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# AS Chemistry- helping each other out! watch

1. (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
2. (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

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?
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?
4. (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
5. (Original post by Coerce)
No the answer is -672 kJ mol^-1 I'll message you a picture of i can plug my phone into laptop two seconds
Well then the mass that was used was 200grams not 100 grams!
6. (Original post by Dylann)
Well then the mass that was used was 200grams not 100 grams!
I'll send you imgs in a sec - can you jus run through it then with numericals exam tommorow and the maths part of chapter 7 is confusing me to death! - I take AS maths too wow
7. (Original post by cerlohee)
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.

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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Woman you are a life saver, thank you so much
I'm so grateful and I'm sorry to say if you were with me right now I'd give you a bone crushing hug
Thank you
8. (Original post by Coerce)
I'll send you imgs in a sec - can you jus run through it then with numericals exam tommorow and the maths part of chapter 7 is confusing me to death! - I take AS maths too wow
Can you send me a picture of it as well?

Thanks
9. Hey got a qualitative isa tomorrow and our teacher wasn't very specific :/ Heres the list : OBSERVATIONS AND BALANCED SYMBOL EQUATIONS.
You need to go through reactions and equations off:
Metal Carbonates + acids
Metal Carbonates heated
Metal oxides + acids
Precipitation reactions and the ionic equations for them.
Displacement reaction and change of oxidation number.

Help plz
10. (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
Of course. I got a U in my first chemistry test and now I'm averaging As. Just gotta understand the content, go over it regularly and do past papers. I usually watch YouTube videos on the topics I find difficult and that helps.

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11. Where about's is everyone content wise? My teacher is panicking about finishing.
12. (Original post by BIOCH3M)
Where about's is everyone content wise? My teacher is panicking about finishing.
on AQA have done about 3/11 topic in unit 2. But it's the big topics with the mechanism and stuff the other chapters shouldn't take long. But I am guessing we are fairly behind where we should be at.
13. (Original post by Super199)
on AQA have done about 3/11 topic in unit 2. But it's the big topics with the mechanism and stuff the other chapters shouldn't take long. But I am guessing we are fairly behind where we should be at.
I'm not on AQA and luckily as it seems a bit harder than OCR. We still have a quarter left and no controlled assessments completed either!
14. Can someone help me out I don't really understand the haloalkane topic.

The following compounds can be made from haloalkanes. For each compound, name a haloalkane which could be used, write an equation and state the essential conditions.

a). Propan-2-ol
15. (Original post by BIOCH3M)
Where about's is everyone content wise? My teacher is panicking about finishing.
I do OCR so I've done 3/5 units and parts of my controlled assessment. Lots to do and not enough time. Think imma be finished by Easter.

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16. (Original post by Super199)
Can someone help me out I don't really understand the haloalkane topic.

The following compounds can be made from haloalkanes. For each compound, name a haloalkane which could be used, write an equation and state the essential conditions.

a). Propan-2-ol
How do you make an alcohol from a haloalkane first of all?

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17. (Original post by Dylann)
How do you make an alcohol from a haloalkane first of all?

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By reacting it with a hydroxide ion. Well by nucleophillic substitution. Well I'm guessing propane is the haloalkane but don't you need a halogen to be attached to that? Does it matter what it is if that is right?
18. Hi , how much revision have you guys done so far ?
19. (Original post by Super199)
By reacting it with a hydroxide ion. Well by nucleophillic substitution. Well I'm guessing propane is the haloalkane but don't you need a halogen to be attached to that? Does it matter what it is if that is right?
You're sort of there. You need a haloalkane because the X-C bond (Where X is a halogen) becomes polarised, meaning C has a slight positive and the halogen has a slight negative charge. Therefore, the negative hydroxide ion could attack the carbon and replace the halogen. The question gave the product of propan-2-ol, meaning the hydroxide ion must have attacked the second carbon. In order for this to have happened, the halogen must have also been on that second carbon. So the original haloalkane was 2-chloropropane (or 2-bromopropane). The condition favoured nuc. sub and not elimination, meaning :OH- acted as a nucleophile which IIRC means the conditions are just room temperature and water...
20. (Original post by Dylann)
You're sort of there. You need a haloalkane because the X-C bond (Where X is a halogen) becomes polarised, meaning C has a slight positive and the halogen has a slight negative charge. Therefore, the negative hydroxide ion could attack the carbon and replace the halogen. The question gave the product of propan-2-ol, meaning the hydroxide ion must have attacked the second carbon. In order for this to have happened, the halogen must have also been on that second carbon. So the original haloalkane was 2-chloropropane (or 2-bromopropane). The condition favoured nuc. sub and not elimination, meaning :OH- acted as a nucleophile which IIRC means the conditions are just room temperature and water...
Right got the majority of that. Why can't the halogen be any of the other halogens? Why specifically 2-chloropropane or 2-bromopropane?

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