Chemistry: Post your doubts here!

[Soldier313]

Aoa wr wb

Can someone please explain:

qn 2 cii and 2 ciii of this paper? and for part 2 c iii, if thee's more than one way of doing it, can you please share all the possible methods?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_4.pdf
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_ms_4.pdf

qn 4 bii
qn 6 b ii) why don't we have to use concentrated solutions and a high temperature?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_ms_4.pdf

qn 1 aii) why is the ms quoting the value of the electrode potential to be negative?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_4.pdf
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_ms_4.pdf

Thank you so much!

 

[knowitall10]

Aoa wr wb

Can someone please explain:

qn 2 cii and 2 ciii of this paper? and for part 2 c iii, if thee's more than one way of doing it, can you please share all the possible methods?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_ms_4.pdf

qn 4 bii
qn 6 b ii) why don't we have to use concentrated solutions and a high temperature?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf

qn 1 aii) why is the ms quoting the value of the electrode potential to be negative?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_ms_4.pdf

Thank you so much!

LOL! Soldier313 you always come up with a bunch of questions all at once Anyway...sorry dude, i'll answer thm in a while..lemme have a gud luk at em first...

 

[Soldier313]

LOL! Soldier313 you always come up with a bunch of questions all at once Anyway...sorry dude, i'll answer thm in a while..lemme have a gud luk at em first...

Haha okayyy thanx sis

 

[Jaf]

Aoa wr wb

Can someone please explain:

qn 2 cii and 2 ciii of this paper? and for part 2 c iii, if thee's more than one way of doing it, can you please share all the possible methods?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_ms_4.pdf

Recall that conjugate acids have one extra proton compared to their conjugate bases. See which one of the two compounds have one extra proton attached to it (H+) = HPO^(2–) and H2PO4^(-)
It's pretty obvious it's the latter: H2PO-. So this will act as you acid (or proton donor). The other one will act as the base (or proton acceptor).
HPO4^(2–) + H+ ------> H2PO4^(-)
H2PO4^(-) + OH- ------> HPO4^(2-) + H+ + OH- -----> HPO4^(2-) + H2O [I'm showing the middle step just for understanding that a proton is being donated and accepted by OH- ion to form water - do not include this step in your answer]

As for c)iii) I'm not too confident with these ionic equilibria questions myself. I've just memorized the formula pH = pKa + log(conjugate base/acid). Works every time.
Also if I'm not wrong this specific question with the same compounds also came in a previous year.

The other member can do the rest lol

 

[deane26]

You guys have any worksheets for A2?? Any help would be highly appreciated by me xD

 

[Soldier313]

Recall that conjugate acids have one extra proton compared to their conjugate bases. See which one of the two compounds have one extra proton attached to it (H+) = HPO^(2–) and H2PO4^(-)
It's pretty obvious it's the latter: H2PO-. So this will act as you acid (or proton donor). The other one will act as the base (or proton acceptor).
HPO4^(2–) + H+ ------> H2PO4^(-)
H2PO4^(-) + OH- ------> HPO4^(2-) + H+ + OH- -----> HPO4^(2-) + H2O [I'm showing the middle step just for understanding that a proton is being donated and accepted by OH- ion to form water - do not include this step in your answer]

As for c)iii) I'm not too confident with these ionic equilibria questions myself. I've just memorized the formula pH = pKa + log(conjugate base/acid). Works every time.
Also if I'm not wrong this specific question with the same compounds also came in a previous year.

The other member can do the rest lol

Ow okies, thanx a lot for that!
erm just a little clarification, when it comes to using that formula, this is what the ms says:
7.2 + log (.002/.005)
so i get the bits for the acid n base, but the 7.2, how is that the pKa value, coz i can't see a Ka value that I can use:/
Sorry for the trouble

 

[MaRyy xD]

Paper 1 Oct/Nov 2007 , Q37.
Why is it B?

 

[xhizors]

Aoa wr wb



qn 4 bii
qn 6 b ii) why don't we have to use concentrated solutions and a high temperature?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
Thank you so much!

For this part:
because we have Phenol dude here,
in phenol due to -OH group its stability is not the same as benzene gang
thats why we didn't use conc Hno3 with h2so4 and high temp(also if temp is higher than 60*C than u'll be getting with benzene 2 2 NO2 (aik kai saath 2 free for high temp))

hope that helps!

 

[xhizors]

Aoa wr wb


qn 1 aii) why is the ms quoting the value of the electrode potential to be negative?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_ms_4.pdf

Thank you so much!

Their are other ways too to hit this milestone but here's the shot:
u use formula E* = E(value of right) - E(Value of left)
at right u can see platinium is chillin alone so electrolyte must only be containing Fe3+ and Fe2+ ions
at at left Chlorine is being Pumped in
so u substract 0.77(ironman Potential) - 1.36(Cl)
u get -0.59

hope it helps
Please pray for me i need your prayers

 

[Tkp]

same ques here

Ow okies, thanx a lot for that!
erm just a little clarification, when it comes to using that formula, this is what the ms says:
7.2 + log (.002/.005)
so i get the bits for the acid n base, but the 7.2, how is that the pKa value, coz i can't see a Ka value that I can use:/
Sorry for the trouble

 

[Tkp]

Electrons flow from least positive to most positive
so electron flows frm Chlroine to Fe
so 0.77-1.36

 

[Alool]

Has any one solved this practical paper?
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_y07_sp_31.pdf
please i need help in knowing how to add all the volumes immediately

 

[Wanderer]

JUNE10/41 Q3
PART c-i
How do we make the structure of CCl2...Carbon will share 1 electron with each Chlorine atom and have 1 lone pair left?
I cant understand the ms...
Please help

 

[Wanderer]

NOV08
Q4 Part b iii
Why have they take 4C-C in bonds broken and 2C=C in bonds formed??

 

[Xeee]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_41.pdf

In 1(b), where did they find atomisation enthalpy of nitrogen from? It's not in the data booklet!
They've given '944' as the value to use while finding enthalpy change for nitrogen. Where did they find that from?? :/

 

[geek101]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_41.pdf

In 1(b), where did they find atomisation enthalpy of nitrogen from? It's not in the data booklet!
They've given '944' as the value to use while finding enthalpy change for nitrogen. Where did they find that from?? :/

The enthalpy change of atomization of N2 would be, the enthalpy change when 1 mole of gaseous N atoms form from N2.
This reaction would then be: 1/2 N2 (g) > N (g), in order to convert N2 to N atoms only you will need to break the 3 bonds in the structure of N2. Therefore you'll need to see the energy of the three bonds, which is in the data booklet as the bond energy for N---N which is 944. and this will be endothermic because bond breaking is endothermic, so itll be +944.

in the mark scheme there showing this reaction: N2 (g) > 2N3- (g)
but the reaction with all the steps will be : N2 > 2N > 2N3- (all gasses)

hope that helped = )

 

[Xeee]

The enthalpy change of atomization of N2 would be, the enthalpy change when 1 mole of gaseous N atoms form from N2.
This reaction would then be: 1/2 N2 (g) > N (g), in order to convert N2 to N atoms only you will need to break the 3 bonds in the structure of N2. Therefore you'll need to see the energy of the three bonds, which is in the data booklet as the bond energy for N---N which is 944. and this will be endothermic because bond breaking is endothermic, so itll be +944.

in the mark scheme there showing this reaction: N2 (g) > 2N3- (g)
but the reaction with all the steps will be : N2 > 2N > 2N3- (all gasses)

hope that helped = )

OMG ofcourse! Thanks a lot! :')
In the same question, (c) part, I have trouble forming the equation. How do we predict the formation of LiOH? :/ I thought it was Li^2O :/

 

[Usman04]

Ammonium sulfate in nitrogenous fertilisers in the soil can be slowly oxidised by air producing
sulfuric acid, nitric acid and water.
How many moles of oxygen gas are needed to oxidise completely one mole of ammonium
sulfate?
A 1 B 2 C 3 D 4
plz explain this

 

[A star]

Ammonium sulfate in nitrogenous fertilisers in the soil can be slowly oxidised by air producing
sulfuric acid, nitric acid and water.
How many moles of oxygen gas are needed to oxidise completely one mole of ammonium
sulfate?
A 1 B 2 C 3 D 4
plz explain this

easy
(Nh4)2SO4 +4O2 -----> H2SO4 + 2HNO3 + 2H2O
hence D

 

[Hassi123]

http://www.novapapers.com/images/PDF/CIE/A-LEVEL/chemistry/2012jun/9701_s12_qp_43.pdf

Q3 part d. I thought benzene only reacts with br2 if its not aq or with a catalyst. Here there is an excess of br2 (aq) and it reacts with the benzene.
Please help?