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Chemistry: Post your doubts here!

iKhaled

iKhaled

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freakybandi said:
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_4.pdf
qs 2 anybody?
Click to expand...
hey there..

a) iodine is purple so the colourless solution will turn purple..

b) only reactants of the slow reaction gets written down in our rate equation so for reaction one we got one mole of hydrogen peroxide and one mole of iodide ion but no moles of hydrogen ions so it will not be written in the rate equation so we have a = 1 b = 1 and c = o .... now for the second equation, IO- is made from H2O2 and iodide ion so u can pretend the IO- in the second equation as one mole of hydrogen peroxide and one mole of iodide ion and we also have hydrogen ion in the second equation so a= 1 b = 1 and c = 1 ... now for the third equation, we have HOI which is made from H+ AND IO- which is made from one mole of hydrogen peroxide and one mole of iodide :p but take care u also have got another mole of iodide ion in the third equation so in total u have 2 moles of iodide and 2 moles of hydrogen ion and one mole of hydrogen peroxide so a = 1 b = 2 c = 2

c) initial rate means u draw a tangent at the start of the graph the second the reaction started and calculate the gradient at that point and that would be ur rate

d) half life means the time the reactants reach its half concentration so 0.001/2 = 0.0005 this will be at 90s and 0.0005/2 = 0.00025 at 180s...180-90 = 90s so see the time is constant it always reaches its half concentration after 90s that proves its a first order reaction

e) the way i do this part of the question is i look at 2 lines and see which one has same values so when i divide the 2 lines over each other one reactant cancels and i would be able to calculate the power the reactant raised to for example...

look at line 1 and 2 when u divide them the concentration of the hydrogen ion will cancel each other so u have (0.05/0.07)^x = 1/1.4 do the math and find x and it will be 1 so the order with respect to h2o2 is 1 and now pick line 2 and 3 and divide them over each other (0.07/0.09)^1(0.05/0.07)^y = 1.4/1.8 do the maths and find y and it will be equal to zero so its order with respect to H ion is zero which means our calculation is consistent with step 1! thats it

i hope u got it!
 
F

freakybandi

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iKhaled said:
hey there..

a) iodine is purple so the colourless solution will turn purple..

b) only reactants of the slow reaction gets written down in our rate equation so for reaction one we got one mole of hydrogen peroxide and one mole of iodide ion but no moles of hydrogen ions so it will not be written in the rate equation so we have a = 1 b = 1 and c = o .... now for the second equation, IO- is made from H2O2 and iodide ion so u can pretend the IO- in the second equation as one mole of hydrogen peroxide and one mole of iodide ion and we also have hydrogen ion in the second equation so a= 1 b = 1 and c = 1 ... now for the third equation, we have HOI which is made from H+ AND IO- which is made from one mole of hydrogen peroxide and one mole of iodide :p but take care u also have got another mole of iodide ion in the third equation so in total u have 2 moles of iodide and 2 moles of hydrogen ion and one mole of hydrogen peroxide so a = 1 b = 2 c = 2

c) initial rate means u draw a tangent at the start of the graph the second the reaction started and calculate the gradient at that point and that would be ur rate

d) half life means the time the reactants reach its half concentration so 0.001/2 = 0.0005 this will be at 90s and 0.0005/2 = 0.00025 at 180s...180-90 = 90s so see the time is constant it always reaches its half concentration after 90s that proves its a first order reaction

e) the way i do this part of the question is i look at 2 lines and see which one has same values so when i divide the 2 lines over each other one reactant cancels and i would be able to calculate the power the reactant raised to for example...

look at line 1 and 2 when u divide them the concentration of the hydrogen ion will cancel each other so u have (0.05/0.07)^x = 1/1.4 do the math and find x and it will be 1 so the order with respect to h2o2 is 1 and now pick line 2 and 3 and divide them over each other (0.07/0.09)^1(0.05/0.07)^y = 1.4/1.8 do the maths and find y and it will be equal to zero so its order with respect to H ion is zero which means our calculation is consistent with step 1! thats it

i hope u got it!
Click to expand...
thanku soooooooo much!! :)
 
Soldier313

Soldier313

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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

knowitall10

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

Jaf

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Soldier313 said:
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
Click to expand...

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
 
Soldier313

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Jaf said:
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
Click to expand...
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 :(
 
X

xhizors

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Soldier313 said:
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!
Click to expand...

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!
 
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xhizors

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Soldier313 said:
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!
Click to expand...
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

Tkp

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same ques here
Soldier313 said:
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 :(
Click to expand...
 
Tkp

Tkp

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

Wanderer

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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
 
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