Chemistry: Post your doubts here!

[Dr.MMM]

Can anyone please explain the following question?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf Q:28

 

[biba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf QUESTION 9 PLEASE ANY HELP ??

answer 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!!!

 

[biba]

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

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!

 

[biba]

Answer is C. If 2 can be correct, why not 1?

NaCl can give u only one chloride so how could u substitute 2 chlorides to ethene? that is y 1 is incorrect!

 

[saadgujjar]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
q2,q5,q21,q31-q34,q38 plz explain
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
question 1,3,,9,10,14,31,32,40 plz expain thanks in advance

 

[cute97]

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

Noo Q19 (B) and Q28 (A) according to the mark scheme

 

[strangerss]

answer 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!!!

thanks a lot

 

[darkxangel]

Can someone please help me with this question ?

 

[strangerss]

Noo Q19 (B) and Q28 (A) according to the mark scheme

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 :/

 

[cute97]

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 ????

 

[Aries_95]

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 ????

The answers are correct.
Look again at the options, they are shuffled. the option C in s_O6 is option A in w_11 hence the answers are C and A respectively.

 

[Dr.MMM]

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!

Thanks

 

[Dr.MMM]

Oh and ive got another question:
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf Q:40th

 

[daredevil]

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.

 

[cute97]

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.

hahahhahahhhaa its seems i must take a break now

 

[daredevil]

hahahhahahhhaa its seems i must take a break now

hahha.... that;s always a gud option

 

[darkxangel]

Can someone please help me with this question?
Nov2011 qp 12

 

[daredevil]

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 ??

subsitute the equation for kc of the first reaction into the kc of the second reaction. product changes into reactant so the Kc (2) will be reciprocalled too and become 1/2.
then in the first equation there were 2 mol of X2Y so it was squared in the equation. in the second reaction is is just one so does not need to be squared so square root is taken on both sides of the eq.
if u don't get it like this then tell me and i'll see if i can write out the eq. here.

 

[daredevil]

Can someone please help me with this question?
Nov2011 qp 12

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.
calculate the moles of ethanol according to the mass given and then use mole ratio (as in the equation) to find moles of aldehyde. use moles and Mr of aldehyde to find mass. use the percentage yield formula and incorporate the theoretical yield and the percentage yield in it to find the practical - or actual - yield of the product i.e. aldehyde.
if u cant solve it then tell me and i'll do it. i don't have a pen and paper handy ryt now

 

[ahmed abdulla]

can u write it in the format : (for second question)
intial:
difference:
at equilibrim : >?
for first question no idea at all (ATTACHED)