[rasklatz]

ok im out see you tomorrow ladies,gents,those in between xxx

[Cath-ay]

(Original post by otrivine)
hi how are you can u ask me questions

Good thanks, you?

Electron configurations of:

a) Ni
b) Cu2+
c) Cr3+
d) Mn

[otrivine]

(Original post by Cath-ay)
Good thanks, you?

Electron configurations of:

a) Ni
b) Cu2+
c) Cr3+
d) Mn

im good how aree you
these are easy i want A* questions

[arvin_infinity]

What is the oxidation number of Cr in SrCro4

[Cath-ay]

(Original post by otrivine)
im good how aree you
these are easy i want A* questions

Idk, I'm not good with coming up with A* questions, that's the examiner's job to do

[KrishanC93]

(Original post by rasklatz)
Yeah it was quite a troll comment from myself I must concede.

I see. My comment is officially retracted!

[otrivine]

(Original post by Cath-ay)
Idk, I'm not good with coming up with A* questions, that's the examiner's job to do

lool hmm any hard topics?

[The Illuminati]

The percentage by mass of iron in a sample of moss killer can be determined by titration against acidified potassium manganate(VII).
• Stage 1 – A sample of moss killer is dissolved in excess sulphuric acid.
• Stage 2 – Copper turnings are added to the acidified sample of moss killer and the mixture is boiled carefully for five minutes. Copper reduces any iron(III) ions in the sample to give iron(II) ions.
• Stage 3 – The reaction mixture is filtered into a conical flask to remove excess copper.
• Stage 4 – The contents of the flask are titrated against aqueous potassium manganate(VII).

(i) Suggest why it is important to remove all the copper in stage 3 before titrating in stage 4.

[otrivine]

(Original post by The Illuminati)
The percentage by mass of iron in a sample of moss killer can be determined by titration against acidified potassium manganate(VII).
• Stage 1 – A sample of moss killer is dissolved in excess sulphuric acid.
• Stage 2 – Copper turnings are added to the acidified sample of moss killer and the mixture is boiled carefully for five minutes. Copper reduces any iron(III) ions in the sample to give iron(II) ions.
• Stage 3 – The reaction mixture is filtered into a conical flask to remove excess copper.
• Stage 4 – The contents of the flask are titrated against aqueous potassium manganate(VII).

(i) Suggest why it is important to remove all the copper in stage 3 before titrating in stage 4.

for recrystallisation and to remove any impurities that will affect the reaction

[Cath-ay]

(Original post by otrivine)
lool hmm any hard topics?

Here try this:

[otrivine]

(Original post by Cath-ay)
Here try this:

the answer to part a) is where both electrons are donated to the transition metal ion forming 4 coordinate bonds

[tallen90]

(Original post by Smiley Face :))
Hi, Do we need to know the reaction that occurs iin a hydrogen fuel cell? Or no?? Please let me know asap, thanks

You'll need to know the overall reaction, which is simply 1/2O₂ + H₂ --> H₂O.

[Bright]

Guys Jan 2012 last question part c)

Where am I am going wrong? (I haven't seen the answer yet)

n(Cr2O7^2-) = 5.3X10^-4

n(Fe2+)= 5.3X10^-4 X 6 = 3.18X10-3 (in 25 cm3) -> x 10 0.0318 mol in 250 cm3
n=m/mr

55.8X0.0318 = 1.77444

1.77444/3.25X 100 =54.4%

It doesn't look right to me, someone help please.

[Cath-ay]

(Original post by otrivine)
the answer to part a) is where both electrons are donated to the transition metal ion forming 4 coordinate bonds

Mark scheme says:
many lone pairs (1) forming co-ordinate bonds (1)

So I'd give you the second mark. Need to be more specific for the first mark.

[otrivine]

(Original post by Cath-ay)
Mark scheme says:
many lone pairs (1) forming co-ordinate bonds (1)

So I'd give you the second mark. Need to be more specific for the first mark.

multi means 2 no?

[LifeIsGood]

(Original post by Bright)
Guys Jan 2012 last question part c)

Where am I am going wrong? (I haven't seen the answer yet)

n(Cr2O7^2-) = 5.3X10^-4

n(Fe2+)= 5.3X10^-4 X 6 = 3.18X10-3 (in 25 cm3) -> x 10 0.0318 mol in 250 cm3
n=m/mr

55.8X0.0318 = 1.77444

1.77444/3.25X 100 =54.4%

It doesn't look right to me, someone help please.

Your working out is perfect, re-type it into your calculator you should get 54.6%

[Killjoy-]

(Original post by Bright)
.

It would have been more effective if you put those in multiple spoilers :P

[Cath-ay]

(Original post by otrivine)
multi means 2 no?

Bidentate ligands donate 2 lone pairs only

Multidentate ligands can donate 3 or more lone pairs

[Smiley Face :)]

(Original post by Bright)
Guys Jan 2012 last question part c)

Where am I am going wrong? (I haven't seen the answer yet)

n(Cr2O7^2-) = 5.3X10^-4

n(Fe2+)= 5.3X10^-4 X 6 = 3.18X10-3 (in 25 cm3) -> x 10 0.0318 mol in 250 cm3
n=m/mr

55.8X0.0318 = 1.77444

1.77444/3.25X 100 =54.4%

It doesn't look right to me, someone help please.

ts right ...

[Smiley Face :)]

Hi Can I just ask, what do we nee to know about feasiblilty of cell reactions, its in the spec, but is it just the usual, that the E cell value may vary depending on the temperature and if non standard concentrations are used? thanks