Can anyone please explain the following question?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf Q:28
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf Q:28
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Can anyone please explain the following question?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf Q:28
Answer is C. If 2 can be correct, why not 1?
question 9 , I don't know any one help??.. question 19 it's A right? draw the displayed formula of the compounds and try to make a chiral center from the elements given to you, if you can't then it won't form an optical isomer , like in iv there's also a double bond and the only carbon with a single bond must have either two hydrogen's or two bromine get ?..in 28 it's C right? if you draw the displayed formula of the compound u will see that the double bond is still there , that's because nabh4 will only reduce the aldehyde group to it's primary alcohol , the other options are insensible because for example in A addition of hydrogen in that method will not reduce the aldehyde , it will only make the double bond single. hope you understood
thanks a lotanswer is B
amount of sulphite use = conc x vol
= 0.1 x (25/1000)
=2.5x10^-3mol
2 electrons are lost
amount of electrons lost = 2 x 2.5 x 10^-3 ==== 5 x 10^-3 mol
amount of electorns gained by metallic salt = amount of electrons lost by sulphite
= 5 x 10^-3 mol
amount of metallic salt used = 0.1 x (50/1000) ==== 5 x10^-3
amount of electrons gained PER MOLE of salt = (5 x 10^-3)/(5x10^-3) = 1 unit
hence oxidation state of metallic salt decreases by 1 unit meaning the oxidation satate becomes +2 from +3!!!
Okay I rechecked and in Q19 the optical isomer of iii is C*H(I)CH3CH2I the Carbon with the asterisk is the chiral center and u figured out i and ii right? in Q 28 I messed up the double IS broken in A with the addition of hydrogen , but I didn't know that an adldehyde is also oxidised to a primary alcohol by hydrgenation I never studied that :/ and in C it's wrong because the double bond id broken by nabh4 which is wrong because it will only reduce the aldehyde to a primary alcohol.. Hope you got it now and sorry for the wrong answers :/Noo Q19 (B) and Q28 (A) according to the mark scheme
GUYZZZ CHECK THIS REPEATED QUESTION THE MARK SCHEME GAVE TWO DIFFRENT ANSWERS !!! HELP
Q12http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
and this is the mark scheme for JUNE 06
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_ms_1.pdf
Q11http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
and this is the mark scheme for NOVEMBER 11
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_ms_11.pdf
what is the correct answer ????
C10H14O must be an ALCOHOL because an ALKENE is formed when this compound is dehydrated! and it must b a TERTIARY ALCOHOL, as only tertiary alcohols are not readily oxidised!!!!!
looking at the options:
A,B and C are secondary alcohols! as none of the carbon on which the double bond is fromed has an alkyl group!! a TERTIARY GROUP has an alkyl group left after the formation of a double bond! hence D is correct!
yaar the options in both the papers are different. option C from Q12 is same as option A from Q11. these are the SAME and CORRECT answers.GUYZZZ CHECK THIS REPEATED QUESTION THE MARK SCHEME GAVE TWO DIFFRENT ANSWERS !!! HELP
Q12http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
and this is the mark scheme for JUNE 06
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_ms_1.pdf
Q11http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
and this is the mark scheme for NOVEMBER 11
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_ms_11.pdf
what is the correct answer ????
yaar the options in both the papers are different. option C from Q12 is same as option A from Q11. these are the SAME and CORRECT answers.
hahha.... that;s always a gud optionhahahhahahhhaa its seems i must take a break now
Two equilibria are shown below.
reaction I 2X2(g) + Y2(g--> 2X2Y(g)
reaction II X2Y(g)---> X2(g) + 1/2Y2(g)
The numerical value of Kc for reaction I is 2.
Under the same conditions, what is the numerical value of Kc for reaction II
ans is 1 / root 2 ??
umm i think an aldehyde will be formed bcz it is distilled (and not refluxed) so write out an eqation for the formation of an aldehyde from this alcohol.Can someone please help me with this question?
Nov2011 qp 12
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