Chemistry: Post your doubts here!

[papajohn]

It should be D. Why not D
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q3 Why not not DDDDDD!

Please Zaq help

 

[ZaqZainab]

It should be D. Why not D
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q3 Why not not DDDDDD!

if you notice it is in the 6th group and 2nd period
ther is only 1 element that is O(8)
1s2 2s2 2p3

Can you tell me how you think it is D?

 

[not.maria]

we

It should be D. Why not D
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q3 Why not not DDDDDD!

for the first four IE the difference is roughly 2000
between the 4th and 5th there is a difference of 3550.but the principal quantum number is still the same i.e the principal quantum no is 2.
the distance between 2p and 2s is small.while distance between 2s and 1s is large

 

[papajohn]

if you notice it is in the 6th group and 2nd period
ther is only 1 element that is O(8)
1s2 2s2 2p3

Can you tell me how you think it is D?

Last trend such a big difference?" 2p6 is stable,

 

[sweet.sugar]

yup i got all of them Hadi Murtaza solved them
thanks anyways

Thats gud
Got some (actually 6 ) q from m/j/06 :/
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q1-D Q2-B Q10-C Q27-B Q38-B

 

[ZaqZainab]

we

for the first four IE the difference is roughly 2000
between the 4th and 5th there is a difference of 3550.but the principal quantum number is still the same i.e the principal quantum no is 2.
the distance between 2p and 2s is small.while distance between 2s and 1s is large

woh what i do is look for a huge change in the ionisation energy like here it is from 13300 to 71000
so basically the outer shell has 6 electron here cause the when we remove the 7th one is is most probably closer to the nucleus
and from the periodic table this element is in the 6th group
and they have mentioned from Li to Ne so 2nd period

Last trend such a big difference?" 2p6 is stable,

What do you mean ? yes it but where will you get those 2 extra electrons from this is for Ne

 

[not.maria]

woh what i do is look for a huge change in the ionisation energy like here it is from 13300 to 71000
so basically the outer shell has 6 electron here cause the when we remove the 7th one is is most probably closer to the nucleus
and from the periodic table this element is in the 6th group
and they have mentioned from Li to Ne so 2nd period

What do you mean ? yes it but where will you get those 2 extra electrons from this is for Ne

the first four ionization energies are for 2p4 electrons
the fifth and 6th are for 2s2
the last one is for 1s2
hence such a BIG difference because it is closest to the nucleus
the closer electron to the nucleus the more the IE

 

[papajohn]

we

for the first four IE the difference is roughly 2000
between the 4th and 5th there is a difference of 3550.but the principal quantum number is still the same i.e the principal quantum no is 2.
the distance between 2p and 2s is small.while distance between 2s and 1s is large

Smart girl!

 

[ZaqZainab]

Thats gud
Got some (actually 6 ) q from m/j/06 :/
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q1-D Q2-B Q10-C Q27-B Q38-B

Q1
Mole=Concentration*Volume
Mole ratio for NaOH and N2O4 is 2:1
Moles of NaOH=0.5*V
Moles of N2O4=0.02
use mole ration cross multiply
2 : 1
(o.5*V):0.02
V=0.08 dm^3
convert to cm^3

 

[♣♠ Magnanimous ♣♠]

Thats gud
Got some (actually 6 ) q from m/j/06 :/
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q1-D Q2-B Q10-C Q27-B Q38-B

1) 1 mol of N2O4 = 2 mol of NaOH
so 0.02mol =? } 0.02*2 = 0.04 mol of NaOH
so now to find volume => moles / conc ===> 0.04/0.5 ==> 0.08 dm^3 and in option it is cm^3 so to convert in it 0.08*1000 ==> 80cm^3 D

2) for Cl+2 there should three peaks ---> one of 35+35 2)35+37 and 3) 37+37
this 35 in one isotope and 37 is another isotope of chlorine! this are it's possibilities so B


27) it is butan-2-ol because as you can see it is secondary alcohol and this type of alcohol always give KETONE and ketones do react with ==> 2,4DNP giving orange precipitate but this is not clear test bcoz it may be aldehyde and in ques it is written is not reacts with fehling reagent which sures us that it is ketone bcoz they do not react with this reagent and remains blue color so B
i hope yout got his three btw idk others sorry for that and regret my english please

 

[ZaqZainab]

the first four ionization energies are for 2p4 electrons
the fifth and 6th are for 2s2
the last one is for 1s2
hence such a BIG difference because it is closest to the nucleus
the closer electron to the nucleus the more the IE

That what i was trying to say

 

[ZaqZainab]

Thats gud
Got some (actually 6 ) q from m/j/06 :/
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q1-D Q2-B Q10-C Q27-B Q38-B

Q10
Initial moles of N2O4 as 1 and that of NO2 as zero.

50% of N2O4 dissociates means that 1 moles of NO2 is formed. So total number of moles = 1.5

PArtial pressure of N2O4 = 0.5/1.5 = 1/3
Partial pressure of NO2= 1/1.5 = 2/3

So Kp = (2/3)^2 divided by (1/3). So 4/3 is the answer.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Can anyone explain question number 13 ans B, 28 ans D and 35 ans B.

Questions are;
In 13 why can't A be correct.
In 28 why is D correct, I think D is correct as tertiary alcohols don't oxidise even by strong oxidising agents right?
In 35 I don't get a point.

And in 19 is NO a reducing agent or not?

Q13 i have problems with equations too >.< but here i wouldn't have gone for A
lets look at other reaction of NaOH something you might be familiar with
like this one NaOH + НСl = NaCl + Н2О
why not NaClOH2?
Sulphur dioxide is acidic its just like HCl, it is weaker though
now you are left with B C or D
when acid+ base water is release
so not D and C isn't balanced
but i guess we need to learn this kind of equations
here is a link http://www.allreactions.com/index.php/group-1a/natrium/sodium-hydroxide just have a quick look
Q28 yes you are right
Q35
Carbon monoxide is neutral, and does not react like the other two acidic gases.

s04p1

Regarding Q28.

Key phrase in the question is "excess NaOH was used"
NaOH + SO3 --> NaHSO3 ---(1)
As there is excess NaOH, the HSO3- (acid) will further react.
NaHSO3+ NaOH --> Na2SO4 + H2O --(2)

Combine eqn (1) and (2), you will have the overall equation.
2NaOH + SO3 -->Na2SO4 + H2O

Q28. Tertiary alcohols can be oxidized by strong oxidizing agents (which is out of the syllabus), reason why they question use the phrase "not oxidized by mild oxidizing agents" is simply to reassure students that we are definitely talking about tertiary alcohols.

Q35 is as explained by ZaqZainab

Q19. Question is referring to NO2, are you asking about NO or NO2?

 

[Jelleh Belleh]

That what i was trying to say

We could also just use the data booklet to determine/verify the element.
..It will be given to us, right? The Ionization Energies page, I mean.

 

[ZaqZainab]

We could also just use the data booklet to determine/verify the element.
..It will be given to us, right? The Ionization Energies page, I mean.

yup but that would take a while?

 

[Metanoia]

I dont get it.
wont the equilibrium shift to the right hand side if ammonia i.e product is removed?

You can try viewing it in this manner.

A<-->B
During equilibrium, the rate of forward reaction = rate of backward reaction.

When B is removed, we understand that the equilibrium will shift right.
This does not necessary mean that rate of forward reaction is increased.
It is actually rate of backward reaction decreased, due to decrease in concentration of B.
And rate of forward reaction > rate of backward reaction till equilibrium is reestablished.

The sketch hopefully makes sense, its nearly 2am here and I can't draw straight.
Maybe I'll redo it in the morning.

 

[MariamMalik]

Q1. Find out the number of moles of O2, since 1 mole of gas occupies 24000 cm^3 at RTP.

moles of O2 = 500/24 000
molecules of O2 = moles x 6.02 x 10^23 = (500/24000) x 6.02 x 10^23 = 1.25 x 10^22

Q2. Comparing the original structure to the resulting structure, we see that 5 C=C double bonds "gone".
So 5 moles of H2 are added.

If unable to visualise the above explanation, the diagram can show how H could be added in these locations.
View attachment 44133

Q7 and Q10

View attachment 44134

Q21. Is the answer really D? I got B.
View attachment 44136

Q22. The C=C undergoes oxidative cleaving, and the ends are oxidised to COOH.

View attachment 44137

Q23.
If confident, this question can be approached in a mathematical way

CnH(2n+2) + (3n+1)/2 O2 --> n CO2 + (n+1)H2O

when n increases, moles of O2 increases linearly. So its a line with a positive gradient.

In Q.2 i don't understand how only 5 double bonds are broken?

 

[FarahMJ]

how come an ionic bond formed?

Both 1 and 2 are correct

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf Q19 ansC , Q22 ansA , Q27 ansD , Q36 ansA please help

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q11
Q13 can someone explain?
Q16 why not D? thats how HCl or HI is formed by breaking I-I bond and Cl-Cl bond
Q18
Q21
Q39 which two alkenes i have problems with ring structures
Q40