You need to know all organic reactions, tests, flame test, brief information about redox and Intermolecular forces, electron configuration, I don't know if there's anything else. Same I haven't revised for As yet it's just so annoying, there is so much to revise for each subject
Thats true but i think boundaries will be very high. As this is the last exam session for this series & many would be repeating OR giving complete chem A levels. So have to take it srsly :\
so you think chem4 and 5 boundaries will be extremely high this year?
Can someone help me with this, I thought It is Phenol because it is slightly soluble in water but very soluble in HCl but that complex with Cu2+ confuses me
Can someone help me with this, I thought It is Phenol because it is slightly soluble in water but very soluble in HCl but that complex with Cu2+ confuses me
The amine because amines form coloured complex with Cu2+ ions. Adding aqueous copper sulphate is a functional group test iirc. And I am guessing phenylamine is more soluble in HCl than it is in water because it is protonated and gains charge, more easily hydrated. But not sure if I'm right. I remember doing that question though
The amine because amines form coloured complex with Cu2+ ions. Adding aqueous copper sulphate is a functional group test iirc. And I am guessing phenylamine is more soluble in HCl than it is in water because it is protonated and gains charge, more easily hydrated. But not sure if I'm right. I remember doing that question though
Can you remind me of this functional group test with copper sulphate, I can't seem to remember it!!
And yeah, that seems like a reasonable explanation, thank you!! And lol you have a good memory I guess
I just don't really know what to do and how to use all the data ive been given (and why ive been given those data).
First of all, you have the volume of the titre and the volume of HCl, find the number of moles for the base and the HCl using n=cv. This is done so as to get the moles of ethanoic acid remaining after reacting with NaOH as it's stated that NaOH reacts with both the acids with the word "both" being in bold. Now, once you get the moles, you take the difference of the moles of HCl and NaOH. Now that you have the moles remaining(for ethanoic acid), this is the no of moles at equilibrium for ethanoic acid. Then check the difference of the initial moles and the moles at equilibrium as well. This is to get the moles that have reacted and hence you'll get the moles of the alcohol and ester. For water, you already have the moles and now just you got to add up simply. Substitute the moles in your Kc expression as volume cancels out!
First of all, you have the volume of the titre and the volume of HCl, find the number of moles for the base and the HCl using n=cv. This is done so as to get the moles of ethanoic acid remaining after reacting with NaOH as it's stated that NaOH reacts with both the acids with the word "both" being in bold. Now, once you get the moles, you take the difference of the moles of HCl and NaOH. Now that you have the moles remaining(for ethanoic acid), this is the no of moles at equilibrium for ethanoic acid. Then check the difference of the initial moles and the moles at equilibrium as well. This is to get the moles that have reacted and hence you'll get the moles of the alcohol and ester. For water, you already have the moles and now just you got to add up simply. Substitute the moles in your Kc expression as volume cancels out!
Thanks i did the question now. Although i still don't quite understand how finding the mol of hcl and naoh and subtracting the two gives me the mol left in equilibrium of ethanoic acid.