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November 01, 2025, 11:36:37 am

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

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Yacoubb

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Re: Chemistry 3/4 2013 Thread
« Reply #480 on: February 17, 2013, 01:06:38 am »
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I think someone noted that generally the first exams under a new study design are set easier to assess its effectiveness. I can't vouch for that though.

I actually heard the complete opposite. Lecturers from TSFX mentioned that VCAA often makes the first practice exams rather tricky. To be honest, I'd prefer a difficult exam because the cut off would be much lower than an exam with a high cut off!

thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #481 on: February 17, 2013, 09:04:07 am »
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I actually heard the complete opposite. Lecturers from TSFX mentioned that VCAA often makes the first practice exams rather tricky. To be honest, I'd prefer a difficult exam because the cut off would be much lower than an exam with a high cut off!

Doesn't matter. :) Just hope that every exam has easy parts so you guys can show what you know, and hard parts because they're sadists.
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #482 on: February 17, 2013, 10:59:44 am »
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Also the mass of solute required? 400 mL of a solution that is 1.00 mol L−1 with respect to Fe3+ ions, using Fe2(SO4)3.9H2O? Could anyone help with this one?  ;)
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Re: Chemistry 3/4 2013 Thread
« Reply #483 on: February 17, 2013, 12:34:15 pm »
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Also the mass of solute required? 400 mL of a solution that is 1.00 mol L−1 with respect to Fe3+ ions, using Fe2(SO4)3.9H2O? Could anyone help with this one?  ;)
So what do you mean? I'm having trouble understanding what the question is asking ?
btw. the question that you posted before was an amazing question! :) Thanks!
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Re: Chemistry 3/4 2013 Thread
« Reply #484 on: February 17, 2013, 12:59:41 pm »
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Also the mass of solute required? 400 mL of a solution that is 1.00 mol L−1 with respect to Fe3+ ions, using Fe2(SO4)3.9H2O? Could anyone help with this one?  ;)

I think they man [Fe3+] = 1.00 M. Work out how many Fe3+ ions there are in solution, then work out how much of hydrated ferric sulfate you need in mols, then work out the mass of it.
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Stick

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Re: Chemistry 3/4 2013 Thread
« Reply #485 on: February 17, 2013, 01:04:03 pm »
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Also the mass of solute required? 400 mL of a solution that is 1.00 mol L−1 with respect to Fe3+ ions, using Fe2(SO4)3.9H2O? Could anyone help with this one?  ;)

I'm not sure if I'm understanding this properly or not, but this is how I think I'd tackle it:
1. 1.00M of Fe3+ ions means the concentration of Fe2(SO4)3.9H2O is 0.500M.
2. Amount of Fe2(SO4)3.9H2O is given by 0.500 * 0.400 = 0.200mol
3. Mass of Fe2(SO4)3.9H2O is given by 0.200 * 561.9812 = 112.3924g

Obviously not correct to the appropriate amount of significant figures and I also googled the molar mass so don't stress if your answer is a bit off. :)
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Re: Chemistry 3/4 2013 Thread
« Reply #486 on: February 17, 2013, 01:09:21 pm »
+1
OMGGG! I just found that INTERNATIONAL MOLE DAY is on the 23rd of October , from 6.02 am to 6.02 pm ! :)
AINT THAT COOL? :) :P
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Re: Chemistry 3/4 2013 Thread
« Reply #487 on: February 17, 2013, 01:10:01 pm »
+2
Yeah, our class celebrated it last year. :P
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Re: Chemistry 3/4 2013 Thread
« Reply #488 on: February 17, 2013, 01:32:06 pm »
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OMGGG! I just found that INTERNATIONAL MOLE DAY is on the 23rd of October , from 6.02 am to 6.02 pm ! :)
AINT THAT COOL? :) :P

that is SO DAMN COOL!  :D
But technically, shouldn't it be on the 6th of Feb? OHh hang on... have they gone with 10^23?
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Re: Chemistry 3/4 2013 Thread
« Reply #489 on: February 17, 2013, 02:29:19 pm »
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that is SO DAMN COOL!  :D
But technically, shouldn't it be on the 6th of Feb? OHh hang on... have they gone with 10^23?

yup :)

We "celebrated" it last year too.... lol.
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Re: Chemistry 3/4 2013 Thread
« Reply #490 on: February 17, 2013, 04:23:43 pm »
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I'm not sure if I'm understanding this properly or not, but this is how I think I'd tackle it:
1. 1.00M of Fe3+ ions means the concentration of Fe2(SO4)3.9H2O is 0.500M.
2. Amount of Fe2(SO4)3.9H2O is given by 0.500 * 0.400 = 0.200mol
3. Mass of Fe2(SO4)3.9H2O is given by 0.200 * 561.9812 = 112.3924g

Obviously not correct to the appropriate amount of significant figures and I also googled the molar mass so don't stress if your answer is a bit off. :)
CORRECT! You got it right! :') yeah, strangely worded questions but anyhow.
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Bad Student

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Re: Chemistry 3/4 2013 Thread
« Reply #491 on: February 17, 2013, 07:39:38 pm »
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Can someone please explain the gradient of a pH curve when you titrate a strong acid with a strong base, or weak acid with a weak base or strong acid with a weak base etc.

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Re: Chemistry 3/4 2013 Thread
« Reply #492 on: February 17, 2013, 09:22:47 pm »
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Can someone please explain the gradient of a pH curve when you titrate a strong acid with a strong base, or weak acid with a weak base or strong acid with a weak base etc.
I'm not entirely sure what your asking, but I hope this helps!

The gradient is the rate of change. A low gradient would mean the pH doesn't change very drastically when solution is added, a larger gradient would cause a more rapid change in pH.

This is a strong acid and weak base. You could find graphs with weak acid etc on the internet or your textbook

From this, you can see that when you start adding base, initially the graph isn't very steep (low gradient), so adding base only increases the pH a small amount.

As you approach the equivalence point, the graph gets steeper (higher gradient), so adding a small amount of base will cause a greater rise in pH. This rapid change in pH is what may lead you to overshoot the endpoint during a titration as the colour of the indicator changes rapidly (as the pH rapidly changes).

The graph then flattens out (low gradient), as adding more base increases the pH by less and less each time. 

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Re: Chemistry 3/4 2013 Thread
« Reply #493 on: February 17, 2013, 10:14:42 pm »
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So I've got a question related to a titration between iron ions (fe2+) and pottasium permanganate.
One of the questions in the prac book (which for the life of me I can't think of an adequate answer) is:
5. The amount of additional potassium permanganate solution required to reach the endpoint is minute. This excess will notionally increase the calculated amount of iron present but, in practice, the extra has no effect on the calculated amount. Why not?
I'm really stumped on this one.
Also, another question asks: In most titration exercises, the titres are averaged before doing the calculations. Why is this
inappropriate in this exercise?
Thanks for helping out!

Bad Student

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Re: Chemistry 3/4 2013 Thread
« Reply #494 on: February 17, 2013, 10:26:21 pm »
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I'm not entirely sure what your asking, but I hope this helps!

The gradient is the rate of change. A low gradient would mean the pH doesn't change very drastically when solution is added, a larger gradient would cause a more rapid change in pH.

This is a strong acid and weak base. You could find graphs with weak acid etc on the internet or your textbook
(Image removed from quote.)
From this, you can see that when you start adding base, initially the graph isn't very steep (low gradient), so adding base only increases the pH a small amount.

As you approach the equivalence point, the graph gets steeper (higher gradient), so adding a small amount of base will cause a greater rise in pH. This rapid change in pH is what may lead you to overshoot the endpoint during a titration as the colour of the indicator changes rapidly (as the pH rapidly changes).

The graph then flattens out (low gradient), as adding more base increases the pH by less and less each time.

Sorry about the bad phrasing.

What I wanted to ask was why is the gradient steep in some places and gentle in other places. Or why isn't the change in pH linear.

Thanks.
« Last Edit: February 17, 2013, 10:30:34 pm by Bad Student »