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AS Chemistry- helping each other out!

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Original post by Dylann
Average titre probably, the average you work out from concordant results.

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That was my guess too, although that isn't really a result that matters in your ISA (obviously it does matter inside your school, but from school to school this will vary).
Original post by Dylann
Yeah that makes sense. I think if you compare the thermodynamic properties as you said, it could suggest cyclohexane is more stable, since cyclohexane has a enthalpy of formation of about -170 and benzene has +50. Though this doesn't necessarily indicate stability as the part you quoted talks about overlapping of orbitals which is actually more to do with stability than enthalpies.


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What that is showing is cyclohexene, but what they were asking was about cyclohexane.. It threw me off too.
Original post by l1lvink
What that is showing is cyclohexene, but what they were asking was about cyclohexane.. It threw me off too.

No. It is cyclohexane at the bottom left, showing cyclohexane is lower in energy than benzene.
Original post by langlitz
No. It is cyclohexane at the bottom left, showing cyclohexane is lower in energy than benzene.

Ah yes, sorry I missed that bit. In only saw the alkenes.
Hello, there I'm in Cambridge Board. Doing A levels Chemistry here!!
Reply 726
Original post by noshahmad
Sure :biggrin:


So I just had a mock exam and got an E, at the end of this year I need a high A, is that possible? If so how ? I feel so stressed and keep imagining myself failing. God forbid
Original post by kira1
So I just had a mock exam and got an E, at the end of this year I need a high A, is that possible? If so how ? I feel so stressed and keep imagining myself failing. God forbid




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Its possible., just work really hard, I'm in your boat too😃.
Reply 728
Original post by Kadak
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Its possible., just work really hard, I'm in your boat too😃.


Thanks
Original post by kira1
So I just had a mock exam and got an E, at the end of this year I need a high A, is that possible? If so how ? I feel so stressed and keep imagining myself failing. God forbid

DEFINITELY! never fear lol i am a student on a gap year (unconditional and conditional offers in biomedical science from Bristol,Manchester,Herts) and i was in the same situation,but i bumped up to an A in f322 in my first year. Just resitting f321 as i keep getting B/C no idea why lol....well actually my silly mistakes keep accumulating :colondollar: My advice is flash cards,posters and plenty of past-papers!:biggrin: Good luck. Btw anyone stuck feel free to ask me about any bio,maths or chem questions in my inbox
Reply 730
Original post by shakeyboy
DEFINITELY! never fear lol i am a student on a gap year (unconditional and conditional offers in biomedical science from Bristol,Manchester,Herts) and i was in the same situation,but i bumped up to an A in f322 in my first year. Just resitting f321 as i keep getting B/C no idea why lol....well actually my silly mistakes keep accumulating :colondollar: My advice is flash cards,posters and plenty of past-papers!:biggrin: Good luck. Btw anyone stuck feel free to ask me about any bio,maths or chem questions in my inbox


Thank you so much
Just went by my biology teacher she told me that she doesn't think I'll be getting a C let alone an A
Anyone here who can answer a AS Chem question to do with Energetics pls
Original post by kira1
So I just had a mock exam and got an E, at the end of this year I need a high A, is that possible? If so how ? I feel so stressed and keep imagining myself failing. God forbid


I got a D in my mocks and 90% im my final exams. It's defo possible, as long as you put the work in and learn from your mistakes (examiner reports are good for this)
PM if you want actual advice haha. :smile:

Edit: my teacher also told me that I wouldn't get above a C in Chen. Belieeeeeve in yourself and work hard you'll be fine 😎
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Original post by Coerce
Anyone here who can answer a AS Chem question to do with Energetics pls


Yeah what's the question? :smile:


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Original post by cerlohee
Yeah what's the question? :smile:


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Working out Enthaplhy chance of combustion;;

0.74 g (0.010 mol) of propanoic acid was burned in the simple calorimeter like that described above for the combustion of methanol. The temperature rose by 8.0 K. What value does this give for the enthalpy change of combustion of propanoic acid?

I understand that the equation to use is q =m x c x temp change so it'd be 0.74? 4.2 x 8
This from where I get stuck - also I believe the mass I'm inputting to be wrong

I have another question too

50.0cm^3 of 2.00cm moldom^-3 sodium hydroxide and 50.0cm^3 of 2.00 moldm^-3 hydrochloric acid were mixed in an expanded ploysterent beaker , temp rising by 11.0k

so it'd be 100 x 4.2 x 11 = 4620 - I had a nosey at the back and its -46.2 kj
do I get this value by working out moles mxv/100 which would be; 2 x 50/1000 which equals 0.1
thus 4620/0.1 = 46200 = 46.2kjmol

Sorry for the essay but I have an exam tommorow and want to be through
Hello people, I'm doing Edexcel AS Chem
Original post by Coerce
Working out Enthaplhy chance of combustion;;

0.74 g (0.010 mol) of propanoic acid was burned in the simple calorimeter like that described above for the combustion of methanol. The temperature rose by 8.0 K. What value does this give for the enthalpy change of combustion of propanoic acid?

I understand that the equation to use is q =m x c x temp change so it'd be 0.74? 4.2 x 8
This from where I get stuck - also I believe the mass I'm inputting to be wrong

I have another question too

50.0cm^3 of 2.00cm moldom^-3 sodium hydroxide and 50.0cm^3 of 2.00 moldm^-3 hydrochloric acid were mixed in an expanded ploysterent beaker , temp rising by 11.0k

so it'd be 100 x 4.2 x 11 = 4620 - I had a nosey at the back and its -46.2 kj
do I get this value by working out moles mxv/100 which would be; 2 x 50/1000 which equals 0.1
thus 4620/0.1 = 46200 = 46.2kjmol

Sorry for the essay but I have an exam tommorow and want to be through



For the first question, we need to know the mass. This was probably given to you earlier in the question, try and find the mass of the calorimeter that was used. Until we know the mass, we can't work out much.

I think for the second question you have it perfect..

q=MxCxdeltaT
q=100grams (volume of both reactants, assuming density is like that of water, where 1ml is 1g) x 4.20 x 11
q=4620J = 4.62kJ of heat evolved.

We can see that equimolar reactants reacted, so the amount of moles is simply (conc x vol) = 2 x 0.05 = 0.1 moles

So 0.1 moles evolved 4.6kJ of heat. We need to work out how much heat was evolved per mole. 4.62kJ/0.1 mol = 46.2kJmol-1

However, we are talking about the system and since the reaction is exothermic (temp rose) it's -46.2kJmol-1 :biggrin:
Original post by Dylann
For the first question, we need to know the mass. This was probably given to you earlier in the question, try and find the mass of the calorimeter that was used. Until we know the mass, we can't work out much.

I think for the second question you have it perfect..

q=MxCxdeltaT
q=100grams (volume of both reactants, assuming density is like that of water, where 1ml is 1g) x 4.20 x 11
q=4620J = 4.62kJ of heat evolved.

We can see that equimolar reactants reacted, so the amount of moles is simply (conc x vol) = 2 x 0.05 = 0.1 moles

So 0.1 moles evolved 4.6kJ of heat. We need to work out how much heat was evolved per mole. 4.62kJ/0.1 mol = 46.2kJmol-1

However, we are talking about the system and since the reaction is exothermic (temp rose) it's -46.2kJmol-1 :biggrin:



There is not one given in the question - its the Nelson Thrones book if by chance oyu have it? I believe the mass would be usually 100cm^3?
Original post by Coerce
There is not one given in the question - its the Nelson Thrones book if by chance oyu have it? I believe the mass would be usually 100cm^3?


It should be there! Look at the previous question about the combustion of methanol and see what mass they used there. If not, just assume 100grams...I will get hold of a book tomorrow and check. What page is it on?

q=MxCxdeltaT
q=100x4.20x8
q=3360J
q= 3.36kJ of heat evolved

0.01 moles of propanoic acid released 3.36kJ of heat energy. Therefore, 1 mole of propanoic should evolve 3.36kJ/0.01mol = 336kJmol-1

Since temp rose it is exothermic, so -336kJmol-1

Another way of thinking about it is:

0.01 moles = 3.36kJ heat energy

We need 1 mol, so we have to make the left side 1 mol. What do we multiply the left side by to get to 1 mol? 1mol/0.01moles = 100

So we multiply both sides by 100

1 mol = 336kJ, exothermic so -336kJmol-1.

Is this the answer in the book?
Original post by Dylann
It should be there! Look at the previous question about the combustion of methanol and see what mass they used there. If not, just assume 100grams...I will get hold of a book tomorrow and check. What page is it on?

q=MxCxdeltaT
q=100x4.20x8
q=3360J
q= 3.36kJ of heat evolved

0.01 moles of propanoic acid released 3.36kJ of heat energy. Therefore, 1 mole of propanoic should evolve 3.36kJ/0.01mol = 336kJmol-1

Since temp rose it is exothermic, so -336kJmol-1

Another way of thinking about it is:

0.01 moles = 3.36kJ heat energy

We need 1 mol, so we have to make the left side 1 mol. What do we multiply the left side by to get to 1 mol? 1mol/0.01moles = 100

So we multiply both sides by 100

1 mol = 336kJ, exothermic so -336kJmol-1.

Is this the answer in the book?


No the answer is -672 kJ mol^-1 I'll message you a picture of i can plug my phone into laptop two seconds

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