VCE Chemistry Question Thread

[kimmie]

hey guys can someone tell me if AAS (atomic absorption spectroscopy) can be used to analyse anions? or is it just cations?

[blacksanta62]

hey guys can someone tell me if AAS (atomic absorption spectroscopy) can be used to analyse anions? or is it just cations?

Anions are non metals in ionic forn since they want to accept electrons. AAS is used only on metals which are cations.

[Elizawei]

Yes, I'd say you can. For instance, the phenyl group (-C6H5). However, I'm not sure how much this relates to VCE Chemistry as you basically don't need to do anything with benzene, just be aware that it exists

Thanks zsteve! There was this question on a commercial practice exam that asked me to circle all the functional groups in a diagram and i didn't know if I should have circled benzene

Ooh one more quick question, for an esterification reaction, are the states of all reactants and products liquid? 
e.g CH3COOH(l) + CH3CH2OH(l) --> CH3COOCH2CH3(l) + H2O(l)

[zsteve]

Thanks zsteve! There was this question on a commercial practice exam that asked me to circle all the functional groups in a diagram and i didn't know if I should have circled benzene

Ooh one more quick question, for an esterification reaction, are the states of all reactants and products liquid? 
e.g CH3COOH(l) + CH3CH2OH(l) --> CH3COOCH2CH3(l) + H2O(l)

That looks good to me. Don't forget to specify H2SO4(l) or 'concentrated H2SO4' as the catalyst. Importantly, H2SO4(aq) doesn't work

[BNard]

Hey guys, sorry to ask such a basic question, but could someone run me through the process for working this out?

A 10 mL solution of hydrochloric acid has a pH of 2.  What volume, in water, in mL, must be added to it to change the pH to 3?

Thanks so much

[zsteve]

Hey guys, sorry to ask such a basic question, but could someone run me through the process for working this out?

A 10 mL solution of hydrochloric acid has a pH of 2.  What volume, in water, in mL, must be added to it to change the pH to 3?

Thanks so much

A increase in pH by 1 means decrease in [H+] by a factor of 10. So your volume must increase by a factor of 10, i.e. to 100mL. You need to add 90mL

[Apink!]

Hello everyone - I'm stuck on a question, and would very appreciate your help!

Carbon monoxide reacts with chlorine to form phosgene, COCl₂.

CO_(g)  +Cl_2(g) <---> COCl_2(g)

I.0 mol of Cl₂ was mixed with 1.0 mol of CO in two separate 1.0L containers at two different temperatures - T1 and T2, and the concentration of COCl₂ was measured. The results of this experiment is given below (I've attached the graph - I drew it and it looks almost identical to the one in the question minus the shaky axis drawing XD)

Which of the temperatures, T1 or T2 is the higher? Give a reason to support your answer.

----

Okay, I thought the higher temperature would be T1 because
a) It produced more COCl₂ over the same time period as the other container, meaning that it had a faster rate of reaction
b) higher rate of reaction usually means higher temperature

But the answer says that it's T2 that is the higher temperature. The reasoning they give me was:
T2, as it has the higher initial rate (most reaction rates increase with increasing temperature) and equilibrium is reached faster)

Can someone please tell me which part of my reasoning was flawed and how a higher initial rate and the fact that equilibrium has reached faster tells us that reaction rate was faster?

Thank you so much everyone!

[sushibun]

Can someone please help me answer this question?
I don't have the solution to it and I'm not sure. I'm guessing D because it seems like a bigger molecule that might be harder to get through the column. I don't know. What would be an acceptable response?
Thanks in advance!

[@#035;3]

I have no clue if I'm right but.. I think peak A corresponds to CH3C(CH3)2OH while peak D represents Butan-1-ol.
My reasoning is that because there's branching for the molecule 2 methyl-propan-2ol, this results in weak intermolecular bonds within the molecule and weaker attractions to the stationary phase.
Someone pls coreect me if i'm wrong

[sweetcheeks]

@Apink!
Where did that question originate from? Was there any indication if forward reaction is exo or endothermic?

I can see a slight flaw in part a) of your answer. This is an equilibrium reaction, just because there is a higher temperature it doesn't mean that more product would be formed. If the forward reaction were exothermic, would a higher temperature give you a larger or smaller yield?

[Swagadaktal]

I have no clue if I'm right but.. I think peak A corresponds to CH3C(CH3)2OH while peak D represents Butan-1-ol.
My reasoning is that because there's branching for the molecule 2 methyl-propan-2ol, this results in weak intermolecular bonds within the molecule and weaker attractions to the stationary phase.
Someone pls coreect me if i'm wrong

That reasoning is correct - but I think the question must tell us whether the stationary phase is polar or non polar.

I think most stationary phases are polar though (idk it's just what most questions have) -- is there some underlying assumption with GLC? But if it is polar, than the branched one would be D coz it has weaker dispersion forces - the straight alkanol has stronger non polar forces therefore will adsorb less to the stationary phase and come out of the chamber quicker (giving A) --- but tbh I think there has to be an indication of what kind of stationary phase it is.

Can anyone tell us whether there's an assumption here? Or is there a different way to work it out we're missing?

[Apink!]

Could I have help with question 9 and 10?

Correct answers are C and A respectively.

I had said D for Q9, and C for Q10. Could you please explain to be why I was wrong?

[HighTide]

Could I have help with question 9 and 10?

Correct answers are C and A respectively.

I had said D for Q9, and C for Q10. Could you please explain to be why I was wrong?

Solids and liquids aren't included in the equilibrium expression. They're taken as 1. So for 9, AgCl isn't included.

For Q10, the only thing I can think of is the equilibrium constants. So for reaction 1 the backward reaction is prevalent and it requires silver and chloride ions. Reaction 2 the forward reaction is more prevalent and requires silver ions. We can eliminate B and D. But for C, if you add HCl, you get chloride ions. Chloride ions would go to the first reaction. However, due to the stoich ratios, you need 1:1, but no matter how much chloride you add, the silver ions are being consumed in the second reaction. So basically, you only have as much as the Ag ions initially in the solution. Not 100% sure, just a guess. But A does seem more right.

[Apink!]

Hi High tide!
Thank you for helping me with these questions! I really appreciate it
Unfortunately,  I still don't understand Q10 I do agree that it feels more "right" but I want to know clearly why

--

Just a quick question:

So liquids and solids don't affect the equilibrium of a system in any way. But could someone give me an idea as to why this is so?
I mean surely, they're chemicals regardless of what states they are in...and presumably affect the system it's in. Why should I consider only the aqueous ones? EDIT: and gaseous ones

EDIT: @sweetcheeks I found the question on a tsfx paper (I think). No it doesn't tell you whether the reaction is exothermic or endothermic. There is actually a later question that asks you to figure it out

Moderator action: Merged double post. You can edit previous posts.

[kimmie]

okay thanks!

Anions are non metals in ionic forn since they want to accept electrons. AAS is used only on metals which are cations.