Chemistry: Post your doubts here!

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf

4 (b) (ii)

someone PLEASE help!

Just draw the mirror diagram of the organic compound

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

Question 3 and 26 please someone help !

 

[NinjaInPyjamas]

Just draw the mirror diagram of the organic compound

Not getting it.

 

[sara kamal]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_42.pdf

Please explain Q2 b....i will be very thankful!

 

[NinjaInPyjamas]

okay now question 26 .. hope I'm not disturbing you

A and C get oxidised with nothing getting reduced. D no reaction. In B, the organic reagent gets oxidised and Ag gets reduced, thus answer is B.

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_42.pdf

Please explain Q2 b....i will be very thankful!

See the data booklet for electrode reactions.
Use Cr+3 + e=Cr+2 value is -0.41
Cr2O7+14H+6e.............
Rest just write the reaction.balance by multiplying eq1by 6.Cr+3 on right will add up.

 

[itallion stallion]

Plz help me with these questions
Octobernovember 2009 paper42 q5 part A.
Octobernovember 2009 paper41 q3 bi and bii
October November 2008 paper4 q2 part b.
THANKS!!

 

[Haya Ahmed]

A and C get oxidised with nothing getting reduced. D no reaction. In B, the organic reagent gets oxidised and Ag gets reduced, thus answer is B.

How A is Oxidised ? can you write the equation ... !?

 

[sara kamal]

See the data booklet for electrode reactions.
Use Cr+3 + e=Cr+2 value is -0.41
Cr2O7+14H+6e.............
Rest just write the reaction.balance by multiplying eq1by 6.Cr+3 on right will add up.

i asked for part b, thanks for the reply though

 

[AbbbbY]

any A2 candidates???

Aye!

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf

4 (b) (ii)

someone PLEASE help!

E is CH3COO-
C is CH2OHCH(OH)COOH

4 bii, esterification will happen.
Just make the ester. I can't write it down cos I can't show the branches via typing. If you don't get it, I can draw it and send a pic.

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf


Q3
31
Question 36 ( why 2 is wrong ?)

please help !

 

[AbbbbY]

i asked for part b, thanks for the reply though

pH = pKa + lg([salt]/[acid]) (or pH = pKa - lg([acid]/[salt]) )
5.5 = -lg(1.79 × 10^-5) + lg([Sa]/[A])
5.5 - 4.75 = lg([Sa]/[A])
lg([Sa]/[A]) = 0.75
[Sa]/[A] = 10^0.75
[Sa]/[A]= 5.62
S = 5.62A
A+S = 100
100-A=5.62A
A = 15.1 = 15
So S = 85.

Acid = 15, Salt = 85. Is this the answer?


Edit: Somehow [ S ] wasn't being posted without the gaps so I just changed it to Sa.

 

[ahmed abdulla]

Posting this for more than once
Paper 5 , http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_53.pdf

Question 2b(iii) ( moles )
Question 3 , how to Complete the table when two rows not given


& Whats is the max. Volume a gas syringe can hold which we are supposed to label here ?

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf


Q3
31
Question 36 ( why 2 is wrong ?)

please help !

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

 

[Haya Ahmed]

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

Loads of Thanks

 

[AbbbbY]

Plz help me with these questions
Octobernovember 2009 paper42 q5 part A.
Octobernovember 2009 paper41 q3 bi and bii
October November 2008 paper4 q2 part b.
THANKS!!

ON09 42 Q5a.
I've answered this before a day or two ago but I'll do it again anyway.
Look. This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question. The examiner wants you to analyze those figures, see whats going on, and apply THAT to the question.

The question shows that all carbons in Benzene can be made to lie on the same plane, but not in cyclohexane. The question also shows that butane Carbons can be made to lie on the same plane, but methylpropane, also an alkane, cannot.
This tells us that if a compound has a ring, OR branching, it cannot be made to lie on the same plane.

All of them are co-planer apart from B.

C is coplaner because you don't have to consider the Oxygen atoms. The others are unbranched and two individual chains lying on the same plane.
Get it?

ON 41

Q3bi
[Cu(H2O)6]2+ = Light blue
[Cu(NH3)4(H2O)2]2+ = Deep blue.

You don't even need to know how to analyze absorption spectrums to know this. Remember the Cu ion test in p3?

Q3biii
I'd make a graph between the two shown. Meaning, with a lower amplitude than (NH3)4 but higher than (H2O)6 and with the maximum point around 700nm.


ON 08
2b:
1 1 0
1 1 1
1 2 2

I won't lie. I got this question wrong too when I attempted it back in my mocks. Asked a teacher to explain it to me.
Basically, you have to consider ALL steps above the slowest step identified.

As in, if you're considering step 3 to be the slowest, you have to use the values from step 2 and 1 as well. If you're considering step 2 slowest, you've to use values from step 1 as well and so forth. I still can't come up with a logical answer as to why we do this. Maybe someone else can help clarify. Namehere

 

[sitooon]

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

Abby can you answer the question of paper5 posted ab ahmed abdullah

 

[AbbbbY]

Abby can you answer the question of paper5 posted ab ahmed abdullah

Will do in a lil while. A bit stuck up + I loathe P5's

 

[sara kamal]

pH = pKa + lg([salt]/[acid]) (or pH = pKa - lg([acid]/[salt]) )
5.5 = -lg(1.79 × 10^-5) + lg([Sa]/[A])
5.5 - 4.75 = lg([Sa]/[A])
lg([Sa]/[A]) = 0.75
[Sa]/[A] = 10^0.75
[Sa]/[A]= 5.62
S = 5.62A
A+S = 100
100-A=5.62A
A = 15.1 = 15
So S = 85.

Acid = 15, Salt = 85. Is this the answer?


Edit: Somehow [ S ] wasn't being posted without the gaps so I just changed it to Sa.

Jazak Allah khair for yr help,May Allah grant you with good grades !