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Author Topic: VCE Chemistry Question Thread  (Read 2313320 times)  Share 

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T-Infinite

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Re: VCE Chemistry Question Thread
« Reply #195 on: January 24, 2014, 10:28:01 pm »
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If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.
Yeah, I just saw that example from the TSFX notes booklet. But I'm still unsure how to find the amounts in mol T_T.
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #196 on: January 25, 2014, 03:33:52 am »
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If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.

Strictly speaking, 2M of sulfuric acid does not mean 4M of H+ as sulfuric acid only completely ionises the first time to form HSO4 -. This acid doesn't completely dissociate (around 10% dissociation here) so the concentration of H+ is closer to 2M than 4M. If you're interested and want to do the calculations, the Ka of HSO4 - is around 0.01. Try a calculation where [H+] = 2M and [HSO4-] = 2M initially and you'll find that the concentration of H+ at equilibrium is still very close to 2M.

Hey guys, quick question. How do you find the amount in mol of the ions, when you're only given the concentration and the molar mass? I may have forgotten some steps. Question is,

For a 0.20M solution of potassium sulfate, K2SO4, calculate the amount, in mol of:
a) potasium ions, K+
b) sulfate ions SO4^2-
c) oxygen atoms



Erm...you can't...because you need a volume. You would have 0.40 M K+ ions but you can't find the number of moles. n=cV remember
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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #197 on: January 25, 2014, 11:27:02 am »
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Yeah this is a textbook missprint.  see this thread for others who fell over the same question:

Missprint Chapter 3 Question 6

It should have given you a volume of 250ml.

If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.
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T-Infinite

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Re: VCE Chemistry Question Thread
« Reply #198 on: January 25, 2014, 11:27:39 am »
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Erm...you can't...because you need a volume. You would have 0.40 M K+ ions but you can't find the number of moles. n=cV remember
Really? Why is this question in my textbook? I thought the same thing when I first saw the question, they didn't give us volume... so I'd assumed there was another way.
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T-Infinite

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Re: VCE Chemistry Question Thread
« Reply #199 on: January 25, 2014, 11:29:03 am »
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Yeah this is a textbook missprint.  see this thread for others who fell over the same question:

Missprint Chapter 3 Question 6

It should have given you a volume of 250ml.
Ohhh, thank you for telling me this! :D I suspected there was something odd about it.


*soz for double post.
« Last Edit: January 25, 2014, 11:31:53 am by T-Infinite »
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survivor

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Re: VCE Chemistry Question Thread
« Reply #200 on: January 25, 2014, 01:24:35 pm »
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3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.


The answer is 11 mol, but i don't understand how they get there. Thanks
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Re: VCE Chemistry Question Thread
« Reply #201 on: January 25, 2014, 01:27:27 pm »
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3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.


The answer is 11 mol, but i don't understand how they get there. Thanks

Ok maybe its just me, but what substance is it referring to. I feel like this question is awfully ambiguous.

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Re: VCE Chemistry Question Thread
« Reply #202 on: January 25, 2014, 01:31:02 pm »
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3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.


The answer is 11 mol, but i don't understand how they get there. Thanks
Limiting reagent is CO (6 mol CO reacts with 2mol Fe2O3). Hence, we have 1 mol of unreacted Fe2O3
2 mol Fe2O3 reacting with 6 mol of CO yields 4 mol Fe (2*2) and 6mol CO2 (2*3)

4+6+1=11 mol
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #203 on: January 25, 2014, 01:37:13 pm »
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Limiting reagent is CO (6 mol CO reacts with 2mol Fe2O3). Hence, we have 1 mol of unreacted Fe2O3
2 mol Fe2O3 reacting with 6 mol of CO yields 4 mol Fe (2*2) and 6mol CO2 (2*3)

4+6+1=11 mol

Thanks for that psyxwar :)

But could you please show how you got the steps? Thanks

psyxwar

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Re: VCE Chemistry Question Thread
« Reply #204 on: January 25, 2014, 01:38:18 pm »
+1
Thanks for that psyxwar :)

But could you please show how you got the steps? Thanks
What do you mean? Isn't it just stoich?
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Re: VCE Chemistry Question Thread
« Reply #205 on: January 25, 2014, 02:00:59 pm »
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If I spilled HCl on the floor, how would I safely clean it?
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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #206 on: January 25, 2014, 02:15:28 pm »
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Tell your teacher.    :P

If they are not there, neutralise it with a weak base (sodium carbonate) then mop it up.  Then tell your teacher.


If I spilled HCl on the floor, how would I safely clean it?
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #207 on: January 25, 2014, 04:15:13 pm »
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Find the amount in mol of

H2O molecules in 20.0 g of CuSO4.5H2O

Cl− ions in 34 g of FeCl3
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #208 on: January 25, 2014, 04:29:32 pm »
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Find the amount in mol of

H2O molecules in 20.0 g of CuSO4.5H2O

Cl− ions in 34 g of FeCl3

1. n(CuSO4.5H2O) = 20/249.6 = 0.08012821mol
n(H2O) = n(CuSO4.5H2O) x 5 = 0.401mol

2. n(FeCl3) = 34/162.3 = 0.2094886 mol
n(Cl-) = n(FeCl3) x 3 = 0.63 mol

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Re: VCE Chemistry Question Thread
« Reply #209 on: January 25, 2014, 04:41:06 pm »
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Quick Q about pH curves

In the image i've attached, the pH curve on the right is a titration between a strong acid and strong base. Why is phenophthalein used as the indicator? I thought it would be bromothymol blue? There's still a sharp end point between 6.0 - 7.6
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