ocr a f325 revision thread

june 2012..Q6d? ....explanation pls?...

Reply 3081

Original post by Nuna

any tips on what will most likely be the big markers?

I am expecting the usual massive hard question in the same kind of style as the last question of Jan 2013. Other than that, it is anyone's guess. Personally I would like a nice large question on rates/using tables to determine Kc.

Reply 3082

Original post by Nuna

june 2012..Q6d? ....explanation pls?...

Cr has a more negative electrode potential, meaning it is more likely to be oxidised. The next mark is unintuitive, and I would not have got it- you had to simply show the direction of the half equations, Cr shifting to the left to form electrons/ion and Cu shifting to the right forming Cu from the ion and electons.

(edited 10 years ago)

Reply 3083

Original post by galacticfear

Cr has a more negative electrode potential, meaning it is more likely to be oxidised. The next mark is unintuitive, and I would not have got it- you had to simply show the direction of the half equations, Cr shifting to the left to form electrons/ion and Cu shifting to the right forming Cu from the ion and electons.

oops..I meant jan 2012, Q6d ..sorry...?

Reply 3084

Original post by Nuna

oops..I meant jan 2012, Q6d ..sorry...?

Whenever there is an entropy question I start by laying out the formula (delta)G = (delta)H - T(delta)S. The equation shows us that we are going from more moles of gas to fewer. Therefore the entropy must be negative. At low temperature the magnitude of T(delta)S is less significant than the value of (delta)H, therefore (delta)G is negative. Opposite occurs at high temperatures.

Reply 3085

Original post by galacticfear

Whenever there is an entropy question I start by laying out the formula (delta)G = (delta)H - T(delta)S. The equation shows us that we are going from more moles of gas to fewer. Therefore the entropy must be negative. At low temperature the magnitude of T(delta)S is less significant than the value of (delta)H, therefore (delta)G is negative. Opposite occurs at high temperatures.

?..

Reply 3086

Original post by Nuna

?..

2SO2(g) + O2(g) -> 2SO3(g)

2______ 1_____ 2

Therefore 3 -> 2 gas molecules.

Reply 3087

Original post by galacticfear

2SO2(g) + O2(g) -> 2SO3(g)

2______ 1_____ 2

Therefore 3 -> 2 gas molecules.

of course...tiredness has taken my common sense out of me, Thank you

Reply 3088

Original post by Nuna

of course...tiredness has taken my common sense out of me, Thank you

Speaking of which I am going to pack it in now, and get 3 hours sleep... I am sure if I just blag my way through I will do OK. Goodnight.

(edited 10 years ago)

Reply 3089

Original post by galacticfear

Speaking of which I am going to pack it in now, and get 3 hours sleep... I am sure if I just blag my way through I will do OK. Goodnight.

I'm sure you will do great...thank you so much for all the help, all the best for tomorrow

Reply 3090

CosmicVengeance

13

Cramming done just need to go over the titrations section and re read that word doc notes, sleep now chem will be hell

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Reply 3091

master y

12

An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
Calculate the percentage yield, by mass, of sodium ferrate(VI).
Show your working.

cN ANYONE HELP

Original post by eggfriedrice

They have to be in standard conditions, so higher temperature would increase rate of reaction (kinetics) making it non standard. Increasing a concentration would shift the equilibrium one way hence increasing or decreasing the Ecell value.

Original post by A-New-Start

yeah exactly sometimes we do have to multiply by 3 and divide by 3 : therefore effectively the no of of I2 is same as no of mol of Cu2+

ok i show it like this

there are two equations

2Cu2+ + 4I- --> 2CuI + I2

and then the titration step:

2S2O3 2- + I2 --> 2I- + S4O6 2-

now here if we know no of mol of I2 in titration step: we just backtrack to the previous step which also has I2 and hence no of mol of Cu2+ = no of mol of I2 - however imagine if in the first step there was a 3I2 - so then do you multiply by 3 and then divide by 2/3 to get no of mol of Cu2+ ?

Original post by AyshaK

Fuel cell produce electrical current as long as fuel and oxidant supplied.
In a hydrogen and oxygen fuel cell:

1/2 O2 + H2 ---> H20
To get this overall equation 2 half equations are involved
2H20 + 2e- ---> H2 + 2OH-
2e- + 1/2 O2 + H2O ---> 2OH-

Some fuel cells use hydrogen rich fuels such as ethanol or methanol and you might need to know the equations:

CH3CH2OH + 3H2O ---> 2CO2 + 12H+ + 12E-
3O2 + 12H+ + 12E- ---> 6H2O
= C2H5OH + 3O2 ---> 2CO2 + 3H20

Reply 3092

alltimeemilyxo

quick questions guys, im really finding pH of buffers hard, which formula should i remember?
[H+] = Ka x [HA]/[A-] or
pH = pKa + log[A-]/[HA] ??

Reply 3093

alltimeemilyxo

Original post by master y

An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
Calculate the percentage yield, by mass, of sodium ferrate(VI).
Show your working.

cN ANYONE HELP

% yield is actual over theoretical isnt it?
so you just do moles of sodium ferrate over moles of iron oxide assuming they are 1 to 1 ratio and times by 100

Reply 3094

Joey952

12

what is the equation when OH is the subject of your formula ?

Reply 3095

Kat12395