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The answer should be D. The -OH on left won't be oxidised by cold, dilute KMnO4, but the alkene would be oxidised to a diol. When hot, conc. KMnO4 is used, that -OH group is oxidised to a ketone and the carbons with the double bond in the ring(alkene group) would be oxidised to Carboxylic acid and a ketone respectively.
C, because a lone pair from the chloride ion would attack the AlCl3 molecule. As Aluminium has an incomplete octet, there should be an empty orbital to accept the lone pair from chloride ion.Can someone explain Q14?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
thank you
No. of moles of NaOH is 0.04 as we calculated earlier meaning (0.1 - x) = 0.04
haha so does your answer match with the marking scheme?No. of moles of NaOH is 0.04 as we calculated earlier meaning x = 0.04
So concentretion of acid and alcohol is found out by
= (0.1 - x)
= (0.1 - 0.04)
= 0.06
This paper reminds me of last year, I gave this very paper in the exam last year in AS
I made a mistake earlier i corrected it check the first post...haha so does your answer match with the marking scheme?
coming back to the question.. i really dont get it. could you please explain why have they taken 0.04 for the acid and alcohol.their equilibrium moles should be 0.06.?
oh okay. thanks a lot!I made a mistake earlier i corrected it check the first post...
Btw, i dont remember my solution from last year but i do remember that i got Kc = 2.25 so i must have done it right there
Yes.
okay thanks.. if we just write alcohol will it be marked as a wrong asnwer?
Probably. I have seen some mark schemes which do cut a mark for not writing the type of alcohol.
i dont get why we use excess air when converting SO2 to SO3 in the contact process......i know it shifts the equilibirium shifts to the right but i dont get why...??????
), so according to the principle if any change in the factors which affect the equilibrium is altered than the reaction behaves in a way which opposes the change and tries to attain eq. Sooo if you add excess O2 the concentration of the reactants increases and so the equilibrium shifts to the right to produce more products so they reaction will come to eq. again, hence increasing the conc. of the products so SO3 increases!
Aoa,
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