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Chemistry: Post your doubts here!

B

bravegirl123

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AbbbbY said:
The answer should be B.

D is showing the overlap of four orbitals, and is hence showing a square-planer arrangement. (Hint hint: XeF4). We all know that XeF4 contains 4 sigma bonds, regardless of what the diagram is showing. Hence, D is not the correct answer, rather a smartly made confusing option.
Click to expand...

i dont get it
 
S

sara kamal

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Aymen Ezazi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
guys , i need help with Q2 and 27.. thnx a lot !!!!
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Q2
first find mass of nitrogen in 14g of fertilizer by
15/100 x 14=2.1g
then find mole=2.1/14(Mr or nitrogen)=0.15mol
then since con=mol/vol=0.15/5=0.03mol/dm3


Q27
all of them wud have had molecular formula C10H14O
so nw read second statement that its unreactive towards mild oxidizing agents...which clearly tells that X is tertiary alcohol
so the only possibility is D,since if u again break the bonds to attach H and OH,the OH can be attached to carbon atom which in turn is attach to 3 other carbon atoms.


hope it helped!
 
R

rawaaaan

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please answer this questionnss pleaseee
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf question31
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w03_qp_1.pdf question 2,23 and 31
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf question 9 and 18
and the last question is http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf question 7
pleasee atleast answer what u know
 
R

rawaaaan

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ques
Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Q3 , 10, 13, 28,31, 38
please
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question 10
at initial stage pressure 1 atm
volume is 100%
at equilibrium h20-->80%
h2-->20%
02-->10%
total------>110%=1.1
partial pressure h20-->80%/110% *1=0.8/1.1
h2-->20%/110% *1=0.02/1.1
02-->10%11% *1=0.1/1.1


question 31 1)is wright because u should watch out how the arrows are going so u will see from u-->R u go from U to S to R so its like -(92)-(-134) u get +42
2) twi is wrong because when u go from t-->s -75-(92)=-167 and they said its endothermic which should be positive ,so its wrong
3)r to t (-134)+(92)-(-75)=33 and in the question its with negative so ofc its wrong
 
R

rawaaaan

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Aymen Ezazi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
2 , 11 , 30 , 39 plzz..
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question 2 10cm3 of cs2 and 30cm3 of 3o2 ,they r equal to the number of moles ,and 10cm3 of co2,and 20 of so2
then they said 50 cm3 of oxygen but only 30 reacted so 20 are left ,this 20cm3 of oxygen left u add them with 10cm3 and 20cm3 of co2 and so2 so u get 50 sm3 of gas after burning
and for the volume of gas after adding naoh im not sure but i guess because of when naoh rects with the products it gives 2NAOH +SO2+CO2+NA2CO3+H2CO3 SO2 and so2 and co2 reacts so 20 is left from naoh

question 30
number of moles of ethanol is 0.65 and for ethanoic acid is 0.5
and the mass for ch3co2ch2ch3 is 0.5*82=44
they want the yield of ester so the mass of ethanoic acid 22/44*100
 
Aymen Ezazi

Aymen Ezazi

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rawaaaan said:
please answer this questionnss pleaseee
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf question31
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf question 2,23 and 31
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf question 9 and 18
and the last question is http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf question 7
pleasee atleast answer what u know
Click to expand...


Nov 2005-- only one bcoz any molecule can contain isotopes so only radius will be the same in all isotopes and nucleon no: and relative isotopic mass will be different..

Oct 2003-- Q23 , its not A bcoz bromine would not react with this alkane bcoz it is saturated.. not B bcoz Cl would substitute Br which is attached to the third group and the removed Br would not take place of any H of the first carbon...not D bcoz we dont need Cl on the first carbon.. so its C ..

june 2004-- Q9 they asked for same conc. of hydrogen as HCL which is a strong acid.. its not A bcoz A is a weak acid , not C bcoz its an alkali so no H+ ions.. now the formula for sulphuric acid is H2so4 so its contains 2 H+ ions and HCL has only one H+ ion.. so it will be nitric acid HNO3 containing one H+ ion..
 
Aymen Ezazi

Aymen Ezazi

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rawaaaan said:
question 2 10cm3 of cs2 and 30cm3 of 3o2 ,they r equal to the number of moles ,and 10cm3 of co2,and 20 of so2
then they said 50 cm3 of oxygen but only 30 reacted so 20 are left ,this 20cm3 of oxygen left u add them with 10cm3 and 20cm3 of co2 and so2 so u get 50 sm3 of gas after burning
and for the volume of gas after adding naoh im not sure but i guess because of when naoh rects with the products it gives 2NAOH +SO2+CO2+NA2CO3+H2CO3 SO2 and so2 and co2 reacts so 20 is left from naoh

question 30
number of moles of ethanol is 0.65 and for ethanoic acid is 0.5
and the mass for ch3co2ch2ch3 is 0.5*82=44
they want the yield of ester so the mass of ethanoic acid 22/44*100
Click to expand...


in Q2 there is 50 cm^3 of cs2 not 30.. thn how will we do it??

and in q 30 why did we multiply 0.5 into 30??
 
biba

biba

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

biba

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Dr.MMM said:
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
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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!
 
cute97

cute97

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strangerss said:
question 9 , I don't know:eek: 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:)
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Noo :( Q19 (B) and Q28 (A) according to the mark scheme
 
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