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March 28, 2024, 11:17:20 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313098 times)  Share 

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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #975 on: June 08, 2014, 11:54:10 am »
+1
In shortcut VCE Chemistry style:

[H+] = [chloroacetate]
1.3 * 10^-3 = [H+][chloroacetate]/[chloroacetic acid]

The equilibrium concentration of chloroacetic acid is given by 1 - [H+]. Think about it. By mole ratios, the number of moles of chloroacetic acid reacting is equal to the number of moles of H+ formed
Now we're going to make an assumption (which I think doesn't work here) that the amount dissociated is negligible in comparison to the initial concentration. AKA 1 - [H+] is still close to 1
Then, we get 1.3*10^-3 = [H+][chloroacetate] = [H+]^2
[H+] = 0.036 M
You can find the others from this.

Now this is an example when you can JUST make the assumption 1- [H+] is still close to 1.
If you were to solve it properly, you would have 1.3 * 10^-3 = [H+]^2/(1-[H+]), which is a quadratic equation deemed by VCAA to be too hard for the average student to solve. Solving the quadratic gives [H+] = 0.0354 M, which is quite close.

In either case, in a VCE exam just do it the first way.
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Reus

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Re: VCE Chemistry Question Thread
« Reply #976 on: June 08, 2014, 02:37:25 pm »
0
Find the % of Oxygen in:
Fe2O3


...49?
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #977 on: June 08, 2014, 02:41:07 pm »
0
I think it's just

MM= 159.7g/mol
Mass of oxygen=48g

48/159.7 *100 = 30%

Reus

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Re: VCE Chemistry Question Thread
« Reply #978 on: June 08, 2014, 04:36:20 pm »
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Ah God! How did that click in my head  :o ::)
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #979 on: June 08, 2014, 10:10:54 pm »
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ok so lets say i have an equilibrium reaction of like

yeah.. and its K value is 1.56M^-1.

Does this mean that it favours the exothermic (back) reaction?

Does the unit corresponding to M (molarity), whether it may be M^2, M^3, or M^-2 relate to whether or not a reaction is exo or endothermic? Or do i just look at the value of delta H..?

Because for some reason, all the exothermic reactions ive seen are like


any clarification would be great :P just trying to get a grip on this new unit haha

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #980 on: June 08, 2014, 10:44:07 pm »
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ok so lets say i have an equilibrium reaction of like

yeah.. and its K value is 1.56M^-1.

Does this mean that it favours the exothermic (back) reaction?

Does the unit corresponding to M (molarity), whether it may be M^2, M^3, or M^-2 relate to whether or not a reaction is exo or endothermic? Or do i just look at the value of delta H..?

Because for some reason, all the exothermic reactions ive seen are like


any clarification would be great :P just trying to get a grip on this new unit haha

Technically, equilibrium constants are meant to be expressed using a dimensionless quantity called a chemical activity, which is for most cases numerically equal to the concentration of the aqueous species in M. That's why we normally use molar concentrations in equilibrium constants. However, equilibrium constants aren't strictly meant to have units, but whatever.

In VCE chemistry, the unit comes from the fact that the equilibrium expression has [SO3]^2 on the top but [SO2]^2 [O2] on the bottom. As concentrations are in M, this means the overall units are M^2/M^3 = M^-1. Just means more gas particles on the left than on the right.

Exo/endothermic means delta H value negative/positive. It turns out that this reaction is exothermic. Really, you're meant to give a temperature when quoting equilibrium constants as these constants change with temperature only.
Exothermicity is not a very good indication of equilibrium position; dissolution of ammonium nitrate is endothermic but you'd hardly say it's disfavoured given the solubility rules that ammonium AND nitrate ions are soluble. Dissolution of salts is generally endothermic. Think about it; that's why you dissolve more at higher temperatures.
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #981 on: June 08, 2014, 10:47:52 pm »
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Thanks for the lzxnl!

melbin123

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Re: VCE Chemistry Question Thread
« Reply #982 on: June 09, 2014, 12:46:02 pm »
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@sense thx A lot buddy :)
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Reus

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Re: VCE Chemistry Question Thread
« Reply #983 on: June 10, 2014, 07:29:15 pm »
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Need help with this question, well I ended up working it out however clueless towards what the answer should be as we weren't provided an answer sheet.
So I'd love to see what you guys get :)

"Anglesite is a lead bearing ore found in large deposits in Australia. One particular deposit of Anglesite was found to contain 80% of lead (II) sulfate.
What mass of lead could be extracted from every one tonne (1000kg) of Anglesite processed?
"

Thanks guys  ;D
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #984 on: June 10, 2014, 08:01:12 pm »
0
Need help with this question, well I ended up working it out however clueless towards what the answer should be as we weren't provided an answer sheet.
So I'd love to see what you guys get :)

"Anglesite is a lead bearing ore found in large deposits in Australia. One particular deposit of Anglesite was found to contain 80% of lead (II) sulfate.
What mass of lead could be extracted from every one tonne (1000kg) of Anglesite processed?
"

Thanks guys  ;D

Complete guess but did you get 800g?
Sorry if this is wrong :(
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Reus

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Re: VCE Chemistry Question Thread
« Reply #985 on: June 10, 2014, 08:20:50 pm »
+1
Complete guess but did you get 800g?
Sorry if this is wrong :(
Not quite...
I ended up with 683.24kg if it were 100% of PbSO4 in the extract.
However considering it is 80%; I went with 800kg instead of 1000kg which resulted in m(Pb)=546.59kg out of the 1000kg
Therefore each tonne of Anglesite contains 547kg of lead.

Yet, positive my calculations for the 80% of lead is incorrect, I have a feeling haha.
(Reason why I posted this :P to make sure if I'm right!)

So no need to be sorry! I'm probably wrong too  ::)
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Bestie

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Re: VCE Chemistry Question Thread
« Reply #986 on: June 11, 2014, 09:06:40 pm »
0
what is the m/e of base peak of propan-1-ol in mass spectrometry after complete fragmentation?

thank you

Reus

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Re: VCE Chemistry Question Thread
« Reply #987 on: June 11, 2014, 09:25:01 pm »
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M(propan-1-ol) = 60g/mol

We begin with C3H8O. Fragmentation of others will occur.
CH3O+ would be graphed with the highest point (Generally graphed for you). However given there is no data, O is most electronegative, compared to other fragmentations.(holds onto electrons the best and will form the base peak)

Molar mass of CH3O+ = 12+(1x3)+16
                                 = 31.
Henceforth m/e base peak is 31.

Correct me if I'm wrong :)
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Bestie

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Re: VCE Chemistry Question Thread
« Reply #988 on: June 11, 2014, 09:28:02 pm »
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thank you but how do you know its CH3O+ and not something like OH+?

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #989 on: June 12, 2014, 10:18:50 am »
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They give you a graph in which you can find the base peak. Then, you can work out an appropriate structure for it.
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