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March 29, 2024, 03:01:02 am

Author Topic: VCE Chemistry Question Thread  (Read 2313246 times)  Share 

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Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8805 on: September 25, 2020, 06:38:44 pm »
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Oh so it is right

For the other question I’m so confused how do you work out the concentration of hydronium ions from the percentage ionisation

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8806 on: September 29, 2020, 01:10:59 pm »
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Oh so it is right

For the other question I’m so confused how do you work out the concentration of hydronium ions from the percentage ionisation

%ionisation=concentration of hydronium ions/concentration of acid

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8807 on: September 29, 2020, 01:54:14 pm »
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would this be right then

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Re: VCE Chemistry Question Thread
« Reply #8808 on: September 29, 2020, 05:38:21 pm »
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Hello,

I don't understand how the group of an element can be determined from its electron configuration. I understand how it might follow for group 1 elements, but for scandium, for instance, I don't understand.

Thanks

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8809 on: September 30, 2020, 02:45:28 pm »
+4
Hello,

I don't understand how the group of an element can be determined from its electron configuration. I understand how it might follow for group 1 elements, but for scandium, for instance, I don't understand.

Thanks

Groups are entirely a human construct, and are based on the order of the periodic table. Like, they could've decided that instead of starting a new row after neon , to instead start a new row after nitrogen, but instead they decided not to. Chemists instead decided to make it so the group you're in depends on your electron configuration, and the reason for this is because your electron configuration essentially controls your reactivity.

would this be right then

Entire first bit is wrong. Check your maths, (3/11)*100 is definitely NOT 83%. Always remember your bullshit check - 3/11 is definitely less than 50%, so why is your answer more than 50%?

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8810 on: September 30, 2020, 05:28:48 pm »
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Hello
would this be right

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8811 on: September 30, 2020, 10:29:20 pm »
0
How come when recharging more voltage has to be applied than the voltage which is produced during discharge?

EDIT question: why does hydrogen gas combustion not produce CO2
« Last Edit: September 30, 2020, 10:47:50 pm by Coolgalbornin03Lo »
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8812 on: October 01, 2020, 12:24:07 am »
+4
Hello
would this be right

I think that's fine - but also, I'm a little tired of just checking homework for you, so I'm not going to help you with those types of questions anymore, sorry. You're welcome to still post them, other users may be more willing/able to check for you, but please don't get offended if I don't answer these questions for you while I still continue to help others. May I suggest speaking to a friend in your class and seeing what they got, and if you both wrote the same thing, you're likely correct? I'm still more than happy to help with questions where you don't understand the concepts, or questions for which you aren't sure how to approach them.

How come when recharging more voltage has to be applied than the voltage which is produced during discharge?

EDIT question: why does hydrogen gas combustion not produce CO2

It's because you lose some energy in the charging process. Next time you charge your phone, I highly recommend turning it off and feeling it while it's charging - you should notice that the phone is heating up slightly, despite being turned off! That's where the energy has gone, into that heat.

As for the hydrogen gas combustion - where does the CO2 come from when you burn a hydrocarbon, say methane? Could hydrogen gas get it from the same source?

p0kem0n21

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Re: VCE Chemistry Question Thread
« Reply #8813 on: October 01, 2020, 10:00:03 am »
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Hi guys! I was wondering if someone could help me with this question.

I can immediately eliminate B as an answer since there is no chemical B at the beginning of the reaction. I think the answer isn't A either, since the the concentrations of A and B would probably be the same once equilibrium is reached, but please correct me if I'm wrong. I can't really go much farther than that. I thought about how the reaction quotient is initially 0, since the concentration of B is initially 0 M, but then raises to some positive number as the system reaches equilibrium, but I'm not sure if that is helpful. I suppose there is something important surrounding how the concentration of B eventually becomes greater than [A], but I can't wrap my head around how that would affect the chemical equation i.e. whether the coefficient of A or B would be greater as a result.

Thanks in advance  :)

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8814 on: October 01, 2020, 11:42:25 am »
+4
Hi guys! I was wondering if someone could help me with this question.

I can immediately eliminate B as an answer since there is no chemical B at the beginning of the reaction. I think the answer isn't A either, since the the concentrations of A and B would probably be the same once equilibrium is reached, but please correct me if I'm wrong. I can't really go much farther than that. I thought about how the reaction quotient is initially 0, since the concentration of B is initially 0 M, but then raises to some positive number as the system reaches equilibrium, but I'm not sure if that is helpful. I suppose there is something important surrounding how the concentration of B eventually becomes greater than [A], but I can't wrap my head around how that would affect the chemical equation i.e. whether the coefficient of A or B would be greater as a result.

Thanks in advance  :)

This is an interesting one, and I'm not convinced we have enough information to answer this - but I'm going to just run through my thinking ANYWAY, and maybe this will help us glean something.

So my first impulse is the answer is A - if it were C and D, I would expect the square factor to make the chemicals A or B to be much smaller. But, we don't know what the equilibrium constant is, so that logic goes out the window.

I then thought - why not make an ICE table? Couldn't hurt. If the equation is A->B:

Code: [Select]
     A    B
I    1    0
C   -x   +x
E   1-x  +x

Okay, so [A]~0.2 M, and conc(B)~0.4 M. If I have the stoichiometry correct, then setting these two equal to E SHOULD give the same x:

A: 0.2=1-x <==> x=0.8
B: 0.4=x <==> x=0.4

Okay, so this would suggest the reaction is NOT A->B. What about 2A->B? Well, the ICE table is:

Code: [Select]
     A    B
I    1    0
C   -2x   +x
E   1-2x  +x

Which would give:

A: 1-2x=0.2 <==> 2x=0.8 <==> x=0.4
B: x=0.4

Which are the same! So this would suggest to me that the answer is C, 2A-->B.

So yeah, with these kinds of questions, it's worth just playing around and trying random things - I didn't know to use an ICE table, I just decided to make one, and noticed that there was a logical progression I could follow.

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8815 on: October 04, 2020, 10:51:44 am »
0

As for the hydrogen gas combustion - where does the CO2 come from when you burn a hydrocarbon, say methane? Could hydrogen gas get it from the same source?

I wanna say it comes from the Cs in the hydrocarbon?

It’s like how carbon combustion doesn’t produce water because there’s no H’s right?
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jnlfs2010

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Re: VCE Chemistry Question Thread
« Reply #8816 on: October 04, 2020, 04:47:28 pm »
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Why do we need to get the endpoint as close to the equivalence point as possible?
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Owlbird83

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Re: VCE Chemistry Question Thread
« Reply #8817 on: October 04, 2020, 05:36:39 pm »
+7
Why do we need to get the endpoint as close to the equivalence point as possible?
The purpose of a titration is to find an unknown concentration, and you want to do this as accurately as possible. You want to get the point that the colour of the indicator changes (endpoint), close to the equivalence point because it is a visual indicator that the acid and base have neutralised and are in molar ratio, which enables you to calculate the concentration of the unknown using the volume it took to neutralise. If the colour changed somewhere away from the equivalence point, it wouldn't be helpful, because you wouldn't know when they are in the molar ratio, so can't do stoichiometry.
Hope that helps!
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8818 on: October 04, 2020, 07:09:41 pm »
+2
I wanna say it comes from the Cs in the hydrocarbon?

It’s like how carbon combustion doesn’t produce water because there’s no H’s right?

Exactly! So, why wouldn't burning H2 produce CO2?

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8819 on: October 08, 2020, 03:41:27 pm »
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For this question
for b why when the volume is halved do all the concentrations stay the same and for a) why are the reactant decreasing shouldn't the reactant increase and the product decreases