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

1. how do you work out the enthalpy change of neutralisation in this question:

25cm^3 of 2moldm^-3 dilute nitric acid is added to 25cm^3 of 2moldm^-3 aqueous potassium hydroxide. The temperature increased from 22 degrees to 35.5 degrees.

specific heat capacity, c=4.18 J/gk and density of solution= 1g/cm^3
2. Are people taking this exam tomorrow afternoon going to sleep properly? There are also various Maths exams on the following afternoon as well...how are people managing their time between F325 and C3?!
3. (Original post by Princess_perfect786)
how do you work out the enthalpy change of neutralisation in this question:

25cm^3 of 2moldm^-3 dilute nitric acid is added to 25cm^3 of 2moldm^-3 aqueous potassium hydroxide. The temperature increased from 22 degrees to 35.5 degrees.

specific heat capacity, c=4.18 J/gk and density of solution= 1g/cm^3
Work out q =-mc(delta T) and then divide that by the number of moles that react (be careful that it's the molar quantity that forms 1 mole of water! I think for sulfuric acid it's a bit different!)
4. (Original post by Princess_perfect786)
how do you work out the enthalpy change of neutralisation in this question:

25cm^3 of 2moldm^-3 dilute nitric acid is added to 25cm^3 of 2moldm^-3 aqueous potassium hydroxide. The temperature increased from 22 degrees to 35.5 degrees.

specific heat capacity, c=4.18 J/gk and density of solution= 1g/cm^3
-56.43? I canexplain if the answer is right LOL
5. (Original post by darkiee)
-56.43? I canexplain if they answer is right LOL
Yep that's right, it's always around -57 kJmol^-1 for any strong monobasic acid because they all dissociate completely in solution so the reaction is basically the same (ionic equations are identical)
6. how would you go about doing this type of question guys?
Attached Images

7. (Original post by Flux_Pav)
cheers
say if the question asked for the diluted h202 concentration. what would you do differently? this time would you do :

1.1725x10^-2 x 1000
___________________

250 (insted of 25)

and thts the answer?
no, the diluted concentration is just the concentration value before you x10

so its 1.59 g dm-3
8. any one got any ideas its driving me nuts!

can some one please go over Question 6e in the June 2010 exam...

I got NH4CrO4 but its (NH4)2Cr2O7?
9. (Original post by BA1)
I can't get an A*, I had 78/90 in F324 (which is just under 90%) and I'm not bothering to resit it as it clashes with my holiday dates end of next week :P I also did badly on the A2 practicals so had less than 90% in total for them. I just need a grade A for university so I'm not bothering to try and get the A*.

PS: That's why you should try and do well in the AS modules, to take the pressure off A2! I'm walking into an A2 maths exam next week needing 23/100 raw marks for a grade A, or 90/100 for an A* lmao!
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!
10. (Original post by Rocks:))
I get CH3OH + 1/2O2 -------> C02 + 4H+ +4e-

Is this right?
No
You have

CH3OH ------> CO2
Balance oxygen with hydrogen then balance hydrogen with H+ then balance charge with e-

So add one H20 to the left to balance oxygen, then we need to add 6H+ to balance hydrogen to the RHS. Then add 6e- to the right to balance charge
11. (Original post by goldlock)
You know how you add two half equations to make an overall equation? To deduce the methanol half equation, you should kind of make a maths sum of it. So write the overall equation and beneath it, write the oxygen half equation and "subtract" it - see if you can work from there (btw, you need to make sure the electrons are balanced, multiply the oxygen half equation by 1.5 to before you subtract)
Thanks.

If you look at it that way it makes it so much simpler.

So what they're asking me is do I have to add to the oxygen half equation to get the overall equation.

Thanks again.
12. (Original post by jlcf)
http://www.ocr.org.uk/qualifications...nts/index.html

Here more in the LEGACY past paper sections. Once you download it, remember to use these files:

2816-01 - Unifying Concepts in Chemistry questions
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions

The questions are probably similar but they may throw in some curveball like they did in Jan 2011 paper. All you can do is prepare and hope something paper would be easy.
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!
Attached Images
13. ocr_41858_pp_09_jun_l_gce_2816_01.pdf (333.4 KB, 486 views)
14. I've got another question
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.

15. (Original post by goldlock)
Work out q =-mc(delta T) and then divide that by the number of moles that react (be careful that it's the molar quantity that forms 1 mole of water! I think for sulfuric acid it's a bit different!)
i got the answer as -54kJ but the actual answer is -54kJ/mol. what am i supposed to be dividing -54 by?
16. (Original post by touran22)
any one got any ideas its driving me nuts!
is this the empirical formulae part?
You do the % give divide them by the molar mass. They you divide through by the smallest number to give the ratios
17. 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?
18. (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

Edit; :O sorry I must've accidentally posted half way through my message lol, full post futher down the page.
19. (Original post by CoffeeStinks)
Thanks.

If you look at it that way it makes it so much simpler.

So what they're asking me is do I have to add to the oxygen half equation to get the overall equation.

Thanks again.
Exactly, yeah the only stumbling block as I said was to make sure that your electrons are balanced correctly - obviously you can't tell since you can't see both half equations, but you can see that you need to times 02 by 1.5 because you don't want it appearing in the methanol half equation - the oxidant can't appear in the fuel's half equation.
20. (Original post by Pandit Bandit)
how would you go about doing this type of question guys?
hey man

this is the solution I posted earlier
http://www.thestudentroom.co.uk/show...postcount=1256

hope this helps
21. (Original post by goldlock)
Yep that's right, it's always around -57 kJmol^-1 for any strong monobasic acid because they all dissociate completely in solution so the reaction is basically the same (ionic equations are identical)
Ok besically the formula is q=mc^t

M= 25+25 = 50, total volume

C= 4.18

and change in temperature = 13.5 from 35.5-22

therefore q= 50x4.18x13.5= 2821.5

we know its an exothermic reaction cause heat was given out so -2821.5

Now this is where you scale up, you fine the moles by using n=cv

therefore n= 25x0.2 divide by 1000= 0.05

You then divide your answer by 0.05 mols so -2821.5 divide by 0.05= -56430j

conert to kJ by dividing that answer by 1000 and you get -56.43kj

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