Chemistry: Post your doubts here!

[zackle09]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

how do you do question 17 and 23?

 

[ahmed abdulla]

intermediate forms only in teritary halogenoalkaens and in electrophilic addition.. intermediate forms in SN1 mechanism ONLY , not SN2

whats the linkage between intermediate and the graph
i dont see the word intermediate in the question
how did u now the graph codes for intermediate rn

 

[h4rriet]

I'm sorry, but I still don't understand q.6, q.29 why is c not acidic(there is co2h bonded to it).

No, that's a -CO2C-, not -CO2H. It's an ester bond.
For 6, all you gotta do is put the numbers they've given you into the formula. Also, one thing: V is found using mass/density.

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

how do you do question 17 and 23?

17: i didn't work out the whole thing but after a look i gues this is how u do it. calculate the no. of mole of O2 then write up an imaginative eq like M instead of the metal identity bcz u dont know it. prob put in the values of each metal in the option and get the answre from the molar ratio

 

[Aymen Ezazi]

whats the linkage between intermediate and the graph
i dont see the word intermediate in the question
how did u now the graph codes for intermediate rn

yes they did not , but one of my friend who has already done A'levels and is in a uni now, expalined this.. that when and intermediate is formed the the energy decreases , one question like this came even in paper 2 with the same graph and had the intermediate mentioned..

 

[MONJUR HASAN NISHAD]

Okay chem dudes, I need some serious help! In my chem prac exam (p34) 2nd question my titration reading was 30.5 and our supervisor got 20.5! How much will I lose in that 2nd question??? In the first question, my point of intersection of that graph was 9.5 but the supervisor got 15. However I think two or three of my temperature readings were close to the supervisor. So how much will I lose in the first question??? Please help! I am having a heart attack here!

 

[MONJUR HASAN NISHAD]

One more query, how much overall marks do I need at AS to get 90th percentile? A is 80th percentile and you calculate the 90th percentile by this formula gt(for A) + gt(for A) - gt(for B)

 

[MONJUR HASAN NISHAD]

I am dying here!!! Someone tell me how much I am gonna lose in prac before it's too late!!!!!!

 

[bravegirl123]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
In Q5, isnt part D also showing a pi bond?

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
In Q5, isnt part D also showing a pi bond?

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.

 

[unseen95]

t

2) Ok, the word to emphasize on is RATE of reaction. Increasing the surface area of catalyst, increasing the temp and increasing the pressure all increases the RATE of reaction for the forward reaction but effects the YIELD of forward reaction differently. Increasing the catalyst SA means more molecules of the reactant can bind to the catalyst hence the rate increases, increasing temp means the reactant molecules have more energy, so the rate of successful collision increases and the rate increases (the yield of forward reaction would decrease because the forward reaction is exo. and the equilibrium will shift to the left), increase in pressure also increases the rate of forward reaction because there are greater no. of moles of reactant on the left. Only D, which involves removing the ammonia as it is formed does not effect the rate, but only effects the yield. So the answer is D.

14) the equation would be-
3 Ba(NO3)2 + 10 Al -------> 5Al2O3 + 3BaO + 3N2

Mr of Ba(NO3)2 = 137 + 14*2 + 16* 6 = 261
looking at the stoichiometry of the reaction, 3mols Ba(NO3)2 gives 3 mols N2. So-
3*261 g Ba(NO3)2 = 3*24 dm^3 N2
.783 g Ba(NO3)2 = .072 dm^3 N2 = 72 cm^3 N2
Your answer is B.

Gemeaux solved 24.

thanks this answer helped me a lot

 

[Raiyan3]

Help me with question number 11 please
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
And question number 9
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf

 

[A star]

Help me with question number 11 please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
And question number 9
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf

11C

 

[felicia tan]

37 Which compounds may result from mixing ethane and chlorine in the presence of sunlight?
1 CH3CH2Cl
2 CH3CH2CH2CH3
3 CH3CHCl CHCl CH3

Can anyone explain to me why is the answer A (1,2 and 3 is correct)

 

[Raiyan3]

11C

Why is it C? please explain bro

 

[A star]

Why is it C? please explain bro

it shows 0.25 mole of X are reduced and acording to molar ratio Y should be increased by (2 times the value of the the amount of X converted ) hence it had to be C or D now put these values in the equation Kc=(0.5)^2/0.25 you will get 1 hence the answer is C

 

[A star]

37 Which compounds may result from mixing ethane and chlorine in the presence of sunlight?
1 CH3CH2Cl
2 CH3CH2CH2CH3
3 CH3CHCl CHCl CH3

Can anyone explain to me why is the answer A (1,2 and 3 is correct)

1 is simple momo substitution
2) two ethane radicals have combined(termination step)
3) two chloroethane radicals combined(termination step)

 

[A star]

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.

awesome bro didnt think about it that way :/

 

[felicia tan]

1 is simple momo substitution
2) two ethane radicals have combined(termination step)
3) two chloroethane radicals combined(termination step)

thank you!

 

[A star]

thank you!

welcome