questions

[bubbles]

Electrolysis
This is probably a very simple question:
A circuit was connected for the electroplating of silver, copper and gold.
CELL A: AgCl (aq)
CELL B: CuCl2 (aq)
CELL C: AuCl3 (aq)
In order to collect 1.97g of gold in 20minutes, the current flowing through the circuit would be: ?
I get 804.16mA

The answer is B. 2413mA

Calculating energy changes
Hydrogen peroxide decomposes according to the following equation:
2H2O2 (aq) -> 2H2O (l) + O2 (g)    change in H = -191 kJ/mol
100ml of a solution containing 10.2g of hydrogen peroxide is placed in a thermally insulated calorimeter. The calorimeter and contents (including the solution) are found to have a calibration factor of 4.96J/degrees K  . Some manganese dioxide is added to the hydrogen peroxide solution to catalyse its decomposition. Calculate the temperature rise that occurs, assuming that no heat loss occurs during the process.
Answer is 57.8

[fusion]

for the first question
you do n(Au) = 1.97/197 =0.01
Au had +3 charge, therefore n(e-) = 0.01 * 3 = 0.03

n(e-) = IT / Faraday constant

0.03 = (20*60 time in seconds)I/96500
(0.03*96500)/(20*60) = 2.413 A, which i guess = to 2413mA

[fusion]

for the second question
n(H2O2) = 10.2/(2+16*2) = 0.3

now we know from equation 2mole H2O2 releases 191KJ
therefore 0.3 mol will release 191 * (0.3/2) = 28.65 KJ

change in h = calibration factor * change in temperature

28.65 = 4.96 * change in temperature

change in time = 5.78, not sure why answer was 57.8 maybe to do with one of them was given in KJ and the other in J, hope this gives you idea tho.

[Mao]

for the second question
n(H2O2) = 10.2/(2+16*2) = 0.3

now we know from equation 2mole H2O2 releases 191KJ
therefore 0.3 mol will release 191 * (0.3/2) = 28.65 KJ

change in h = calibration factor * change in temperature

28.65 = 4.96 * change in temperature

change in time = 5.78, not sure why answer was 57.8 maybe to do with one of them was given in KJ and the other in J, hope this gives you idea tho.

in that case, temperature rise would be ~5780 K.

... not exactly pleasant to have a calorimeter almost the temperature of the sun's outer layer.

[fusion]

lol i ment different units in the question somewhere not in my answer , anyway i didnt know hence 'maybe'!

[Mao]

I was referring to the question itself, not your workings.