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July 17, 2025, 04:14:29 pm

Author Topic: VCE Chemistry Question Thread  (Read 2833968 times)  Share 

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HopefulLawStudent

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Re: VCE Chemistry Question Thread
« Reply #5445 on: July 22, 2016, 07:13:04 pm »
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So the solvent will give a peak too? Didn't know that.

Is that the same for HPLC? Will the solvent for HPLC give a peak also?

zsteve

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Re: VCE Chemistry Question Thread
« Reply #5446 on: July 22, 2016, 08:06:54 pm »
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So the solvent will give a peak too? Didn't know that.

Is that the same for HPLC? Will the solvent for HPLC give a peak also?

Well, I'm pretty sure that for HPLC if you're dissolving your substance in a solvent A and then using another solvent B as the mobile phase for HPLC, you'll get a peak for the solvent.

If your solvent is also used for mobile phase, I guess you'll get constant background readings due to solvent :P
« Last Edit: July 22, 2016, 08:08:47 pm by zsteve »
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HasibA

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Re: VCE Chemistry Question Thread
« Reply #5447 on: July 23, 2016, 01:49:01 pm »
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when we change pressure, why do the concentrations of reactants and products sharply increase then follow le chatelier's principle to go back into equilibrium? like an explanation of how this works, not just why- thanks :)

edit: why do concentrations change, more so, than just increase- thanks :) (this related to equilimbrium conc./time graphs)
« Last Edit: July 23, 2016, 01:56:50 pm by HasibA »
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HighTide

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Re: VCE Chemistry Question Thread
« Reply #5448 on: July 23, 2016, 02:22:34 pm »
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when we change pressure, why do the concentrations of reactants and products sharply increase then follow le chatelier's principle to go back into equilibrium? like an explanation of how this works, not just why- thanks :)

edit: why do concentrations change, more so, than just increase- thanks :) (this related to equilimbrium conc./time graphs)
When you change the pressure, you change the volume. The lower the volume (higher pressure), the more likely particles are to react. Consider putting many people in a large room then moving all of them to a smaller room, they'd bump into each other more often. So basically, the more particles you have, the more likely for that reaction to occur. Hence why increased pressure favours the side with fewer particles.
So considering the reaction: A + D  --> 2C
If I have A + D in a container, and change the pressure (eg increase), then A + D have less room to move around and more chance of colliding with the right orientation. This would result in an increase in [C] and a decrease in [A] and [D]. Over time, as [C] increase, it's more likely to react, 2 C's would collide to make A + D, and then A and D would react again and then this would continue until they reached equilibrium. This part corresponds to the flattening out. Overall though, a new equilibrium is established. That's because, with lower pressure, you have more [C] and less of [A] and [D].
You can see this through the equilibrium constant as it stays...constant.
K= [C]^2/[A][D] --> If C increases, A and D must decrease.
Alternatively, if you decrease pressure, C decreases and the concentration of reactants will increase accordingly.

« Last Edit: July 23, 2016, 02:28:44 pm by HighTide »
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HopefulLawStudent

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Re: VCE Chemistry Question Thread
« Reply #5449 on: July 23, 2016, 03:11:40 pm »
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For the reversible equation:

N2 + 3H2 <----> 2NH3

Let's say that we've increased the amount of N2. Would I be correct if I were to say:

"There will be a net forward reaction so that the concentration of N2 will decrease and equilibrium reestablished. However, the concentration of N2 when equilibrium is reestablished will still be more than the concentration of N2 at equilibrium before N2 was added to the solution."

Is that last bit (underlined) too ambiguous + is that second sentence even true?

NB: This wasn't to an actual question or anything; I'm just throwing around potential things that could be asked and making sure my equilibrium theory is on point. :P

Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #5450 on: July 23, 2016, 05:42:10 pm »
+1
For the reversible equation:

N2 + 3H2 <----> 2NH3

Let's say that we've increased the amount of N2. Would I be correct if I were to say:

"There will be a net forward reaction so that the concentration of N2 will decrease and equilibrium reestablished. However, the concentration of N2 when equilibrium is reestablished will still be more than the concentration of N2 at equilibrium before N2 was added to the solution."

Is that last bit (underlined) too ambiguous + is that second sentence even true?

NB: This wasn't to an actual question or anything; I'm just throwing around potential things that could be asked and making sure my equilibrium theory is on point. :P
Hey,

 I feel that your understanding is correct, however I would challenge the underlined section. This is due to the fact that there are multiple different concentrations of N2 that are present as the reaction progresses prior to the addition of more N2. If you were to mention this, you would need to be more specific in your recognition of the time. If you see the image attached, you can see that the concentration of N2 after equilibrium is established for a 2nd time is still higher than the concentration of N2 immediately prior to the addition of the extra N2.

As the book states:
Note that the concentration of N2O4 at the new equilibrium position is still higher than it was at the original equilibrium position.

Hope this helps :)

 
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blacksanta62

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Re: VCE Chemistry Question Thread
« Reply #5451 on: July 23, 2016, 07:37:09 pm »
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Hey guys, could someone explain part b to me please? I have the worked solutions but it doesn't seem to make sense. Weird thing is I had that moment of revelation when the answer popped into my head but it disappeared before I could write it down.
Thanks :)
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HighTide

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Re: VCE Chemistry Question Thread
« Reply #5452 on: July 23, 2016, 07:43:38 pm »
+1
Hey guys, could someone explain part b to me please? I have the worked solutions but it doesn't seem to make sense. Weird thing is I had that moment of revelation when the answer popped into my head but it disappeared before I could write it down.
Thanks :)
Hey mate,
The forward reaction is exothermic. So decreasing heat by cooling means that the forward reaction occurs to a lesser extent. Hence the backward reaction is favoured. That means you get more carbon dioxide in gaseous form--> better fizz.
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blacksanta62

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Re: VCE Chemistry Question Thread
« Reply #5453 on: July 23, 2016, 09:04:13 pm »
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Thanks hightide :)
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blacksanta62

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Re: VCE Chemistry Question Thread
« Reply #5454 on: July 23, 2016, 09:26:01 pm »
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More of a query: Would we ever be asked to write a balanced chemical equation for the reaction between oxalic acid dihydrate and sodium hydroxide? I can't seem to work it out :(

I think it's because it's given to me in the form H2C2O4.2H20
So far I have: NaOH(aq) + H2C2O4.2H2O(aq) ------> ?

Any help would be appreciated :)

Edit: In the end of year exam :)
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Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #5455 on: July 23, 2016, 09:33:23 pm »
+1
More of a query: Would we ever be asked to write a balanced chemical equation for the reaction between oxalic acid dihydrate and sodium hydroxide? I can't seem to work it out :(

I think it's because it's given to me in the form H2C2O4.2H20
So far I have: NaOH(aq) + H2C2O4.2H2O(aq) ------> ?

Any help would be appreciated :)

Edit: In the end of year exam :)
Hey,
 To make it easier, the dihydrate part of the reactant does not actually take part in the reaction, so you could easier write the equation with or without the dihydrate.
The equation without would read:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
I would recommend staying away from the reaction with the dihydrate part involved, due to the fact that sodium oxalate would occur has a trihydrate, so you would need to consider this in the equation when oxalic acid is dihydrate.
H2C2O4.2H2O + 2NaOH → Na2C2O4.3H2O + H2O
I'd recommend staying away from that.

I'm unsure if anything of the sort is involved in your curriculum, but I'm sure that it's not involved in SACE.
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geminii

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Re: VCE Chemistry Question Thread
« Reply #5456 on: July 23, 2016, 09:46:40 pm »
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Could anyone give an explanation as to why metals have lower core charges? ???
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blacksanta62

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Re: VCE Chemistry Question Thread
« Reply #5457 on: July 23, 2016, 09:48:15 pm »
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So the H2 and the Na simply switch locations? Then balance? I've understood what you've done perfectly but I'm looking for a shortcut in case it ever comes up. Making those neural pathways baby!
see what I did there   ;) ;)
Thank you :)
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Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #5458 on: July 23, 2016, 09:57:41 pm »
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So the H2 and the Na simply switch locations? Then balance? I've understood what you've done perfectly but I'm looking for a shortcut in case it ever comes up. Making those neural pathways baby!
see what I did there   ;) ;)
Thank you :)
Basically, but remember it's an acid + base reaction so H2O is always formed, and the leftovers just form together (that's how I remember it)
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blacksanta62

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Re: VCE Chemistry Question Thread
« Reply #5459 on: July 23, 2016, 10:21:35 pm »
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Cool, no carbonate, just water and the salt. Got it :D
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