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# charco(Online)

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1. otrivine
09-06-2013
30-05-2013
Hi Charco,

Can you please have a look at my thread and see if perhaps you may be able to help?

http://www.thestudentroom.co.uk/show....php?t=2367444
3. yonewt
08-05-2013
hey um is it ok if i post multiple questions on the physics forum at a time?
4. originaltitle
08-05-2013
I DON'T MEAN TO CAUSE ANY SORT OF INCONVENIENCE WHATSOEVER BUT I REALLY DO NEED AN ANSWER ASAP TO THIS QUESTION: http://www.thestudentroom.co.uk/show....php?t=2344479. IT WOULD BE REALLY VERY AWFULLY NICE INDEED IF YOU HELPED. IT ISN'T A VERY DIFFICULT QUESTION EITHER. SO WOULDJA PLEASE. MANY THANKS.
5. tammie94
07-05-2013
Hey charco can you help me with this question... is the mark scheme wrong?

A mixture of 480 g of iodine and 600 cm3 of water was put into a flask. The mixture was stirred and hydrogen sulphide gas, H2S(g), was bubbled through for several hours.The mixture became yellow as sulphur separated out. The sulphur was filtered off and the solution was purified by fractional distillation. A fraction of HI(aq) was collected containing 440 g of HI in a total volume of 750 cm3. Determine the percentage yield of hydroiodic acid.
I2 + H2S => 2HI + S

The working out in the mark scheme:
amount I2 reacted = 1.89 mol / HI formed = 3.44 mol (1)
theoretical amount HI produced = 3.78 mol/484 g (1) (<-- why did they multiply the mol of I2 by 2, shouldn't it be the mol of HI they multiply by 2?)
% yield = = 91.0 % (1)

EDIT: never mind I got my answer
6. Egos
25-04-2013
Hi there!

Do you think you could help me with these questions?

http://www.thestudentroom.co.uk/show...418&p=42373917

Thank you very much
7. otrivine
03-04-2013
Hi Charco did u please get my quote for the Kc question please!
8. NiceToMeetYou
01-04-2013
Hi, I saw your posts in http://www.thestudentroom.co.uk/showthread.php?t=76693 and was wondering if you could help me out with the same investigation?

I'm a little confused as to how to get the order of reactions, I have concentrations at different times, and if I plot these on a graph it's like a backwards 'S' meaning the inital rate is very close to zero

The bit im confused about is what point do i take the rate at, if it's at any point won't it be different at different points?
9. scorpio22
17-03-2013
hi, i was wondering if you could help with this question please.the threas is ...

http://www.thestudentroom.co.uk/show...3#post41830723

i worked out that the moles of 1.48 X 10-3 mol of mno4- after scale up from 10cm3 to 50cm3 and 2.475 X 10-3 mol of V solid.

(if i divide mols 2.5/ 1.5 = 1.6 for ratio,wasnt sure if i was going to the right so i decided to get the moles to the nearest whole number).

1.5 mol of mno4 /1 X 6 =9 simplified is 3mno4-
2.5 mol of V / 1 X 6 is =15 simplified is 5v(?)

I don't understand how to find the charge of V and wasnt sure what the next step was, or if i am even right so far.Thanks
09-03-2013
Could you suggest some good physical chemistry textbooks or websites which start at or just after A-level but go to a decently advanced level (undergraduate coverage at least) on the main physical topics (e.g. thermodynamics, equilibria, kinetics, electrochemistry, surface chemistry, gas/liquid/solid state, solutions)?

Thanks a ton for the help.

Name
charco
Where I study
University of Life
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