AQA A2 CHEM4 & CHEM5 June 2015 [official discussion thread]

Where do people usually get higher marks, CHEM4 or CHEM5?

Well, I'm getting around 90 marks or above for chem4, haven't quite cracked chem5 as we've only just finished yet but I'm getting there.

Does anyone else prefer A2 to AS?

Well, I'm getting around 90 marks or above for chem4, haven't quite cracked chem5 as we've only just finished yet but I'm getting there.

Does anyone else prefer A2 to AS?

Hey Jan 2010 Chem 4 paper question 1a) i)
So I worked out the moles of propanoic acid and moles of water right but got the moles of ethanol wrong.
What I did was the ratios so since 0.54 of ester ethyl propanoate is in the equilibrium mixture that means there will be 0.54 propanoic acid since 1:1 ratio and so since ethanol is a 1:2 ratio I doubled it to get 1.08. Then to work out equilibrium moles of ethanol I did 2.00 - 1.08 to get 0.92 which is wrong. Can someone explain to me please how it is supposed to be done?

Hello, this maybe an easy 1 marker, but I still can't figure it out. Jan 13 Chem4, Q 1aii, i subbed in all the values and somehow got it wrong. Thanks

Here is the solutions to 1ai) and 1aii)

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Does this help? I've not checked the mark scheme for it so let me know if its wrong.

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Thanks! I realised i didn't put any brackets in. Silly error, problem is I don't know when to put them in and when not too. Do you put them in based in the equation? So like in that q, theres a bracket on b & c, so shall i put a bracket on b & c?

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Thanks so much for your help!

How do you know how many ligand's can join onto a metal ion? Are you usually told?

I don't understand how complex formation occurs?!
( I understand that it forms co-ordinate bonds, but how do you know how many-if that makes sense?!)

Any help would be much appreciated

The brackets in the rate equation just mean concentration of b, concentration of c, and so on.
What kind of calculator do you have?
I recommend highly that you get a Casio with a natural display so you can type in easily what you want to work out. When you type in equations into your calculator just bracket everything and it should work out nicely. Particularly exponents like the squared part in the equation before because it means that your calculator will work out the bits in the brackets first. I'm probably going on a bit now so I'll shut up haha

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H2O molecules are small and so up to 6 can fit around a metal ion. Chloride ions are larger than H2O molecules and so only four can fit around a metal ion.

E.g. [Co(H2O)6] + 4Cl- ---> [CoCl4]2- + 6H2O.

Ligand substitution has occurred and only four chloride ions have formed co ordinate bonds with the cobalt ion.

Hope that was somewhat helpful

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Hello, can anyone please help me out on Jun11 Chem5. Q 3ai) I don't get why you divide 944 by 2 & how you get the value for H-H bond. Q8bii) I understand how to do the calculation like how you actually go about it, but I don't understand how you get the equation (which i got wrong and lost marks, as I then got wrong moles of fe2+)

Thanks

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you've got one mole of 2N so use that bond value and add it to 3 x the bond value of hydrogen. So do 944 + 3x((436) = 2252 (you times the hydrogen value by 3 because theres 3 moles).

Then do the same for the product, theres 2 moles of NH3 which means theres 2x the amount of bonds found in one mole of NH3 in other words, the bond value for N-H is 338, 2 moles of NH3 means theres 6 of these bonds so do 6 x 388 = 2328

then do bonds broken - bonds formed = 2252- 2328 = -76 but the question asked for ONE mole of NH3 so do -76 divided by 2 to get the value for one mole, which is -38 kJ mol-1