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October 13, 2025, 11:14:17 am

Author Topic: VCE Chemistry Question Thread  (Read 2920322 times)  Share 

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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #2760 on: February 11, 2015, 09:13:47 pm »
+1
Therefore n(H3O+) required to neutralise is also = 0.027603 mol
H2SO4 (aq) --> 2H3O+

Where'd you get these two steps from?

Quick question in terms of 'Concordant titres'...

Say if I did six trials and the results of those six (in chronological order) were:

18.70mL, 18.73mL, 18.57mL, 18.65mL, 18.65mL, 18.65mL

Which results would I use when calculating the average titre?

In practice, you should use all of them - but I'd be sneaky and only take the last three... :P

IndefatigableLover

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Re: VCE Chemistry Question Thread
« Reply #2761 on: February 11, 2015, 09:17:19 pm »
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In practice, you should use all of them - but I'd be sneaky and only take the last three... :P
Haha same here I would just take the last three though I've been told it was also calculate it when it is first concordant as well... I'll just take the last values just to be lazy LOL xD

zyzz101

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Re: VCE Chemistry Question Thread
« Reply #2762 on: February 11, 2015, 09:20:07 pm »
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Quick question in terms of 'Concordant titres'...

Say if I did six trials and the results of those six (in chronological order) were:

18.70mL, 18.73mL, 18.57mL, 18.65mL, 18.65mL, 18.65mL

Which results would I use when calculating the average titre?

So, Concordant titres are titres that are within a difference of +0.1ml or-0.1ml.
therefore,we can use all of em except 18.57ml.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #2763 on: February 11, 2015, 09:22:17 pm »
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So, Concordant titres are titres that are within a difference of +0.1ml or-0.1ml.
therefore,we can use all of em except 18.57ml.

Actually, they're all within +/-0.1 mL if we center at 18.65.

zyzz101

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Re: VCE Chemistry Question Thread
« Reply #2764 on: February 11, 2015, 09:28:16 pm »
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Actually, they're all within +/-0.1 mL if we center at 18.65.

Wouldnt it be the case that,there is a difference of 0.16 ml between 18.57 and 18.73?

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #2765 on: February 11, 2015, 09:30:36 pm »
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Wouldnt it be the case that,there is a difference of 0.16 ml between 18.57 and 18.73?

Yes - but by making sure they're all within +/- 0.1 mL, you've allowed for 0.2 mL of uncertainty in your calculations. Hence, the top and bottom point should be within 0.2 mL.

Eiffel

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Re: VCE Chemistry Question Thread
« Reply #2766 on: February 11, 2015, 11:45:46 pm »
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i remember my teacher saying to me it is +- 0.10 from the smallest to largest titre

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #2767 on: February 11, 2015, 11:51:53 pm »
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i remember my teacher saying to me it is +- 0.10 from the smallest to largest titre

Eh, depends how you define your uncertainties, really. :P There's no "hard and fast" way to do it, just gotta focus on what makes sense statistically for your results. (/what'll get you nicer results, but that's frowned upon)

EDIT: Looking back at what I wrong, this sounds a lot like I'm just covering my arse... Mainly because I forgot when writing it that statistical freedom isn't really given at VCE. :S Trust your teacher over me in the numbers you write down in a practical, better to please them than argue over what is "statistically right" and what isn't.

Eiffel

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Re: VCE Chemistry Question Thread
« Reply #2768 on: February 12, 2015, 10:32:21 pm »
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quick q

after standardisation, the conc of HCL was found to be ... a mass of 7.71 of grease buster was weight into flask.... and made up to 250ml with de ionised water.

does this volume  = volume of grease buster? how does it affect our results

aliquots of 20ml of the diluted sol were titrated with the standardised HCL and the ave titre was found to be 21.10ml.

what does it mean to be standardised, and what diluted sol are they talking about?

« Last Edit: February 12, 2015, 10:42:45 pm by Eiffel »

Eiffel

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Re: VCE Chemistry Question Thread
« Reply #2769 on: February 12, 2015, 10:36:22 pm »
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Eh, depends how you define your uncertainties, really. :P There's no "hard and fast" way to do it, just gotta focus on what makes sense statistically for your results. (/what'll get you nicer results, but that's frowned upon)

EDIT: Looking back at what I wrong, this sounds a lot like I'm just covering my arse... Mainly because I forgot when writing it that statistical freedom isn't really given at VCE. :S Trust your teacher over me in the numbers you write down in a practical, better to please them than argue over what is "statistically right" and what isn't.

:D that doesnt matter.
« Last Edit: February 13, 2015, 01:04:18 am by Eiffel »

bts

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Re: VCE Chemistry Question Thread
« Reply #2770 on: February 13, 2015, 05:40:01 pm »
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hello,
I was wondering whether this statement in my textbook is correct?
'covalent bonds tend to be stronger than ionic bonds'

How does NaCl dissociate in water? what bonds are broken and how?

thank you

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #2771 on: February 13, 2015, 06:41:45 pm »
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'covalent bonds tend to be stronger than ionic bonds'

lolwut. I think there's more context in this that you might need to provide? Can you give us the full sentence?

Eiffel

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Re: VCE Chemistry Question Thread
« Reply #2772 on: February 13, 2015, 07:45:01 pm »
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can someone explain 2 and 3

2) v(hcl) = 0.5L , c(hcl) = 10^-1 hence n(hcl) = 0.05 sub into n = v/v(stp) gets 1.12?

3) i dont know how this is done

« Last Edit: February 13, 2015, 07:48:04 pm by Eiffel »

bts

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Re: VCE Chemistry Question Thread
« Reply #2773 on: February 13, 2015, 08:02:53 pm »
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hello,
I was wondering whether this statement in my textbook is correct?
'covalent bonds tend to be stronger than ionic bonds'

How does NaCl dissociate in water? what bonds are broken and how?

thank you
lolwut. I think there's more context in this that you might need to provide? Can you give us the full sentence?

'covalent bonds tend to be stronger than ionic bonds but this means that, to break these bonds, help is needed in the form of special proteins called enzymes' ?

Cogglesnatch Cuttlefish

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Re: VCE Chemistry Question Thread
« Reply #2774 on: February 13, 2015, 08:14:17 pm »
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can someone explain 2 and 3

2) v(hcl) = 0.5L , c(hcl) = 10^-1 hence n(hcl) = 0.05 sub into n = v/v(stp) gets 1.12?

3) i dont know how this is done
For question 2, you were right up until n(hcl). What you would do then is:
n(Al)= (2/6)*0.05= 1/60 mol. Then use the mole ratio to find the mol of H2; n(H2)= n(Al)* 3/2= 0.0250.
Finally v(H2)= 0.0250*22.5=0.562

EDIT: Im not entirely sure about question 3 but I was thinking that you would find n(Na2CO3) and n(acid) which are 0.15 and 0.30 mol respectively; so the ratio of acid to sodium carbonate is 2:1 which immediately caused HCl to come into my mind as Na2CO3 + 2HCl= CO2 + 2H2O + 2NaCl. Thus, a is the answer as HCl is a monoprotic acid. But again im not confident that is the way to approach such a question.
« Last Edit: February 13, 2015, 08:26:35 pm by Cogglesnatch Cuttlefish »
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