Any chemists? I need some help with simple mole calculations. - TennisForum.com

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post #1 of 84 (permalink) Old Mar 21st, 2006, 08:52 PM Thread Starter
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Any chemists? I need some help with simple mole calculations.



OK, so I have this practical tomorrow but aren't 100% sure on a couple of things. I need to get a pretty high A, or else I'm fucked.



So anyway, the first one seems OK, you just do

5.200 (the mass I used) X1000 / 204.2 X 250... right? (I'm not sure of the units, though.

For 2, you need the molar concentration. Now, the formula we use to get concentration is moles/volume in dm cubed. BUT... I don't know what the moles are. Are the MOLES what I had for the first question? If so, I could do:

*moles in first question*/ volume (which was an average of 23.35cm3)/1000

= mol dm-3

and for question 3 it would just be the same as question 2 but with my second volume from titration 2 (which was 25.35cm3)

I think thats right... but I'm not sure if the concentration for the first question is the same as the moles. I have no other way of working out the moles for the C= N/V thing.

I hope you're following me.

So basically I just want to know how to work out the second question. Someone else suggested it could be... moles of KHP= molar concentravion X the titation volume (23.35) / 1000. Thus, that is the moles of NaOH, so, concentration NaOH would = moles/volume of NaOH (25cm3)/1000.

It's so stupid but I need to get it right.

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post #2 of 84 (permalink) Old Mar 21st, 2006, 10:21 PM
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oh man i wish i could help; i'm "supposedly" taking a high chem class right now at school, but your paper is like a foreign languages to me. sorry.
you're probably more advanced in chem than i will ever be though.

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post #3 of 84 (permalink) Old Mar 21st, 2006, 10:38 PM
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I'm not a chemist by any means ( still a uni student) but I might be able to help with some of it. What level chem is this btw? I don't want to totally screw you up if you're working on something way over my head.

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post #4 of 84 (permalink) Old Mar 21st, 2006, 10:40 PM
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I haven't taken a chem class in 7 years, so I can't help you.

But is that Tim handwriting I see? How cute.


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post #5 of 84 (permalink) Old Mar 21st, 2006, 10:58 PM
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I'm not sure about the equation you're given in question 1, but here's how I would find the molar concentration:

Take the amount of grams you have and multiply it by 1 mol over the molar mass, that will give you mols. In any of the chem courses I've taken, molar concentration has been mols/L so if that's how you should find it, divide the mols by 0.250L which is what you put the 5g into. If you have to use cm^3, divide by 250 and if you have to use dm^3, just convert the cm^3 into dm^3 and divide by that.

For question 2, just remember that the mols of NaOH used to neutralize the solution is the same amount of mols in KHP in the solution. You already know the amount of mols of NaOH because you already found the mols of KHP. To find the molar concentration, do it the same way as question 1. For the volume, use the amount needed to neutralize the solution.

Question 3 is done much the same way as question 2.

Question 4 is a bit tricky but if you need help with that too, I could probably help you out. Hope what I said above makes sense... I'm just a little confused because I've never seen or used the equation in question 1.

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post #6 of 84 (permalink) Old Mar 22nd, 2006, 07:37 AM Thread Starter
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Quote:
Originally Posted by RainyDays
oh man i wish i could help; i'm "supposedly" taking a high chem class right now at school, but your paper is like a foreign languages to me. sorry.
you're probably more advanced in chem than i will ever be though.
US system.



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post #7 of 84 (permalink) Old Mar 22nd, 2006, 07:38 AM Thread Starter
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Quote:
Originally Posted by angele87
I'm not a chemist by any means ( still a uni student) but I might be able to help with some of it. What level chem is this btw? I don't want to totally screw you up if you're working on something way over my head.
It's A-level. I'm guessing you're from the US... it's basically what people do here aged 17/18.

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post #8 of 84 (permalink) Old Mar 22nd, 2006, 07:38 AM Thread Starter
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Quote:
Originally Posted by decemberlove
I haven't taken a chem class in 7 years, so I can't help you.

But is that Tim handwriting I see? How cute.
That's messy handwriting. I have poor handwriting at the best of times.

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post #9 of 84 (permalink) Old Mar 22nd, 2006, 07:41 AM Thread Starter
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Quote:
Originally Posted by angele87
I'm not sure about the equation you're given in question 1, but here's how I would find the molar concentration:

Take the amount of grams you have and multiply it by 1 mol over the molar mass, that will give you mols. In any of the chem courses I've taken, molar concentration has been mols/L so if that's how you should find it, divide the mols by 0.250L which is what you put the 5g into. If you have to use cm^3, divide by 250 and if you have to use dm^3, just convert the cm^3 into dm^3 and divide by that.

For question 2, just remember that the mols of NaOH used to neutralize the solution is the same amount of mols in KHP in the solution. You already know the amount of mols of NaOH because you already found the mols of KHP. To find the molar concentration, do it the same way as question 1. For the volume, use the amount needed to neutralize the solution.

Question 3 is done much the same way as question 2.

Question 4 is a bit tricky but if you need help with that too, I could probably help you out. Hope what I said above makes sense... I'm just a little confused because I've never seen or used the equation in question 1.
Hmmm.... well it's just question one. It tells you how to do it with the formula, I don't understand the way you worked it out, it's nothing I've ever come across before.

I just want to know... when you've worked out the concentration for question one, is that also the MOLES of KHP, or, to get the moles, do you use the equasion:

concentration= moles/volume?

THEN you have the moles of NaOH, too, so use

concentration= moles/volume.

It's just this bit.

Question 4 is fine.

I'll do both my calculations so you can see what I've got.

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post #10 of 84 (permalink) Old Mar 22nd, 2006, 07:47 AM Thread Starter
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post #11 of 84 (permalink) Old Mar 22nd, 2006, 07:50 AM
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Bloody hell Tim, I wish I could help...!!! But I know everything about nothing, so anything useful is too difficult to me! Good luck though

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post #12 of 84 (permalink) Old Mar 22nd, 2006, 07:54 AM Thread Starter
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Quote:
Originally Posted by PamShriverRockz
Bloody hell Tim, I wish I could help...!!! But I know everything about nothing, so anything useful is too difficult to me! Good luck though
Blah. You got my hopes up then.

Thanks anyway.

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post #13 of 84 (permalink) Old Mar 22nd, 2006, 01:16 PM Thread Starter
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Bump.

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post #14 of 84 (permalink) Old Mar 22nd, 2006, 02:11 PM
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not ignoring it, tim --- but this does not ring a bell.



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post #15 of 84 (permalink) Old Mar 22nd, 2006, 02:19 PM Thread Starter
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Quote:
Originally Posted by No Name Face
not ignoring it, tim --- but this does not ring a bell.


Oh... I remember your thread about balancing equasions.

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