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AQA CHEM5 A2 Chemistry - 19th June 2013

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Reply 1960

Can anyone explain how the equation in question 3 (c) has been rearranged to find T.... it's been a while since I've done proper maths!! ImageUploadedByStudent Room1371579202.703766.jpg

ImageUploadedByStudent Room1371579202.703766.jpg

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Reply 1961

PhoenixSeeker

MgO how is it sparingly soluble in water? surely it forms Mg(OH)2
June 2012

Reply 1962

jqian

Original post by xodani

Can anyone help with June 2011 3ai) Why is the answer -38 and not -76?

It says formation of one mole of NH3 not 2.

Reply 1963

jqian

Original post by Ren_9

Can anyone explain how the equation in question 3 (c) has been rearranged to find T.... it's been a while since I've done proper maths!! ImageUploadedByStudent Room1371579202.703766.jpg

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They've just collected the terms with T on one side and the terms with no T on the other.

Reply 1964

Ren_9

Original post by xodani

Can anyone help with June 2011 3ai) Why is the answer -38 and not -76?

You have to calculate it per mol. Just divide your answer by 2 since two mols of ammonia is produced

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Reply 1965

xodani

Original post by jqian

It says formation of one mole of NH3 not 2.

Ah how silly of me! Thank you very much :smile:

Reply 1966

darryalar

can some one please tell me whether to use (products - reactants) and (reactants - products) when calculating enthalpy change. Thanks in advance.

Reply 1967

Ren_9

Original post by jqian

They've just collected the terms with T on one side and the terms with no T on the other.

I don't mean to be annoying but if you take T to the other side wouldn't it just cancell out (Eg -T+T=0 ) ?? *confused*

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Reply 1968

Adam020

Original post by tilownii

Why does it say purple or yellow in January 2013 :frown:

oh crap, my mistake. [Fe(H2O)6]3+ IS yellow, just all the complexes it makes are brown :colondollar:

Reply 1969

jqian

Original post by darryalar

can some one please tell me whether to use (products - reactants) and (reactants - products) when calculating enthalpy change. Thanks in advance.

If you're given enthalpy of combustion data, use reactants - products (use cr'a'p, combustion reactants- products).
If you're given enthalpy of formation use products - reactants.
If you're given the bond enthalpies then you use sum of the bonds broken (i.e. the reactants) - the sum of the bonds formed (the products).

Reply 1970

jonnyb123

Whew, done 7 past papers today (everything besides the specimen). This is like my third time going through them so I think I'm just about as ready as I'll ever be...

Probably just going to go over random questions for the next couple of hours.

How is everyone feeling about tomorrow?

Reply 1971

JamilMCR

Can anyone give me the equation for when H202 is used as an oxidising agent? Thanks

Reply 1972

jqian

Original post by Ren_9

I don't mean to be annoying but if you take T to the other side wouldn't it just cancell out (Eg -T+T=0 ) ?? *confused*

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On the left hand side you have -542.6T, on the right hand side you have -15.2T, if you add 542.6T to both sides the left hand side now has no T terms as -542.6T + 542.6T = 0T = 0. The right hand side becomes -15.2T + 542.6T which is 527.4T.