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March 29, 2024, 05:21:59 am

Author Topic: VCE Chemistry Question Thread  (Read 2313324 times)  Share 

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Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9105 on: May 11, 2021, 04:57:45 pm »
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for this question
a) if the temperature is decreased would it shift to the left
if the volume of the reaction vessel is decreased, increasing pressure would it shift to the left
c) if an inert gas(neon) is added at constant volume would it shift to the right where there are more moles of gas

beep boop

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Re: VCE Chemistry Question Thread
« Reply #9106 on: May 13, 2021, 12:45:51 pm »
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does anyone have any tips on making a scientific poster. my research investigation task is coming soon and im a bit worried for it.  ive done plenty of scientific posters, however, i always get a mediocre grade.

any help is appreciated!

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Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9107 on: May 13, 2021, 03:40:44 pm »
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Why does only temperature affect kc

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9108 on: May 13, 2021, 04:09:43 pm »
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Hi Chocolatepistachio,
I suggest using a variable to represent the unknown concentrations. As the experiment is investigating the decomposition of NOCl(g) and the initial amount of NOCl(g) is given, the initial:



As it is unknown how much NOCl(g) is reacted to form NO(g) and Cl2(g), we can use a variable to represent the change in mole. This change will be proportional to the NOCl:NO:Cl2 ratio of 2:2:1.



meaning that since the volume of the flask is 2.00L, equilibrium moles will be



Now the equilibrium constant. Following

and the given equilibrium constant (at 35oC, presumably the temperature of this experimental setup)
we can construct the equation
from which we can get

Applying polynomials, you would eventually come to
which can be substituted back into the original equilibrium mole values to get our final answers of:



at equilibrium.

Hope this helps, please let me know if there are any errors :)

hey sorry how did you work out x using polynomials from the x3/(0.5-x)2= 2.20 x 10-5

wingdings2791

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Re: VCE Chemistry Question Thread
« Reply #9109 on: May 14, 2021, 09:27:31 am »
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hey sorry how did you work out x using polynomials from the x3/(0.5-x)2= 2.20 x 10-5
Expand the denominator of \(\frac{x^3}{(0.5-x)^2}\) and move everything to one side to obtain the equation \((x^2-x+0.25)2.20\times10^{-5}-x^3=0\). From there, use factorisation of polynomials (from Methods- this usually takes some trial and error) to find the value of \(x\). According to my tutor, questions requiring this much calculation are likely never going to come up on the exam, so it's probably not a big deal to save some time and just punch this into a CAS instead. :)
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Re: VCE Chemistry Question Thread
« Reply #9110 on: May 14, 2021, 01:12:20 pm »
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For this question at point A would the volume have decreased (increasing pressure) and reactions shifts to the left

At point B would the temperature have decreased, and the reaction shifts to the left

At point c would some C2H6 have been added and the reactions shifts to the left

Apologies for the late reply

Volume changes - but if you're compressing the container, concentration will increase, not decrease. Hence, volume increases for the first question.

The system has to move to react to change (be very careful of this!). When the system moves one way or another, it is as a reaction to some external change in the other direction. So for the second question, when the system shifts left yet again, it is because the temperature has been modified in the other direction (ie. it has been increased). Had the temperature decreased, by LCP the system would have shifted to the right to minimise the change.

Last bit is 100% correct.

for this question
a) if the temperature is decreased would it shift to the left
if the volume of the reaction vessel is decreased, increasing pressure would it shift to the left
c) if an inert gas(neon) is added at constant volume would it shift to the right where there are more moles of gas

Try giving this one a bit more thought, a few of the same misconceptions are popping up again here. Use the advice I gave to your previous question :)

Why does only temperature affect kc

The equilibrium constant is constant for a constant temperature. Consider the equilibrium constant for any generic gaseous exothermic reaction - the constant will remain the same for most external changes (compression, expansion, introduction/removal of particles, etc.). The one thing that cannot be accounted for is a temperature change because a temperature change for such an exothermic reaction will cause a shift in either direction as defined by LCP (thus altering the constant, because the concentrations are different) without any original change to the concentrations. Think about it like this; if everyone wants a particular item, said item's price spikes because of the demand (sort of like LCP reacting to the change in attitude) but a change in temperature is more like jacking up the price artificially. It's not the best example but it's the first one I thought of. It will definitely help if you look at the equation for the constant more closely.

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Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9111 on: May 16, 2021, 09:56:53 pm »
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Thanks
I don’t get why temperature has increased though

If an inert gas is added then would there be no change

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9112 on: May 22, 2021, 07:33:06 pm »
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If you are calculating the pH taking the log when looking at significant figures why are only the numbers after the decimal point significant

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9113 on: May 27, 2021, 05:50:45 pm »
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how would you solve this isnt this unsolvable if you put it into the quadratic formula

SmartWorker

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Re: VCE Chemistry Question Thread
« Reply #9114 on: May 27, 2021, 08:16:20 pm »
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how would you solve this isnt this unsolvable if you put it into the quadratic formula

I solved it on the CAS, for x=1.24 x 10^-4 M so there is a solution (positive)
When I used the quadratic equation on the scientific calculator it rounds down to 0 tho?

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aspiringantelope

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Re: VCE Chemistry Question Thread
« Reply #9115 on: May 29, 2021, 02:02:03 pm »
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Just wondering if calculations of pH have been removed from the 3/4 study design?
Thanks! :)

ArtyDreams

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Re: VCE Chemistry Question Thread
« Reply #9116 on: May 29, 2021, 02:25:30 pm »
+8
Just wondering if calculations of pH have been removed from the 3/4 study design?
Thanks! :)
Yep! They're not on the current study design  :D

aspiringantelope

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Re: VCE Chemistry Question Thread
« Reply #9117 on: May 29, 2021, 02:38:02 pm »
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Yep! They're not on the current study design  :D
Great!
Thanks for your reply!

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #9118 on: June 09, 2021, 04:17:26 pm »
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if someone could help with this question

Corey King

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Re: VCE Chemistry Question Thread
« Reply #9119 on: June 09, 2021, 04:19:51 pm »
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Hey guys,
Could you have the C=O group of a hydrocarbon (lets say alkane) undergo addition reaction with some other substance, so that the double bonded oxygen is replaced by two elements with single bonds?
Many thanks,
Corey