Chemistry: Post your doubts here!

[darks]

View attachment 62435
9701/12/M/J/12

Solve for each equation.
I will only show for D, which is the right answer.
Initial moles of P = 2 mol
mol at equilibrium of R = x
so for equation D,
mol at equilibrium of P = 2 - x
mol at equilibrium of Q = x/2
Sum of all mol at equilibrium expressions
x+(2-x)+(x/2)
Simplifies to [2+(x/2)] which should be the total moles according to the question.

 

[Holmes]

Go to the data booklet, there you'll find 4. Standard Electrode potential ................................, there you'll find the equation for oxidation of Sn2+ to Sn4+ and reduction of KMNO4, just balance them so that the mole ratio of electrons is the same on both sides and then add the two equations, the electrons will cancel out to give you the equation, and then you know the rest!

Appreciations and a thunderous Applause for you.
Thank you teacher.

 

[Mstudent]

Thanks!

March 2017 P1 for you all

Thanks bro!

 

[Mstudent]

Appreciations and a thunderous Applause for you.
Thank you teacher.

Yo are very welcome, anytime

 

[Holmes]

N-N bond breaks as it is very weak (energy 160 Kj/mol). NF2 is the single product. Statement 1 is incorrect. Statement 2 is correct as 160 is the energy of the bond that breaks. Molecules non linear as a lone pair present on N of NF2

Thanks bro

 

[Holmes]

Solve for each equation.
I will only show for D, which is the right answer.
Initial moles of P = 2 mol
mol at equilibrium of R = x
so for equation D,
mol at equilibrium of P = 2 - x
mol at equilibrium of Q = x/2
Sum of all mol at equilibrium expressions
x+(2-x)+(x/2)
Simplifies to [2+(x/2)] which should be the total moles according to the question.

can you plz be more clear

 

[Metanoia]

could you plz explain why you left out the fifth carbon?

That would create an identical structure as the 2nd structure (imagine rotating the 2nd structure 120 degrees counter-clockwise).

 

[Metanoia]

Q30 View attachment 62417

Oh, I have left out the other part of the information. There is indeed a chiral carbon in Option D, but option D is a "Cis" isomer (check the C=C).

Option C is a trans-isomer. The question is looking for an optically active trans-isomer

 

[Metanoia]

Holmes
For papers from 2013 and 2012, I have created some videos to explain the MCQs. You can search through the playlists

 

[Metanoia]

A.
Why is 3 correct?

Increasing the pressure also increases the concentration of the gaseous reactants, therefore the rate of backward is also expected to increase.

However, the equilibrium will still shift to the right as the increase in the rate of the forward reaction is more than the increase in the rate of backward reaction.

 

[Metanoia]

C

I can't see the right side of the question, should I assume that the charge is -3?

Sodium metal will react with carboxyl groups COOH and hydroxyl OH groups to form COO- and O-. So if we have a total of 3 such groups, it will form a -3 charge

Sodium hydroxide will react with COOH and not OH

 

[amina1300]

M2017 p12 Q 22 anyone? http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_m17_qp_12.pdf
Ans A can anyone explain why isnt it the other options??


Q27 ??????????
Q 10 19
HELP!!

 

[Holmes]

Holmes
For papers from 2013 and 2012, I have created some videos to explain the MCQs. You can search through the playlists

What paper is this??
Variant???

 

[amina1300]

What paper is this??
Variant???

June 2013, Paper 11

 

[amina1300]

can you plz be more clear

A- Inital: 2 : 0 : 0
Final : 2-x: 2x : x
Total moles: 2-2+2x+x = 3x so A is incorrect.

B-
Initial: 2 : 0 : 0
Final: 2-2x : 2x : x
Total moles: 2-2x+2x+x = 2+x so B is incorrect

C-
Initial: 2 : 0 : 0
Final 2-2x:x:x
Total = 2-2x+x+x = 2 incorrect

D-
Initial: 2 : 0 : 0
Final: 2-x : x/2 : x [x = 2R]
Total moles = 2-x+x/2+x =2+x/2 mols!

 

[darks]

M2017 p12 Q 22 anyone? http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_m17_qp_12.pdf
Ans A can anyone explain why isnt it the other options??
Q26 Why isnt it A? doesnt Kmno4 oxidise secondary alcohol to ketone??

Q27 ??????????
Q 28 29
Metanoia
Holmes
Mstudent
HELP!!

Q22) C-Br bond longest and weakest (Br atomic radius largest) so C-Br bond will be the first to undergo homolytic fission.

Q26) You said it yourself, "ketone"... The product shown in A is not ketone, but an aldehyde. Ketone would be CH3COCH3.

 

[amina1300]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s16_qp_13.pdf
Q30, ???
Q 33 What does statement 2 mean?
Q24 How can D be correct? Doesnt hot kmno4 oxidise to aldehydes and cooh?
Q23 Hiw do v know what radicals form?
q 21 plz show diagram..while explaining
Q14 ??

 

[amina1300]

Q22) C-Br bond longest and weakest (Br atomic radius largest) so C-Br bond will be the first to undergo homolytic fission.

Q26) You said it yourself, "ketone"... The product shown in A is not ketone, but an aldehyde. Ketone would be CH3COCH3.

THANKYOUU n SRRY ABOUT Q 26 THIS IS WHEN I SOLVE A Q.P AT 3:00 AM.
CAN U PLZZ EXplain Q27??
if u hv time 19 n 10 too?

 

[darks]

THANKYOUU n SRRY ABOUT Q 26 THIS IS WHEN I SOLVE A Q.P AT 3:00 AM.
CAN U PLZZ EXplain Q27??
if u hv time 19 n 10 too?

Q 27)
5 possible oxidation products will contain groups this way..
acid & alcohol
acid & acid
acid and aldehyde
alcohol and aldehyde
aldehyde and aldehyde
This equals to acid and aldehyde both being used 4 times each.

Q10) As the question states Mn2+ (product) Catalyses the reaction, so rate gets higher as reaction proceeds. So faster decrease in MnO4-(reactant) conc. with time. So ans B. At the end, the graph is less steep as the reactant conc. very low so less collisions.
Sorry for brief explanations.. all the rest i can explain tomorrow.. i have bio p1 tomorrow.

 

[Metanoia]

What paper is this??
Variant???

You can check my youtube playlist, there are a few years available.