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

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your wrong :biggrin:

if you work out the concentration of H2O2 in the 250, you know the diluted concentration of H2O2

the amount of H2O2 in the dilute sample is 25cm3 in 250cm3, which is the same as 1/10th of the original sample, so if you x10 you get the amount in the original sample

you know the answer to that question is 40 x n[H202] x 10. do you know why we are multiplying by 40?

Thanks

Reply 1301

arvin_infinity

Original post by SmartFool

Thank you! but do you have the mark scheme to these questions anywhere? i need to go over quite a lot on that past paper especially the buffers bit! i can't seem to find it on the ocr website!

Someone will be a life-saver and list all the use of transition metal we need to know! :rolleyes:

Chromium---> stainless steel
...

Reply 1302

Steven

Original post by touran22

any one got any ideas its driving me nuts!

First you divide the % composition of each compound by its Mr, so you get:

11.11/14 : 3.17/1 : 41.27/52 : 44.45/16

This gave you 0.79 : 3.17 : 0.79 : 2.78

Then divide all ratios by the smallest number, so 0.79. This gives you:

1 : 4 : 1 : 3.5

Double everything to make 3.5 up to 7.

2 : 8 : 2 : 7

Which gives you N2H8Cr2O7

Reply 1303

goldlock

Original post by Princess_perfect786

i got the answer as -54kJ but the actual answer is -54kJ/mol. what am i supposed to be dividing -54 by?

To get to -54, you will have divided the value q (in kJ) by a molar quantity (moles), right, so your units are kilojoules/moles :smile:

Reply 1304

MarieLyon

I am pretty sure I am going to fail this exam tomorrow.

Reply 1305

viksta1000

Original post by threerose92

you know the answer to that question is 40 x n[H202] x 10. do you know why we are multiplying by 40?

Thanks

yeah number of c = n x 1000 / v

v = 25

so c = n x 1000/25 = n x 40

Reply 1306

INeedToRevise

When you draw isomers using the ligand 'en' do you have to make sure that the carbons in CH2 are joined to the nitrogen, I mean does this have to be made clear. Could you lose marks?

Reply 1307

KnuckleheadNinja

REDOXplease help. Me. !

so it says

"Hydrogen Iodide, HI, is oxidised to Iodine, I2, by conc. H2SO4 which is reduced to H2S"

so why is the equation (before balancing ox numbers)

2HI + H2SO4 -> H2S + I2

i get the equation but where the hell did that 2 infront of the HI come from? i thought you were supposed to balance AFTER doing oxidation numbers.

(question is from page 183, worked example)

(edited 12 years ago)

Reply 1308

BA1

Original post by KnuckleheadNinja

wow. that must be such a relief. what did you get on the AS mods for maths? thats exactly what i was trying to do. work REALLY hard on C1 and C2 and then can afford to lose a few for C3 and C4. but i dont think i could go in there and get a U and come out with an A!
:biggrin:

C1: 99
C2: 100
S1: 98
Total for AS: 297/300

C3: 79
C4: Next week
M1: 82
Total needed for A = 480 - 297 - 79 - 82
= 22 !!! :-)

(edited 12 years ago)

Reply 1309

SmartFool

for the love of god can anyone tell me the colours of compounds/transition metals we need to know please!

Reply 1310

yesioo

jan 11 mark scheme please?

Reply 1311

goldlock

Original post by KnuckleheadNinja

I've got another question :s-smilie:

i SUCK at redox.
like REALLY badly.

so i was going over the examples on page 183 to try and help and .... no. the worked example on page 182 is fine. but
when it got to step 2 on page 183, where RANDOMLY they've got 2HI and i dont see where that came from. at all. like, its not even balanced? why did they do that! seriously!?

Help much appreciated.
:smile:

What they've done is, they said "balance any atom that changes the oxidation number" - now, iodine changes oxidation number from minus 1 to 0, so they have to balance the iodine atoms :smile:

(S changes oxidation number too but it's already balanced)

Reply 1312

jlcf

Original post by SmartFool

Thank you! but do you have the mark scheme to these questions anywhere? i need to go over quite a lot on that past paper especially the buffers bit! i can't seem to find it on the ocr website!

Here they are in the same folder you downloaded.

Page 63

First question is about kP - partial pressure, do we need to learn?

ocr_40998_ms_09_l_gce_jun[1].pdf362.1KB

Reply 1313

CoventryCity

I have a feeling the carbonic acid and hydrogencarbonate buffer system in the blood will be on the paper

Reply 1314

threerose92

Original post by viksta1000

yeah number of c = n x 1000 / v

v = 25

so c = n x 1000/25 = n x 40

thanks

Reply 1315

goldlock

Original post by CoventryCity

I have a feeling the carbonic acid and hydrogencarbonate buffer system in the blood will be on the paper

Yeah I agree with you, they put the haemoglobing stuff and platin stuff in january which was the kind of biology material - got a good feeling about this one :smile:

Reply 1316

touran22

Original post by Steven

thanks for your help...i see how they got the 07 but no were in the question does it hint that its ment to be doubled? is that just something we have to apply knowledge to?

Reply 1317

threerose92

Original post by cws121

When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is
–15 kJ mol–1.
The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of –24 kJ mol–1.
Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF

can someone explain the answer to the question thanks?