of the R groups, causing the protein to lose its function.

[Coolgalbornin03Lo]

The 'R' groups are definitely affected!! - So you have to keep that in mind when drawing the structures when it asks for a drawing at a high ph etc.
Its like in denaturation, a high pH affects the structure of the R groups, causing the protein to lose its function.

Thanks so much! I drew this in a company exam but it was wrong- but I don’t really trust company exams so just came to ask

[everythangcoZ]

Hey guys,
I often see this sort of thing come up, but how do you know the states for carboxylic acids. My teacher says they're usually aqueous and same with the alcohol, but in the exam report for this question they are liquid?
Do you take it literally in real life setting as liquids (don't know if that makes sense lol) just unsure.

[Moonblossom]

Inert gases have no effect on equlilibrium. Whilst more collisions would occur if adding an inert gas, there would also be a greater proportion of unsuccessful collisions so they are seen as having a neutral effect, where the equalibrium system is not altered as well as the r.o.r

Thanks for clearing that up!

[Moonblossom]

I had a completely different method to a question compared to the answers and I don't know whether I'll get full marks.

The question asked for concentration of C₆H₈O₆ in the original sample.

My method was: 3.13 x 10^-6 ÷ 0.025
= 1.252 x 10^-4 M
1.252 x 10^-4 x (250/20)
= 1.565 x 10^-3 M

The answers had: 3.130 x 10^-6 x (250/25)
= 3.13 x 10^-5
3.13 x 10^-5 ÷ (20/1000)
= 1.565 x 10^-3 M

20 ml sample. Diluted to 250 ml. 25 ml aliquots. 3.13 x 10^-6 mol of substance in question.

Sorry I'm not able to attach pictures

[ActivationEnergy]

Hey guys,
I often see this sort of thing come up, but how do you know the states for carboxylic acids. My teacher says they're usually aqueous and same with the alcohol, but in the exam report for this question they are liquid?
Do you take it literally in real life setting as liquids (don't know if that makes sense lol) just unsure.

What exam is this on? I would assume this reaction is occurring at SLC, thus both are at a liquid state.

I had a completely different method to a question compared to the answers and I don't know whether I'll get full marks.

The question asked for concentration of C₆H₈O₆ in the original sample.

My method was: 3.13 x 10^-6 ÷ 0.025
= 1.252 x 10^-4 M
1.252 x 10^-4 x (250/20)
= 1.565 x 10^-3 M

The answers had: 3.130 x 10^-6 x (250/25)
= 3.13 x 10^-5
3.13 x 10^-5 ÷ (20/1000)
= 1.565 x 10^-3 M

20 ml sample. Diluted to 250 ml. 25 ml aliquots. 3.13 x 10^-6 mol of substance in question.

Sorry I'm not able to attach pictures

Yes it is fine. You have just used a dilution factor instead.