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September 18, 2025, 04:37:16 pm

Author Topic: VCE Chemistry Question Thread  (Read 2900438 times)  Share 

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zsteve

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Re: VCE Chemistry Question Thread
« Reply #5190 on: June 10, 2016, 01:16:28 pm »
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Apparently the answer is C...?

Oops! I didn't realise at first. Note that in the reaction mixture, the spot roughly corresponding to A is slightly shifted. Thus A has somehow changed in the reaction mixture and that's preventing the reaction from progressing.
Sorry I didn't pick that up initially.
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jyce

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Re: VCE Chemistry Question Thread
« Reply #5191 on: June 10, 2016, 05:20:56 pm »
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Could someone please explain this Q to me?

It's a pretty dodgy question. I wouldn't stress over it too much  :)

zsteve

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Re: VCE Chemistry Question Thread
« Reply #5192 on: June 10, 2016, 05:22:00 pm »
+1
It's a pretty dodgy question. I wouldn't stress over it too much  :)

Agreed. Highly unlikely VCAA would pull this.
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HopefulLawStudent

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Re: VCE Chemistry Question Thread
« Reply #5193 on: June 11, 2016, 03:06:16 pm »
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Thanks guys! :D

smiley123

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Re: VCE Chemistry Question Thread
« Reply #5194 on: June 13, 2016, 05:53:50 pm »
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For the combustion of methane, the process of breaking bonds uses 3380 kj of energy and 4270 kJ of energy is released when new bonds are formed. I know that for exothermic reactions the enthalpy has to be negative, but doesn't change in enthalpy=products-reactants? So that would be 4270-3380=+890?

zsteve

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Re: VCE Chemistry Question Thread
« Reply #5195 on: June 13, 2016, 05:59:22 pm »
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For the combustion of methane, the process of breaking bonds uses 3380 kj of energy and 4270 kJ of energy is released when new bonds are formed. I know that for exothermic reactions the enthalpy has to be negative, but doesn't change in enthalpy=products-reactants? So that would be 4270-3380=+890?
Yes, but the equation 'enthalpy=products-reactants' doesn't apply to the values you have. (this equation only applies if you use values, which aren't covered in VCE)
Think about what the numbers mean physically: breaking bonds = consuming energy, I use up 3380kJ. But I get 4720kJ back, so overall I've gained 4720-3380kJ = +890kJ. I get 890kJ out of my reaction, so =-890kJ (negative because I get energy)
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smiley123

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Re: VCE Chemistry Question Thread
« Reply #5196 on: June 13, 2016, 06:03:52 pm »
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ohh okayyy! thank you! :D

geminii

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Re: VCE Chemistry Question Thread
« Reply #5197 on: June 13, 2016, 09:22:50 pm »
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Hey everyone!
I was just doing my chemistry homework about water (I'm in unit 1/2) when I had a thought - it's probably got a very simple answer but I'm not the best at chemistry so hopefully someone could explain it to me!
So my question is, when salt (NaCl) is put into water (H2O), how come it dissolves in the water, instead of NaOh and HCl forming?
And what is actually formed when salt dissolves in the water?
Thanks so much!  :D
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zsteve

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Re: VCE Chemistry Question Thread
« Reply #5198 on: June 13, 2016, 09:32:37 pm »
+1
Hey everyone!
I was just doing my chemistry homework about water (I'm in unit 1/2) when I had a thought - it's probably got a very simple answer but I'm not the best at chemistry so hopefully someone could explain it to me!
So my question is, when salt (NaCl) is put into water (H2O), how come it dissolves in the water, instead of NaOh and HCl forming?
And what is actually formed when salt dissolves in the water?
Thanks so much!  :D

NaCl dissolves in water because the solvent molecules disrupt the ionic lattice of NaCl. So you will end up with the free ions (Na+/Cl-) in solution. (In reality, they are solvated - so you will have Na+(H2O)x, Cl-(H2O)x, but this is beyond VCE level chem)
As you know, self dissociation of H2O yields OH- and H+.
Theoretically: yes, there would form tiny (basically zero) amounts of NaOH/HCl simply from the equilibrium between Na+ and NaOH given presence of OH- ions.

Practically: Na+ and Cl- are way too weak to become NaOH/HCl again. There is practically no NaOH/HCl in solution.
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HopefulLawStudent

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Re: VCE Chemistry Question Thread
« Reply #5199 on: June 13, 2016, 10:21:28 pm »
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Hey... so dumb question. What can affect the Rt value for HPLC and GC? I routinely get these questions wrong and it's so frustrating...

Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #5200 on: June 13, 2016, 10:37:23 pm »
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Hey... so dumb question. What can affect the Rt value for HPLC and GC? I routinely get these questions wrong and it's so frustrating...
The only stupid questions are the ones that do not get asked my friend.

One variable that affects the Retention Time is the solubility of the compound in the liquid phase, a higher solubility will result in a higher retention time -> as it spends less time being carried by the gas.

The boiling point of the compound also changes the Rt. Higher boiling point results in a higher Rt.

The temperature of the column is the only variable which you can really control yourself, a higher temperature will tend to excite the molecules into the gaseous phase. This occurs because the molecules are so energetic that the attractions in the liquid phase are not able to hold them together or because evaporation occurs more readily. A higher column temperature results in a shorter retention time because of this.

I hope that this answered your question!
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Swagadaktal

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Re: VCE Chemistry Question Thread
« Reply #5201 on: June 13, 2016, 10:56:42 pm »
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Hey... so dumb question. What can affect the Rt value for HPLC and GC? I routinely get these questions wrong and it's so frustrating...
I'd also like to add in that larger molecules will also take longer to go through the chamber due to their higher mass and therefore higher retention time.

I think it was also mentioned above but attraction to the stationary phase - if its polar and the molecule is polar it would attract and therefore inrcease Rt.. and decreasing levels of polarity have decreasing Rts.

I dont think temperature is a part of study design but that's actually fascinating - thanks for sharing Jakey :)

If it is apart of SD can some lmk to make sure i dont fail and get accepted into uni ;)
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zsteve

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Re: VCE Chemistry Question Thread
« Reply #5202 on: June 14, 2016, 01:12:13 pm »
+1
I'd also like to add in that larger molecules will also take longer to go through the chamber due to their higher mass and therefore higher retention time.

I think it was also mentioned above but attraction to the stationary phase - if its polar and the molecule is polar it would attract and therefore inrcease Rt.. and decreasing levels of polarity have decreasing Rts.

I dont think temperature is a part of study design but that's actually fascinating - thanks for sharing Jakey :)

If it is apart of SD can some lmk to make sure i dont fail and get accepted into uni ;)

Well higher temperature means molecules move around faster, which intuitively leads to the conclusion that retention time is shorter
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Swagadaktal

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Re: VCE Chemistry Question Thread
« Reply #5203 on: June 15, 2016, 09:34:06 pm »
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Hey guys - Chapters 18-24 in heinemann chemistry (global chemistry production, renewable energy sources and production of random crap - is that a part of the end of year exam?
We have a sac on it right? Whats the sac going to be like? Is it going to ask how to make shit more renewable?
Is it worth going over it myself or should i just wait for directions by my teacher and move onto the other topic? Coz i think that stuff is actually on the exam right?
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jyce

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Re: VCE Chemistry Question Thread
« Reply #5204 on: June 15, 2016, 10:42:13 pm »
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Hey guys - Chapters 18-24 in heinemann chemistry (global chemistry production, renewable energy sources and production of random crap - is that a part of the end of year exam?
We have a sac on it right? Whats the sac going to be like? Is it going to ask how to make shit more renewable?
Is it worth going over it myself or should i just wait for directions by my teacher and move onto the other topic? Coz i think that stuff is actually on the exam right?

Renewable (and non-renewable) energy sources are DEFINITELY examinable on the end-of-year VCAA exam. They'll make an appearance.

As for "production of random crap", you'll have a SAC one of four selected chemicals. Most schools do sulfuric acid. It's just an example for you to apply your knowledge of maximising reaction rates and yields, etc. and to consider a practical/industrial application of chemistry. The SAC will not focus on renewability; rather, it'll focus on principles of reaction rate and yield, plus some other minor things like uses of your selected chemical, safety procedures used in the production your selected chemical, etc. Note that none of these selected chemicals will appear on the end-of-year exam, at least not to the extent where you require any specialised knowledge.

As for "global chemistry production", I have no idea what that is.
« Last Edit: June 15, 2016, 10:46:00 pm by jyce »