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you deleted the post histephenson007 ....
And this skeletal formulae thingie is not working for me ..
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q5(c)
please help
histephenson007 can u answer me this one pls!?Can someone please tell me how to do Nov 2011, p43, question 3(e)(iv)!!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
im using the equation pH= pKa + log [base/acid] and im gettingg 4.36...
to find the no. of carbon atoms, always use this formula---> 100/1.1 * (M+2/M)7c too ... mass spectometry and NMR uhhhhhhhhhhh .... i'm unable to attempt these questions at all despite of going through the topics in OCR and application booklet ... it was tedious studying them maybe that's why i couldn't grasp much ...
7c too ... mass spectometry and NMR uhhhhhhhhhhh .... i'm unable to attempt these questions at all despite of going through the topics in OCR and application booklet ... it was tedious studying them maybe that's why i couldn't grasp much ...
histephenson007 can u answer me this one pls!?
Think of it as a free-radical substitution reaction and add chlorine on each Carbon atom. Beware with the left ones.check out part c of Q5... they've asked to draw skeletal formulae... i'm unable to attempt it ... all my skeletal formulae went wrong
By comparing 1 and 2, we get [CH3CHO] is 1st order.
why did u subtract the no.of moles??First, we need to get the Ka value from the pKa ..
pKa=-log(Ka)
So, Ka = 10 ^(-4.76)
The acid equation is : CH3COOH <---> (H+) + CH3COO-
So, Ka = ([H+]*[CH3COO-]) / [CH3COOH]
Re-arranging the equation: [H+] = (Ka*[CH3COOH]) / [CH3COO-]
But, we don't know what [CH3COO-] is,
So, we hav to find it with the formula C = {n(H+) - n(OH-)} / V
So, C = (0.1*0.01-0.01*0.25) / 0.02 = 0.075
Then, we can calculate for the H+ with the above reaction. And then pH. My teacher just used some pH = 4.76 + log(0.05/0.075), but I don't understand why.
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