How to do these questions

[onlyfknhuman]

[IMG]http://img380.imageshack.us/img380/6813/cemnsu0.th.png[/img][IMG]http://img380.imageshack.us/images/thpix.gif[/img]

From stav, why isnt Question nine. D answer said B

And for question 10. is A no idea how to do this.

[fusion]

For first one should be
n(e-) = IT/Faraday
N(e-) = converting mA to A = (0.0001* 7257600)/96500 = 725.26/96500 = 0.00752029

Ag2 -> 2AG + 2e- (i think, bad at equations lol)

therefore n(Ag2o) = 0.00752029 / 2 =0.00376 which is the same as 3.76 * 10^-3 which is B

[chewybacca]

For first one should be
n(e-) = IT/Faraday
N(e-) = converting mA to A = (0.0001* 7257600)/96500 = 725.26/96500 = 0.00752029

Ag2 -> 2AG + 2e- (i think, bad at equations lol)

therefore n(Ag2) = 0.00752029 / 2 =0.00376 which is the same as 3.76 * 10^-3 which is B

Stab in the dark here...'cos I'm stumped too.

9) Isn't Ag₂ the same as 2Ag⁺? In which case why would you divide by two (as they've done), as ratios are equivalent (2Ag⁺ --> 2e^-)? I can understand where they're coming from though...sort of.
10) Where are they pulling the 1.50V figure from (refer to solutions)...that's my main question

[shinny]

9. But they're asking for mol of Ag2O, not Ag. So divide by 2.

10. The 1.50V is pulled from the question stem at the top of the page stating 'The cell potential is 1.50V'...which was even conveniently bolded :]
+0.34-E=1.5
E=0.34-1.5=-1.16

[chewybacca]

9. But they're asking for mol of Ag2O, not Ag. So divide by 2.

10. The 1.50V is pulled from the question stem at the top of the page stating 'The cell potential is 1.50V'...which was even conveniently bolded :]
+0.34-E=1.5
E=0.34-1.5=-1.16

lol @ 10... makes sense now!

[onlyfknhuman]

For first one should be
n(e-) = IT/Faraday
N(e-) = converting mA to A = (0.0001* 7257600)/96500 = 725.26/96500 = 0.00752029

Ag2 -> 2AG + 2e- (i think, bad at equations lol)

therefore n(Ag2) = 0.00752029 / 2 =0.00376 which is the same as 3.76 * 10^-3 which is B

that makes alot of sense, same as shinji thx

[onlyfknhuman]

9. But they're asking for mol of Ag2O, not Ag. So divide by 2.

10. The 1.50V is pulled from the question stem at the top of the page stating 'The cell potential is 1.50V'...which was even conveniently bolded :]
+0.34-E=1.5
E=0.34-1.5=-1.16

FUCK MAN I HATE IT WEN ITS CHAIN QUESTIONS I TOTALLY DIDNT NOTICE THE 1.5 AT THE TOP _- FK i find stav teh hardest out of all the papers

and one more thing sorry.

Q1.b short answer
ch3cooh is a weak acid therefore ch3coo- cannot be directly calculated? wtf? and then the solutions calculate it

and at the end it says ' consequential marks can be awarded if student incorrectly assumes its a strong acid and calcualtes the ph" wth is  stav talking abotu ~_~

[onlyfknhuman]

bump

[shinny]

Directly calculated as in assuming it fully ionises. The consequential mark is for people who assume it was a strong acid and calculate it how they normally would for HCl. What you're meant to do (as I assume you're doing) is to calculate it by using the method where you approximate the final concentration of the acid to not change much, since the Ka is negligible.

[onlyfknhuman]

okay thx for that