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March 29, 2024, 12:43:13 am

Author Topic: VCE Chemistry Question Thread  (Read 2313158 times)  Share 

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Snorlax

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Re: VCE Chemistry Question Thread
« Reply #450 on: March 07, 2014, 07:38:42 pm »
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Looks like there might be a couple of things you are looking over here:

1) The Sodium Carbonate was added to a 250ml volumetric flask.
2) The ratio of HCl to Na2CO3 is not 1:1
b) n(Na2CO3) = 1.236/106.0 = 0.01166mol
    c(Na2CO3) = 0.01166/0.250 = 0.04664 M
*I check if b) was right*

c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol
   c(HCl) = 0.02332/0.02099 = 1.111 M


It's c) that doesn't come off right. Am I doing a step incorrectly?

Thanks
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thushan

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Re: VCE Chemistry Question Thread
« Reply #451 on: March 07, 2014, 08:59:32 pm »
+1
b) n(Na2CO3) = 1.236/106.0 = 0.01166mol
    c(Na2CO3) = 0.01166/0.250 = 0.04664 M
*I check if b) was right*

c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol
   c(HCl) = 0.02332/0.02099 = 1.111 M


It's c) that doesn't come off right. Am I doing a step incorrectly?

Thanks

"c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol"

Your issue is here :( the n(HCl) in the titre is indeed twice the amount of Na2CO3 in the aliquot. However, the amount of Na2CO3 in the aliquot is not 0.01166 mol, this is the amount of Na2CO3 in the 250.0 mL volumetric flask. Remember you are taking 20.00 mL aliquots of the Na2CO3 from the 250.0 mL sample.

Alternative solution:
n(Na2CO3)250.0 mL = 1.236/106.0 = 0.01166mol
n(Na2CO3)20.00 mL aliquot = 0.01166 x 20.00/250.0 = 0.0009328 mol
n(HCl)titre = n(Na2CO3)20.00 mL aliquot x 2 = 0.001866 mol
[HCl] = 0.001866/0.02099 = 0.08888 M

I bet you anything the 0.08892 M answer is because they rounded off too early.
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Snorlax

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Re: VCE Chemistry Question Thread
« Reply #452 on: March 07, 2014, 10:41:10 pm »
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"c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol"

Your issue is here :( the n(HCl) in the titre is indeed twice the amount of Na2CO3 in the aliquot. However, the amount of Na2CO3 in the aliquot is not 0.01166 mol, this is the amount of Na2CO3 in the 250.0 mL volumetric flask. Remember you are taking 20.00 mL aliquots of the Na2CO3 from the 250.0 mL sample.

Alternative solution:
n(Na2CO3)250.0 mL = 1.236/106.0 = 0.01166mol
n(Na2CO3)20.00 mL aliquot = 0.01166 x 20.00/250.0 = 0.0009328 mol
n(HCl)titre = n(Na2CO3)20.00 mL aliquot x 2 = 0.001866 mol
[HCl] = 0.001866/0.02099 = 0.08888 M

I bet you anything the 0.08892 M answer is because they rounded off too early.

Ahh yess, I was using the amount in the 250.0mL flask than the 20.0mL titre!
Thanks for that!
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #453 on: March 07, 2014, 11:13:54 pm »
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While titrating Iodine against Wine (to determine the wine's Sulfur Dioxide content) why is the Iodine titrated QUICKLY?

Because sulfur dioxide (SO2) is produced during the titration, and you want to carry out the titration BEFORE the gas leaves the conical flask.

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Re: VCE Chemistry Question Thread
« Reply #454 on: March 08, 2014, 08:10:53 pm »
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how do I split this overall ionic equation into reduction and oxidation half equations
2 S2O42-(aq) + H2O (l)  ->  S2O32-(aq) + 2 HSO3-(aq)
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #455 on: March 08, 2014, 09:41:48 pm »
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how do I split this overall ionic equation into reduction and oxidation half equations
2 S2O42-(aq) + H2O (l)  ->  S2O32-(aq) + 2 HSO3-(aq)

Firstly, what's being oxidised? Check oxidation numbers first.
In the left hand side, the sulfur has an oxidation number of +3, in thiosulfate it has +2 while in hydrogen sulfite it has +4. This is a bit of a trippy question because S2O42- is disproportionating; i.e. it is oxidising AND reducing itself. Try balancing S2O42- => S2O32-
and S2O42- => HSO3-
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Snorlax

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Re: VCE Chemistry Question Thread
« Reply #456 on: March 09, 2014, 09:42:43 pm »
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In a typical chromatogram for HPLC/GLC how do I read it? What does it actually mean?

What does the relative peak height tell you about a component?
What does the area under the peak tell you about the component?

Why doesn't the slowest component (one with the highest retention time) have the most area all the time? - (so I'm wondering about the relationship between retention time and each component on a chromatogram)

ALSO:
http://puu.sh/7oHzs.png
Is eluent the same as waste?


Thanks
« Last Edit: March 09, 2014, 09:58:27 pm by Snorlax »
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #457 on: March 09, 2014, 09:53:13 pm »
+1
In a typical chromatogram for HPLC/GLC how do I read it? What does it actually mean?

What does the relative peak height tell you about a component?
What does the area under the peak tell you about the component?

Why doesn't the slowest component (one with the highest retention time) have the most area all the time? - (so I'm wondering about the relationship between retention time and each component on a chromatogram)

Thanks

Ok:

(1.) Area under peak = Concentration of that component in the sample.
(2.) Relative peak height I'm pretty sure means the same as area under peak, and thus means concentration. Could someone please clarify?!
(3.) Retention time depends upon how strongly adsorbed the component is to the stationary phase, and how soluble the component is in the mobile phase. If a component has a high Rt, this means it is strongly adsorbed to the stationary phase. Just because the component is strongly adsorbed does not mean it is highly concentrated in the sample.

Scooby

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Re: VCE Chemistry Question Thread
« Reply #458 on: March 10, 2014, 02:59:55 am »
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Ok:

(1.) Area under peak = Concentration of that component in the sample.
(2.) Relative peak height I'm pretty sure means the same as area under peak, and thus means concentration. Could someone please clarify?!
(3.) Retention time depends upon how strongly adsorbed the component is to the stationary phase, and how soluble the component is in the mobile phase. If a component has a high Rt, this means it is strongly adsorbed to the stationary phase. Just because the component is strongly adsorbed does not mean it is highly concentrated in the sample.

The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide
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Re: VCE Chemistry Question Thread
« Reply #459 on: March 10, 2014, 03:44:18 am »
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The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide

Thanks for clarifying that Scooby. :)

Snorlax

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Re: VCE Chemistry Question Thread
« Reply #460 on: March 10, 2014, 10:34:41 am »
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The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide
Uh, does the peak height mean anything? I don't think stating that a peak height is the peak height and peak area is the pear area is going to help me in any way..
There has to be a reason why different components have differing peak heights
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #461 on: March 10, 2014, 10:58:53 am »
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For Q 4biv of Exam 2004 (Exam 1.. Short Answer ),  - Sorry guys couldn't printscreen the whole q!

But why do we use the absorbance of 7.5mg/L? I used 1.5mg/L and although is is wrong, it was still awarded 1 mark.. for some reason...?

It says 'Q mol In was used up at the point the absorbance reached zero' but I don't understand why this value is used.
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #462 on: March 10, 2014, 11:08:20 am »
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For Q 4biv of Exam 2004 (Exam 1.. Short Answer ),  - Sorry guys couldn't printscreen the whole q!

But why do we use the absorbance of 7.5mg/L? I used 1.5mg/L and although is is wrong, it was still awarded 1 mark.. for some reason...?

It says 'Q mol In was used up at the point the absorbance reached zero' but I don't understand why this value is used.

Have u drawn the calibration curve for it?
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #463 on: March 10, 2014, 11:12:08 am »
+1
Uh, does the peak height mean anything? I don't think stating that a peak height is the peak height and peak area is the pear area is going to help me in any way..
There has to be a reason why different components have differing peak heights

The only questions that you're going to get relating to peak height is to construct a callibration curve based on this data. The peak height basically tells you the absorbance/concentration of a certain compound in the sample. To find what these compounds are, the peak times are compared to a different graph where the Rt value of compounds are known.

So if a compound is known to show a peak at 10 mins at certain conditions and there is a peak at 10 mins in the unkown compound in the same conditions we can assume that this peak is of the same compound.

Have u drawn the calibration curve for it?

Yes :)
« Last Edit: March 10, 2014, 11:16:33 am by Blondie21 »
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #464 on: March 10, 2014, 11:19:25 am »
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Help pls :)
1-propylamine may be produced by a reaction between 1-chloropropane and one other molecular compound. When 2.17g of 1-chloropropane is completely converted to 1-propylamine, what mass of the other compound would be expected to react?
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