Hi,

I was wondering if others may be in the position where they feel they 'know' a sufficient amount, but are worried about this 'application' of science hinted at in the reports.

Does anyone have any interesting predictions of how information may be woven into an 'apllication' question tommorow?

(for instance the question on peroxide and hair in january)

Reply 1224

HEC14

if they asked us for a ratio of A:B
For example, they could do this when proportioning [HA]:[A-] in a buffer system..
Would you have to A/B.. or just leave it as A:B? :smile:

Reply 1225

Wor Carroll

Original post by hey-hey-hey

OK simple Qs
Construct the equation for the oxidation of acidified iron(II) ions by oxygen..How do I start to do this?
i get
Fe2+ + O2 + 4H+--------> Fe3+ + 2H2O
But the answer is this, How?
4Fe2+ + O2 + 4H+-------------> 4Fe3+ + 2H2O

Why the 2 in front of the Fe?

Fe2+ ---------->Fe3+ + e-
O2 + 4e- + 4H+ --------> 2H2O

Then balance the e- :smile:

Reply 1226

viksta1000

15

Original post by Julian_valensi

Hi,

I was wondering if others may be in the position where they feel they 'know' a sufficient amount, but are worried about this 'application' of science hinted at in the reports.

Does anyone have any interesting predictions of how information may be woven into an 'apllication' question tommorow?

(for instance the question on peroxide and hair in january)

that was a stupid question and the reports basically admitted it

Reply 1227

potentialmedicalstudent

Original post by azz92

how did you get the 8 electrons ?

both sides need to have equal charges (in this case it's +4)

Reply 1228

INeedToRevise

8

Original post by raj16

wenever i ave a titration question i got through the same stages
1. MOLES
look for wen they give u a conc and a volume, so in this qu they give u 0.02moldm3 of KMnO4 and the volume 23.45cm3.
Use n=cv and u shuld get 4.69x10-4 mols of KMn04

2.Ratios
By using the equation given u can compare the number of moles. So Mn04 2:5 H2O2.
so divide the answer we got above by 2, thats so u get 1 mole and then times by 5. 1.1725x10-3

1 thing to look out for in the volumes, the answer we've worked out is in 25cm3, so in 250cm3 its 10X as more, 1.1725x10-2. This is the moles of H2O2 in 250cm3.

3.Units
Now they want the answer in g/dm3. So i think about this in terms of units now. Where can i get grams from? mols=mass/Mr

So 1.1725x10-2 /34= 0.39865g
Right at the beginnin of the qu they tell u that u diluted the soultion in 25cm3.
So we ave grams and cm3, divide cm3 by 1000 u get answer in dm3.
25/1000=0.025DM3
Using the units again g/dm3, i see the / meanin divide.
FINALLY!!

0.39865/ 0.025= 15.97g/dm3

Really hope i didnt confuse u further :biggrin:

I was stuck on this question and I get your explanation (+rep). What I don't understand is why do you do 1.1725x10-2/34= 0.39865g :confused:

Do you mean multiply? So 1.1725 x 10-2 x 34. I think this would make more sense..

Reply 1229

INeedToRevise

8

Guys, what equations do we need to learn?
I know the MnO4- with Fe and the iodine thiosulphate one.
Do we need to know any for the hydrogen fuel cell?
Could someone list them pleaaase
+rep :smile:

Reply 1230

hey-hey-hey

10

Original post by Wor Carroll

Fe2+ ---------->Fe3+ + e-
O2 + 4e- + 4H+ --------> 2H2O

Then balance the e- :smile:

Thankyou

Reply 1231

CoventryCity

When we are linking ionic size to lattice enthalpy is this sufficient

The ionic size of ion A is smaller than ion B, this means it has a greater charge density and so the ions have a greater attraction to each other.
This results in a more exothermic value for lattice enthalpy

First is this sentence correct?
Second is this sufficient if this comes up in the exam?

Reply 1232

HEC14

Original post by INeedToRevise

Guys, what equations do we need to learn?
I know the MnO4- with Fe and the iodine thiosulphate one.
Do we need to know any for the hydrogen fuel cell?
Could someone list them pleaaase
+rep :smile:

You need to know what happens at each electrode of the fuel cell,
well, only of the hydrogen fuel cell.
The overall reaction iss...
H2 + 0.5 O2 --> H2O

Reply 1233

Josh0806

Could somebody please explain to me how to balance this equation...

FeO42(–) + H(+) ? Fe(3+)+ O2+ H2O

...to give this...

4FeO42(–) + 20H(+) ? 4Fe(3+)+ 3O2+ 10H2O

Reply 1234

raj16

Original post by INeedToRevise

I was stuck on this question and I get your explanation (+rep). What I don't understand is why do you do 1.1725x10-2/34= 0.39865g :confused:

Do you mean multiply? So 1.1725 x 10-2 x 34. I think this would make more sense..

Woops, sorry! yh i did mean that. was typin it fast

Reply 1235

dunnicare

12

Original post by potentialmedicalstudent

...

Yep got it before you quoted me. I thought the oxidation state of cr was +8 for some reason. Thanks.

Reply 1236

CoventryCity

Original post by INeedToRevise

Guys, what equations do we need to learn?
I know the MnO4- with Fe and the iodine thiosulphate one.
Do we need to know any for the hydrogen fuel cell?
Could someone list them pleaaase
+rep :smile:

Yes you need to know the hydrogen fuel cell one they are

1/2 O2 + H20 +2e--------> 2OH- +0.4V
2H2O +2e- ----------> H2 + 2OH- -0.83V

These are the half equations

Reply 1237

SmartFool

ANYONE PLEASE! How do you work out where the equivalence line is on titration curves as Ive seen on a past paper where the line is at 33cm3 and not 25cm3

Reply 1238

blush.ox

15

Original post by CoventryCity

When we are linking ionic size to lattice enthalpy is this sufficient

The ionic size of ion A is smaller than ion B, this means it has a greater charge density and so the ions have a greater attraction to each other.
This results in a more exothermic value for lattice enthalpy

First is this sentence correct?
Second is this sufficient if this comes up in the exam?

Yes more exo/neg.

Also i doubt it would be any more than 2/3 (max) marks :/ so that should be enough imo. Maybe you could mention greater electrostatic forces of attraction and stronger ionic bonds to be more clear.

Reply 1239

azz92

thanks

this f325 exam is a confusing one it makes you feel like its easy but there are little tricks they dont show or tell you :mad:

Equilibria, Energetics and Elements (F325) - June 2011 Exam.

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