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Chemistry: Post your doubts here!

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Someone help me please for these 3 questions from june 02
ThanksView attachment 15254
For question 23, i would have gone for C because the conversion of 1st compound to the second compound occurs as the C--------------Br and C-OH. you will need to add a further carbon atom, which is achieved by the CN. and finally convert to carbocylic acid by hydrolysis.

For question 24 i would have gone for A because the complete combustion of C is CO2 not CO and NO is not a result for incomplete combustion of Hydrocarbon.

For the last question i would have gone for A as the central carbon would bear 4 different constituents upon reaction with CN-

Hope my replies helped. :)
 
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For question 23, i would have gone for C because the conversion of 1st compound to the second compound occurs as the C--------------Br and C-OH. you will need to add a further carbon atom, which is achieved by the CN. and finally convert to carbocylic acid by hydrolysis.

For question 24 i would have gone for A because the complete combustion of C is CO2 not CO and NO is not a result for incomplete combustion of Hydrocarbon.

For the last question i would have gone for A as the central carbon would bear 4 different constituents upon reaction with CN-

Hope my replies helped. :)

Thank you very much Amy Bloom ! That's sooo humble from you :)
 
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Can anybody help me with this. i need an answer by tomorrow.
 

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Can anybody help me with this. i need an answer by tomorrow.

PART A: NaOH and HCl are in 1:1 ration hence you calculate moles for HCl and use it to find the conc of NaOH
so for HCl, n = (0.1 x 25.45)/1000 = 2.55 x 10 ^-3
so for NaOH, 2.55 x 10 ^-3 = ( c x 25)/1000 = 0.102 moldm-3

PART B: H+ and OH- in 1:1 ratio
so for HCl, n = (0.1 x 17.3)/1000 = 1.73 x 10 ^-3
so for OH-, 1.73 x 10^-3 = ( c x 25)/1000 = 0.0692 = 6.92 x 10 ^-2 moldm-3

PART C:

PART D: (total OH- conc calcualted in B) - (conc of OH- from NaOH calculated in C)-----> (6.92 x 10 ^-2) - (0.051)
= 0.0182 moldm-3

PART E:

PART F: Ksp = [Ca2+] [OH-]^2
= (9.1 x 10 ^-3) (6.92 x 10 ^-2)^2
= 4.36 x 10 ^-5 mol3 dm-9

sorry couldn't do c and e :unsure: but i have used their answers for the next qns.

Hope it helped inshaAllah:)
 
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PART A: NaOH and HCl are in 1:1 ration hence you calculate moles for HCl and use it to find the conc of NaOH
so for HCl, n = (0.1 x 25.45)/1000 = 2.55 x 10 ^-3
so for NaOH, 2.55 x 10 ^-3 = ( c x 25)/1000 = 0.102 moldm-3

PART B: H+ and OH- in 1:1 ratio
so for HCl, n = (0.1 x 17.3)/1000 = 1.73 x 10 ^-3
so for OH-, 1.73 x 10^-3 = ( c x 25)/1000 = 0.0692 = 6.92 x 10 ^-2 moldm-3

PART C:

PART D: (total OH- conc calcualted in B) - (conc of OH- from NaOH calculated in C)-----> (6.92 x 10 ^-2) - (0.051)
= 0.0182 moldm-3

PART E:

PART F: Ksp = [Ca2+] [OH-]^2
= (9.1 x 10 ^-3) (6.92 x 10 ^-2)^2
= 4.36 x 10 ^-5 mol3 dm-9

sorry couldn't do c and e :unsure: but i have used their answers for the next qns.

Hope it helped inshaAllah:)
Ameen. Yeah thanks mate, at least for trying. the parts for which i was stuck you solved it. thanks loads.
 
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hey so i have a question here of november 2007 question 4 part b
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf

can anyone explain me why in part(a) the sign is negative
and in part(b) why we are not taking the value of cl2
thankyou
part a.) sign is negative coz bond formation is exothermic and releases energy, hence the negative sign
part b.) we use hess's law so basically we have the 2 enthalpy of formation for the PbCl2 and PbCl4 so we make and equation
so we get a triangle kind of thing,

Screen shot 2012-09-18 at 2.06.45 AM.png

soooo, you follow the directions of the arrows, arrows in one direction, you add 'em, those in diff directions, you put 'em on the other side of the equal sign:)
-359 + x = -329
hence, x = 30
Hope that helped:) :)
 
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Does anyone have or know where I can find lots of organic chemistry worksheets? Need it Urgently!!
 
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part a.) sign is negative coz bond formation is exothermic and releases energy, hence the negative sign
part b.) we use hess's law so basically we have the 2 enthalpy of formation for the PbCl2 and PbCl4 so we make and equation
so we get a triangle kind of thing,

View attachment 15488

soooo, you follow the directions of the arrows, arrows in one direction, you add 'em, those in diff directions, you put 'em on the other side of the equal sign:)
-359 + x = -329
hence, x = 30
Hope that helped:) :)

I don't know why ! But I never seemed to understand chemical energetics with cycles, etc... ! pffffffffff :'(
 
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