Chemistry: Post your doubts here!

[♣♠ Magnanimous ♣♠]

explain me ques 5 and 6

 

[ZaqZainab]

w07qp1

Qn 9 :
moles of SO3 2- : moles of electrons : moles of metal
0.0025 : ? : 0. 005
1 :? : 2
1 : 2 : 2

Q14. IE should be decreasing down the Group.
2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q26. We use HCN (with trace NaCN) to step up carbonyls. We use NaCN or KCN to step up halogen alkanes.

Q19. Put the iodine atoms on 1st and 2nd carbon. CH2ICHICH3, chiral carbon is the middle carbon.

Q28. Ketones and aldehydes can be reduced by H2 to alcohols. Only COOH wouldn't be reduced by H2.

Q32. Burns readily means CO --> CO2 exothermic.
1. CO2 more stable than CO in terms of energy
2. as explained, exothermic reaction
3. ratio of CO2 is high compared to CO, therefore Kc is high

Q33.
Statement 1: (False) The gas is exerting the same pressure as the atmosphere outside, otherwise the plunger will either move left or right.
Statement 2: (True) When plunger moves inwards, it causes the pressure to increase, the eqm will shift to the left to reduce pressure (produce less moles of gas).
Statement 3: (True) For every one mole of PCl5 dissociated, two moles of gases were formed (1 mol of PCl3 and 1 mol of Cl2). So total volume of gases after dissociation is more than original volume of gas without dissociation.

Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2

Thanks
i am still confused about
Q9 how did you find the moles? shouldn't metallic salt be 0.10/(50*10^-3)=2??
SO3 2- be 0.10/(25*10^-3)=4??

14: Ionisation energy decreases down the group.
19: The molecule is CH3-CHI-CH2I
26: Halogenalkane react with NaCN not HCN
28: H2 in presence of Ni catalyst also reduces aldehyde to alcohol.
32: As CO burns readily to form CO2 so CO2 is more stable in which carbon is+4.
Formation of CO2 is formation of CO and also combustion of CO so combustion of CO is negative so formation of CO2 is more negative.
As more CO2 is more produced so equilibrium constant is high.
33: More moles of gas are produced so one mole occupies 24dm^3 so more volume.
37: Replace D with H and then look at the reactions.

Thanks
Q19 why not
CH3-CH2-CHI2??

 

[Metanoia]

Thanks
i am still confused about
Q9 how did you find the moles? shouldn't metallic salt be 0.10/(50*10^-3)=2??
SO3 2- be 0.10/(25*10^-3)=4??


Thanks
Q19 why not
CH3-CH2-CHI2??

Moles = conc x volume

you did it as moles = conc/volume ?

 

[Metanoia]

View attachment 44961 explain me ques 5 and 6

Q5. Fix P4O10 as 1 mole

P4O10 + 6CaO + --> 2Ca3(PO4)2

Q6.
Just use proportion if its easier to visualise.

mass of N in 100 g of fertiliser = 15 g

mass of N in 14 g of fertiliser = (15/100) x 14= 2.1 g

moles of N in 14 g of fertiliser = 2.1 /14 = 0.15 mol

conc of N in 5 dm^3 of solution = 0.15/5 = 0.03 mol/dm^3

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q4 D is the answer
Q8 D is the answer what about A? PV=nRT n and R and T constant so P=1/V so P increases V decreases? and i do understand why D is correct but why is A wrong?
Q22 i have a problem with angles any notes will do?
Q26 C or B isnt it the same thing C is the actual answer
Q29 what about C? D is the answer
Q38 why int 2 correct
Q39 it is a primary alcohol so shouldn't it be forming an aldehyde

 

[ZaqZainab]

Moles = conc x volume

you did it as moles = conc/volume ?

my bad Thanks again

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q4 D is the answer
Q8 D is the answer what about A? PV=nRT n and R and T constant so P=1/V so P increases V decreases? and i do understand why D is correct but why is A wrong?
Q22 i have a problem with angles any notes will do?
Q26 C or B isnt it the same thing C is the actual answer
Q29 what about C? D is the answer
Q38 why int 2 correct
Q39 it is a primary alcohol so shouldn't it be forming an aldehyde

s08qp1

Q4. BaO2 anion is two oxygen atoms with a total of -2 charge, hard to type out......(O2)2-
We have one single bond among the two atoms and 1 extra electron to each atom.

Q8. If you link it to y = 1/x , you should remember its a curve, with the axis as asymptotes.

Q22. The carbons are all saturated with 4 single bonds, so tetrahedral, 109.5

Q26.
Increase in the 2 oxygen means that the 2 OH has been oxidized to 2 COOH groups.
The two OH must be primary alcohols. Since is an unbranched butanol, the OH groups are on carbon 1 and 4.
If you choose B, the OH attached to the 3rd carbon becomes a ketone instead of the required acid.

Q29. C is ester group, not acid group.

Q38. hmm.. maybe you can explain why you think 2 is correct?

Q39. Its a secondary alcohol.

 

[Ahmed Aqdam]

Thanks
i am still confused about
Q9 how did you find the moles? shouldn't metallic salt be 0.10/(50*10^-3)=2??
SO3 2- be 0.10/(25*10^-3)=4??


Thanks
Q19 why not
CH3-CH2-CHI2??

The second carbon has two H which are same and last one has two I which are same so not optical isomer.

 

[sadiaali]

Please anyone Q 4.Thanks a lot for giving your time explaining me.
Link:http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w13_qp_13.pdf
MS: http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w13_ms_13.pdf

 

[Ahmed Aqdam]

Please anyone Q 4.Thanks a lot for giving your time explaining me.
Link:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf
MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_ms_13.pdf

X will be in Group 7 as it forms X- ions and can be oxidised. Option D has ionisation energies of group 7 element (iodine) so it will be D.

 

[Suchal Riaz]

If anyone has any problem ask me tag me 'suchal riaz' or text me if i am not available.

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

can anyone please explain Q27

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

can anyone please explain Q27

it has a chiral carbon. so think of a single center carbon with one H attached on OH attached and two other groups attached. to make it chiral these two must be difficult. to make n smallest one must be CH3 and other must be C2H5

 

[MiniSacBall]

it has a chiral carbon. so think of a single center carbon with one H attached on OH attached and two other groups attached. to make it chiral these two must be difficult. to make n smallest one must be CH3 and other must be C2H5

Well sir you have got this one wrong, the question says that it doesn't react with hot,acidified KMnO4. Thus, it has to be a tertiary alcohol.
So to make n smallest i has to CH3, C2H5 and C3H7.
So it would be C

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

can anyone please explain Q27

I have answered above which is C

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

Q3,31,

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
Q8
B and D the same thing?

 

[Ahmed Aqdam]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
Q8
B and D the same thing?

Combustion of PbO is when it reacts with O2. This is not the combustion of Pb but PbO.
Formation of PbO is when Pb and O2 react to form PbO.

 

[Suchal Riaz]

Well sir you have got this one wrong, the question says that it doesn't react with hot,acidified KMnO4. Thus, it has to be a tertiary alcohol.
So to make n smallest i has to CH3, C2H5 and C3H7.
So it would be C



I have answered above which is C

i didn't read that part. sorry. i was watching pokemon trying to resurrect my sad old memories :'(

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

Q3

moles of TlNO₃ = (10 × 0.3)/1000 = 0.003 moles
moles of NH₄VO₃ = (20 × 0.1)/1000 = 0.002 moles

So it takes 0.002 moles of V to oxidize Tl(+1) to Tl(+3)
or it takes 0.003 moles of Tl to reduce V(+5) to V(unknown)
Now cross multiply
V(5 - unkown) * 0.002 = 0.003 * Tl(3 - 1)
V(5 - unkown) = (2 * 0.003)/0.002
V(5 - unkown) = 3

So change in oxidation state is - 3
Anwer: B

 

[..sacrifice4Revenge..]

What factor decides that in finding enthalpy change of a reaction do we need the formation data or combustion data?!