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November 01, 2025, 11:38:31 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448755 times)  Share 

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Math Nerd

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Re: Chemistry 3/4 2013 Thread
« Reply #345 on: January 26, 2013, 11:32:03 pm »
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This question is interesting, kind of blank on how to approach it.

When a particular alkyne is burnt in excess air, the reaction which occurs is: 2Cn.H2n-2(g)+11O2(g)->8CO2(g)+6H2O(g). The value of n in this alkyne is?

HenryP

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Re: Chemistry 3/4 2013 Thread
« Reply #346 on: January 26, 2013, 11:50:12 pm »
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This question is interesting, kind of blank on how to approach it.

When a particular alkyne is burnt in excess air, the reaction which occurs is: 2Cn.H2n-2(g)+11O2(g)->8CO2(g)+6H2O(g). The value of n in this alkyne is?

Ok well what I would do is try to balance the equation. As you can see, on the right hand side of the equation there are 8 Carbon Atoms. While on the left hand side there are 2*n carbon atoms coming from the alkyne.
That means that 2*n=8, therefore n=4

Hopefully that sort of makes sense to you!
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Re: Chemistry 3/4 2013 Thread
« Reply #347 on: January 26, 2013, 11:54:25 pm »
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Look at you guys. Asking and answering questions at midnight.

I salute you.
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KevinooBz

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Re: Chemistry 3/4 2013 Thread
« Reply #348 on: January 26, 2013, 11:54:49 pm »
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Alright, thanks everyone. :) Also, when do we learn to use units like ppm, w/w etc? Was I supposed to have already learnt it?
Learning it this year would be beneficial. Some schools cover it but not all. Usually you need to change units in order to obtain the information needed to answer questions. It's an advanced part of the course but if you're decent at maths it should be easy for you. Mainly it requires converting concentrations in different units. If you're converting from mass to volume you require the density which will be given to you.

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Re: Chemistry 3/4 2013 Thread
« Reply #349 on: January 26, 2013, 11:55:28 pm »
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Look at you guys. Asking and answering questions at midnight.

I salute you.

lol  :P agreed!
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Math Nerd

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Re: Chemistry 3/4 2013 Thread
« Reply #350 on: January 26, 2013, 11:56:48 pm »
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Awkward turtles :S. SHOULD HAVE BALANCED IT WHAT WAS I THINKING.

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Re: Chemistry 3/4 2013 Thread
« Reply #351 on: January 27, 2013, 01:00:46 am »
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when they say 'high resolution NMR specroscopy', is it the highest resolution possible? Is is possible for the resolution to be increased so that peaks split even more due to the influence of further away H1 atoms. Or is it just a fact that the peaks can't be split by anything more than 2 atoms away?
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thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #352 on: January 27, 2013, 01:06:39 am »
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when they say 'high resolution NMR specroscopy', is it the highest resolution possible? Is is possible for the resolution to be increased so that peaks split even more due to the influence of further away H1 atoms. Or is it just a fact that the peaks can't be split by anything more than 2 atoms away?

Splitting doesn't generally occur beyond three bonds (i.e. nearest carbon). There are some exceptions - we call this long-range coupling - which occur in benzene-containing molecules and molecules containing alternating single and double bonds in a chain. You won't have to know about long range coupling. :)
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Re: Chemistry 3/4 2013 Thread
« Reply #353 on: January 27, 2013, 08:52:12 am »
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Learning it this year would be beneficial. Some schools cover it but not all. Usually you need to change units in order to obtain the information needed to answer questions. It's an advanced part of the course but if you're decent at maths it should be easy for you. Mainly it requires converting concentrations in different units. If you're converting from mass to volume you require the density which will be given to you.

When should I learn it? In gravimetric analysis, almost everything is taken back to mass. Would it be a good idea to learn it for volumetric analysis?
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KevinooBz

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Re: Chemistry 3/4 2013 Thread
« Reply #354 on: January 27, 2013, 10:09:10 am »
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When should I learn it? In gravimetric analysis, almost everything is taken back to mass. Would it be a good idea to learn it for volumetric analysis?
I would learn it now. It only takes a few questions to get the hang of it so you can spend 15 minutes doing some unit conversions and it'll become very easy.

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Re: Chemistry 3/4 2013 Thread
« Reply #355 on: January 27, 2013, 11:41:43 am »
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I think you should start learning volumetric analysis.
I took me a lot of questions into understanding how titrations work and doing calcs relating to that. There are also lots of definitions to learn , for example, standard solution, equivalence point, etc. !
But after doing it a lot of times, its really fun!! :)
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Re: Chemistry 3/4 2013 Thread
« Reply #356 on: January 27, 2013, 08:22:44 pm »
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No, I've decided I will not be working ahead this year. I'd much rather do it once and do it right, than do it now quickly, failing to understand things properly and having to do it again (like Further last year -.-). I also have a gravimetric analysis SAC when school starts again and I don't want to confuse myself or cram in a lot of information in such a short space of time. I also did a fair amount of volumetric analysis last year (not necessarily to Year 12 standard but we spent a few weeks on it).

I'll make sure I have a look at calculating in other units though. :)
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Re: Chemistry 3/4 2013 Thread
« Reply #357 on: January 28, 2013, 04:09:01 pm »
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Hey guys I was just wondering how you guys find which one is the oxidant and reductant from half equations? I have two half equations which are:

2I^- => I2 + e
And Cr2O7^2- + 14H^+ + 6e => 2Cr^3+ +7H2O

Sorry if it's a bit messy :(

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Re: Chemistry 3/4 2013 Thread
« Reply #358 on: January 28, 2013, 04:21:08 pm »
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Hey guys I was just wondering how you guys find which one is the oxidant and reductant from half equations? I have two half equations which are:

2I^- => I2 + e
And Cr2O7^2- + 14H^+ + 6e => 2Cr^3+ +7H2O

Sorry if it's a bit messy :(

Equation 1:
oxidation reaction therefore the reductant is I-
In oxidation, the reductant loses electrons (i.e. electrons are on right hand side of half equation)
Equation 2:
reduction reaction therefore the oxidant is Cr2O72-
In reduction, the oxidant gains electrons (i.e. electrons are on left hand side of half equation)
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Re: Chemistry 3/4 2013 Thread
« Reply #359 on: January 28, 2013, 09:01:50 pm »
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A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium
sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride
was added to precipitate barium sulfate.
What was the total mass of barium sulfate produced?

I have got the answer and it is 1.84g but the answer says 1.85 .
And do u round the answers to the least dp since the last working out is +/- ? or follow the least sig fig?

Thanks:)
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