Chemistry: Post your doubts here!

[salvatore]

3c) You'll have to draw a double hump diagram for this. It's essentially the same as the simpler energy diagram except we break the single hump into 2 to represent the intermediate steps. There should be two peaks and the first one should be higher than the second. If you recall free radical substitution mechanism you know the intermediate steps. The first hump represents the formation of methyl free radical and HCl and the second represents the chloromethane formation. The reason the second hump is lower is because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one. The activation energy is labeled for the first hump.

4c) Ok, the final product is a ketone. Ketones come from the oxidation of secondary alcohols, so the intermediate must be a secondary alcohol. As you start with an Alkene you have to introduce an -OH group into the molecule so you use steam with conc. H3PO4. The result is CH3CH2CH(OH)CH3.
(For solving synthetic routes it helps if you go backwards, try thinking of the source of the product)

Hope this helps!

Thank a lot.. much appreciated

 

[Ahmad Zaeem]

May June 2012 ....Paper 42 ...Q 1 (c)(1)......What is the concept here ...????...why are we subtracting (Delta H formation of MgCl2(s) and MgCl2(aq) ) to find (deltaH sol).....????? http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

 

[littlecloud11]

Thank a lot.. much appreciated

No problem

 

[littlecloud11]

May June 2012 ....Paper 42 ...Q 1 (c)(1)......What is the concept here ...????...why are we subtracting (Delta H formation of MgCl2(s) and MgCl2(aq) ) to find (deltaH sol).....????? http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

Enthalpy change of solution is essentially the energy change when an ionic solid dissolves and that's exactly the change here. The state changes from solid to a (aq) so if you just find the energy change between these two you get your answer. You could have done it the usual hess's cycle way but the hydration energy for Cl- was not given so you have to use the direct route.

 

[Ahmad Zaeem]

littlecloud11...

Enthalpy change of solution is essentially the energy change when an ionic solid dissolves and that's exactly the change here. The state changes from solid to a (aq) so if you just find the energy change between these two you get your answer. You could have done it the usual hess's cycle way but the hydration energy for Cl- was not given so you have to use the direct route.​

...Thank you !!!.....well after you get the concrpt ...it seems really easy ......lol

​

 

[sweetjinnah]

V is CH3CH2CHOHCH2OH because cold KMnO4 acts on it remember the reaction of alkenes where the double bond breaks to give and two OH attach ! thats what happens

W is CH3CH2COCHO this because now the oxidizing agent causes V having at one end primary alcohol and in the centre the secondary alcohol to oxidize and you know the products are aldehyde and ketone respectively

So nice of u, thanks alot!

 

[Twinks05]

Can anyone tell me how the papers are going to be marked (their percentages)?
I want to know how much each paper weighs, and how CIE will add AS and A level marks together, so I can estimate my marks.
Please can anyone tell me, how the A2 papers weigh, and how the final mark will be given?
Please&Thank you =)

 

[Muhammad Bin Anis]

Can anyone tell me how the papers are going to be marked (their percentages)?
I want to know how much each paper weighs, and how CIE will add AS and A level marks together, so I can estimate my marks.
Please can anyone tell me, how the A2 papers weigh, and how the final mark will be given?
Please&Thank you =)

check out the syllabus and grade threshold of the relevant year

 

[tanmaydube]

When 0.60 moles of H2 (g) and 0.18 moles of I2 (g) were heated to a constant temperature in a sealed container with a volume of 1 dm3, an equilibrium was set up
H2(g) + I2(g) ⇄ 2HI(g)
At equilibrium, 0.16 mol I2(g) had reacted.
(a)(i) Determine the equilibrium concentrations of H2, I2 and HI
(ii) Calculate the equilibrium constant Kc, for this reaction. (Ans 11.6)

Please help thank you!

 

[Ahmedraza73]

I need Unsolved AS Level Chemistry Past papers?
Can Someone Help me?

 

[aalmuhannadi]

How would you determine the change in pressure in this question?

 

[Twinks05]

When 0.60 moles of H2 (g) and 0.18 moles of I2 (g) were heated to a constant temperature in a sealed container with a volume of 1 dm3, an equilibrium was set up
H2(g) + I2(g) ⇄ 2HI(g)
At equilibrium, 0.16 mol I2(g) had reacted.
(a)(i) Determine the equilibrium concentrations of H2, I2 and HI
(ii) Calculate the equilibrium constant Kc, for this reaction. (Ans 11.6)

Please help thank you!

1. Moles before equilibrium:
H2 - 0.60
I2 - 0.18
HI - 0
Moles at equilibrium:
I2 - 0.18 - 0.16 = 0.02 (since 0.16 have reacted, the number of moles of I2 remaining will be the difference between .18 and .16 moles.)
H2 - 0.60 - 0.16 = 0.44 (ratio of I2 to H2 is 1:1 so if 0.16 moles of I2 was used up, then 0.16 mole of H2 was also used up.)
HI - 2 * 0.16 = 0.32 (originally HI was 0 so at equilibrium its mole will be .16 * 2 since ratio of I2 to HI is 1:2. so if .16 moles of I2 was used up to produce the product, then .32 mole of HI will be produced.)

Hence concentrations at equilibrium are:
H2: .44mol/1dm^3 = .44 molperdm3 (A)
I2: .02mol/1dm^3 = .02 molperdm3 (A)
HI: .32mol/1dm^3 = .32molperdm3 (A)

2. Kc = [HI]^2 / [H2][I2]
Kc = .32^2 / .44 * .02
Kc = 11.6 (A)

Hope this helped.

 

[tanmaydube]

OMG!!!! Thanks a lot Twinks05 !!

 

[strangerss]

hey everyone , I have a doubt , well actually lots of doubts in MAY/JUNE 2009 P1 . In question 2 how is the answer D? I thought it would be B because the number of moles of oxygen is equal to the number of moles of sodium chlorate so 0.0350 multiplied with 25.0/1000. So how come it's wrong? , any help please??

 

[littlecloud11]

How would you determine the change in pressure in this question?
View attachment 21673

The answer would be D. Glucose and sucrose have different chiral centers so can't be optical. They have the same molecular formula but different structures hence are structural isomers and you can eliminate A and C. 'n' represents the number of solute molecules in the solution, one molecule of sucrose is hydrolyzed to give two molecules of solute. So, 'n' increases, 'V' 'R' 'T' are constant, so as 'n' increases the pressure increases.

 

[littlecloud11]

hey everyone , I have a doubt , well actually lots of doubts in MAY/JUNE 2009 P1 . In question 2 how is the answer D? I thought it would be B because the number of moles of oxygen is equal to the number of moles of sodium chlorate so 0.0350 multiplied with 25.0/1000. So how come it's wrong? , any help please??

Concentration is mol PER dm^3. You divide the number of moles (.035) by .025 dm^3 (25/1000= .025dm^3) not multiply it.

 

[aalmuhannadi]

The answer would be D. Glucose and sucrose have different chiral centers so can't be optical. They have the same molecular formula but different structures hence are structural isomers and you can eliminate A and C. 'n' represents the number of solute molecules in the solution, one molecule of sucrose is hydrolyzed to give two molecules of solute. So, 'n' increases, 'V' 'R' 'T' are constant, so as 'n' increases the pressure increases.

But the answer is B for some reason? :/ EDIT: oh no sorry you're right it's D, thanks!!

 

[strangerss]

Concentration is mol PER dm^3. You divide the number of moles (.035) by .025 dm^3 (25/1000= .025dm^3) not multiply it.

Oh my GOD , what a stupid mistake I'm such an idiot! thanks a lot for the help

 

[strangerss]

Hy every one there a question in p1 May/June 2009 question 19 I don't understand isn't a chiral center supposed to be attached to four different groups of atoms , but most of the carbon atoms are bonded to other carbon atoms :S

 

[PhyZac]

Hy every one there a question in p1 May/June 2009 question 19 I don't understand isn't a chiral center supposed to be attached to four different groups of atoms , but most of the carbon atoms are bonded to other carbon atoms :S

This is nicely explain in this link http://www.chemguide.co.uk/basicorg/isomerism/optical.html