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Chemistry OCR want to help you with your Q's

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Original post by Jimmy20002012


Density is p = m/v. You must have been given volume? If you have volume, you can work out moles via n = v/24. Then you could work out the mass via p x v = m, then work out Mr via Mr = m/n. Can I see the whole question?
Reply 21
Add @assisttutor on twitter and you can send the question for an instant answer. Very helpful


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Original post by ChemistryOCR
Density is p = m/v. You must have been given volume? If you have volume, you can work out moles via n = v/24. Then you could work out the mass via p x v = m, then work out Mr via Mr = m/n. Can I see the whole question?


No, there is no volume given only density. Sure here is the question: Q8

http://www.thestudentroom.co.uk/attachment.php?attachmentid=366969&d=1424697585


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Original post by Jimmy20002012
No, there is no volume given only density. Sure here is the question: Q8

http://www.thestudentroom.co.uk/attachment.php?attachmentid=366969&d=1424697585


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I have done it! Wow that is hard.. I will attach my method
Original post by ChemistryOCR
I have done it! Wow that is hard.. I will attach my method


I still have no clue, much appreciated.


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Original post by Jimmy20002012
I still have no clue, much appreciated.


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Okay, I know that this is technically only 1 mark because this part is out of 3, but this method seemed to work. It is overly complicated I assume, but the answer is correct. I looked up the density and comparing it to real life densities, this gas is the correct gas. I don't know any easier way to get to the answer. Here goes:

1) From step 2, we know that we have 0.437g of BaFeO4. In moles, (n = m/M) this is (0.437g/ 257.17) = 1.699x10-3 mol of BaFeO4.
2) Ferrate ions is this, dissociated, to give equal number of moles of ferrate ions, 1.699x10-3 mol
3) I know the gas is a 1:1 ratio, so the moles of gas produced is the same number of moles of ferrate ions, 1.699x10-3 mol.
4) To work out the volume of the gas, use n = v/24 (you should remember this) Rearrange for v to make v = n x 24.
5) 1.699x10-3 mol x 24 = 0.040776 dm3
6) Use this volume in the density equation to work out mass. p = m/v, where we know p = 1.333x10-3 gcm-3. Therefore m = p x v (cm3)

1.333x10-3 x (0.040776 x 1000) < (because it needs to be in cm3 and we have it in dm3) = 0.0544g of the gas.
7) n = m/M. To get M which will identify the gas, rearrange so M = m/n

0.0544/1.699x10-3 = 32. That's oxygen!

Now the important bit, look at the equation now. It says in non alkaline conditions, meaning it is acidic. So H+ is present, and it works:

FeO4- + H+ + H20 ---> Fe(OH)3 + O2. All balanced, 1:1 ratio, 3 marks for equation, precipitate and gas formed.. just a seriously complicated, OP way of doing it, but hey it worked so have a look! :smile:
Original post by ChemistryOCR
Okay, I know that this is technically only 1 mark because this part is out of 3, but this method seemed to work. It is overly complicated I assume, but the answer is correct. I looked up the density and comparing it to real life densities, this gas is the correct gas. I don't know any easier way to get to the answer. Here goes:

1) From step 2, we know that we have 0.437g of BaFeO4. In moles, (n = m/M) this is (0.437g/ 257.17) = 1.699x10-3 mol of BaFeO4.
2) Ferrate ions is this, dissociated, to give equal number of moles of ferrate ions, 1.699x10-3 mol
3) I know the gas is a 1:1 ratio, so the moles of gas produced is the same number of moles of ferrate ions, 1.699x10-3 mol.
4) To work out the volume of the gas, use n = v/24 (you should remember this) Rearrange for v to make v = n x 24.
5) 1.699x10-3 mol x 24 = 0.040776 dm3
6) Use this volume in the density equation to work out mass. p = m/v, where we know p = 1.333x10-3 gcm-3. Therefore m = p x v (cm3)

1.333x10-3 x (0.040776 x 1000) < (because it needs to be in cm3 and we have it in dm3) = 0.0544g of the gas.
7) n = m/M. To get M which will identify the gas, rearrange so M = m/n

0.0544/1.699x10-3 = 32. That's oxygen!

Now the important bit, look at the equation now. It says in non alkaline conditions, meaning it is acidic. So H+ is present, and it works:

FeO4- + H+ + H20 ---> Fe(OH)3 + O2. All balanced, 1:1 ratio, 3 marks for equation, precipitate and gas formed.. just a seriously complicated, OP way of doing it, but hey it worked so have a look! :smile:


Wow, that is a lot for 3 mark finally seem to have got the density part, think the MS got a slightly different equation from you. But thanks for all this help :smile:

ImageUploadedByStudent Room1430755464.313188.jpg


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Original post by Jimmy20002012
Wow, that is a lot for 3 mark finally seem to have got the density part, think the MS got a slightly different equation from you. But thanks for all this help :smile:

ImageUploadedByStudent Room1430755464.313188.jpg


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Yeah it is different, but they both balance don't they? hmm, hope I helped in some way..
Reply 28
Hi! Just wondering how you would go about balancing this equation? It's from the June 2014 F325 paper. Thanks :smile:

Fe2O3 + Cl2 + OH- \rightarrow FeO4 + Cl- + H2O
Original post by ChemSoph
Hi! Just wondering how you would go about balancing this equation? It's from the June 2014 F325 paper. Thanks :smile:

Fe2O3 + Cl2 + OH- \rightarrow FeO4 + Cl- + H2O



Don't forget the excess hydroxide ions:

Fe2O3 + 3Cl2 + 10 OH- ---> 2FeO4 2- + 6Cl- +5H20
(edited 8 years ago)
Reply 30
Original post by ChemistryOCR
Don't forget the excess hydroxide ions:

Fe2O3 + 3Cl2 + 10 OH- ---> 2FeO4 2- + 6Cl- +5H20


How was this balanced? I'm unsure with the steps needed.
Original post by ChemSoph
How was this balanced? I'm unsure with the steps needed.


You need all the atoms to balance, and the charge. We have excess OH- so we can use as many as we need to. All that's required is to know that you balance the remaining oxygens with water. So we start with:

Fe2O3 + Cl2 + OH- ---> FeO42- + Cl- + H2O.
You need 2 on the FeO4 straight away:

Fe2O3 + Cl2 + OH- ---> 2FeO42- + Cl- + H2O


Now just balance oxygens, the rule is adding water:

Fe2O3 + Cl2 + 10OH- ---> 2FeO42- + Cl- + 5H2O

Hydrogens balance automatically, now balance the charge using Cl-

Fe2O3 + 3Cl2 + 10OH- ---> 2FeO42- + 6Cl- + 5H2O

Done:smile:
Reply 32
Original post by ChemistryOCR
You need all the atoms to balance, and the charge. We have excess OH- so we can use as many as we need to. All that's required is to know that you balance the remaining oxygens with water. So we start with:

Fe2O3 + Cl2 + OH- ---> FeO42- + Cl- + H2O.
You need 2 on the FeO4 straight away:

Fe2O3 + Cl2 + OH- ---> 2FeO42- + Cl- + H2O


Now just balance oxygens, the rule is adding water:

Fe2O3 + Cl2 + 10OH- ---> 2FeO42- + Cl- + 5H2O

Hydrogens balance automatically, now balance the charge using Cl-

Fe2O3 + 3Cl2 + 10OH- ---> 2FeO42- + 6Cl- + 5H2O

Done:smile:


Aaah, great. I get it! Thank you!!! :biggrin:
Original post by ChemSoph
Aaah, great. I get it! Thank you!!! :biggrin:


More than welcome! Glad I could help :smile: Good luck with your exams!!!
Reply 34
Original post by ChemistryOCR
Hey guys I want to start this thread to help any of you guys out there studying OCR chemistry at A level. Hopefully this will turn out to be a really good revision thread!

I am simply an enthusiastic chemistry student who is going on to study it at university. I love helping others, and I am interested in teaching at some stage!
1) Try and relate your questions, e.g, page references, a past paper question, an example etc
2) It may be easier to post pictures of our working, so we'll see how that goes

Just want to help as many people as I can! :smile: Let's get as many people on this as possible, link your other classmates the more chemists the more revision! :smile::smile::smile:


On the identification questions at the end of Unit 2 papers and the reaction mechanism questions which have the QWC mark beside them are you meant to write in continuous prose or can you do it in a sort of note form (ie write down the stage, and have notes over it saying 'heterolytic fission' or 'peak at blah indicates blah')?
Reply 35
Original post by ChemistryOCR
NmcC yeah less of a temperature change, but then the mass has gone up, so read my answer and see what you think, do you agree? Enthalpy cannot change, the temperature of the water can, but it has gone up in the first place. mc delta T. Mass up, T down, do you think the answer would be the same? I think it should be. Very interesting though


I do agree with you, yes. I should have mentioned, The actual enthalpy change would remain constant because it's the burning of the fuel is what the actual enthalpy change is. You're measuring it using the water. So theoretically, there should be no change in enthalpy by burning the fuel. But because the mass of the water is increasing, it will appear to be less exothermic than it actually is. Good answer, quite a thought provoking problem.
Original post by ThatPerson2
On the identification questions at the end of Unit 2 papers and the reaction mechanism questions which have the QWC mark beside them are you meant to write in continuous prose or can you do it in a sort of note form (ie write down the stage, and have notes over it saying 'heterolytic fission' or 'peak at blah indicates blah')?


It isn't necessarily about solid writing, its organisation and spelling. You can do note form as long as it is organised, in order and you put detailed notes next to them and title them as you suggested.
Reply 37
Original post by ChemistryOCR
It isn't necessarily about solid writing, its organisation and spelling. You can do note form as long as it is organised, in order and you put detailed notes next to them and title them as you suggested.


Ah, ok. Thanks! Bit nervous for these exam malarkeys :O
Original post by ThatPerson2
Ah, ok. Thanks! Bit nervous for these exam malarkeys :O


You'll be fine. Lay it all out, explain everything. You can't over explain. If you are worried you haven't written enough, explain it in more detail, as if they don't know anything, it's good to try and cover all the marking points. Obviously make sure you have mentioned at least 2 points over the required points required in the marks available, don't leave a single stone unturned or those examiners will have you!
Does anyone actually know the different colours for all the halogens in gas, solution (hexane or water) and what happens when they are displaced? Can't understand at all

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