P1 MCQ's preparation thread for chemistry ONLY!!!!

[daisyy]

Q20 i cant think of any reasonable explanation for this question other than the OH- being a nucleophile

as for Q12 ... they said 1 mole of Mg , Al and S so write down all 3 combustion equations and balance but keeping 1 mole of those 3 .. 1) Mg + 1/2 o2--> MgO 2) Al +3/4O2--> Al2O3 3) S+3/2O2-->SO3 notice the moles of O2 in those combustion eqns are increasing but not equally thus the only graph representing this is D

Isn't SO3 formed from SO2 ONLY in the presence of vanadium V oxide and the question says "BURNED" in excess oxygen ie combustion....

 

[Razo513]

Isn't SO3 formed from SO2 ONLY in the presence of vanadium V oxide and the question says "BURNED" in excess oxygen ie combustion....

That's true but theoretically when burning in *excess* SO3 is produced yet tbh I totally overlooked the fact that SO2 should be given not SO3

 

[queen of the legend]

Lool! How cool! Okay umm in the nov 05 question in this case isnt the answer supposed to be A? Cuz the difference between them is constant 0.5...

hmm no coz sulfur can form another oxide thats is SO3 ..and which requires 3 moles of O2

 

[daisyy]

That's true but theoretically when burning in *excess* SO3 is produced yet tbh I totally overlooked the fact that SO2 should be given not SO3

Gotchaa! Makes sense thank you! ;D

 

[daisyy]

hmm no coz sulfur can form another oxide thats is SO3 ..and which requires 3 moles of O2

Thaanks! got it

 

[Pwincessajwa]

can someone help me out plx

 

[iKhaled]

can someone help me out plx

what do u need help with ?

 

[iKhaled]

can someone please explain to me question 7 october november 2007 q9

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

 

[Pwincessajwa]

what do u need help with ?

i need expaination of many queshtions in may june 2010 ppr varient 12 ??

 

[Pwincessajwa]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_12.pdf q2,3,4,6,11 actuallly most of the questions of the ppr bcz this ppr is kinda hard for me

 

[6Astarstudent]

can someone please explain to me question 7 october november 2007 q9

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

first you find the mol ratio of the metal ion and sulphite ion which is 2:1 because same concentration and 50:25 volume so you know 2 metallic ions are required to oxidise 1 sulphite ion.
then you find the change in oxidation number of sulphite ion to sulphate +4->+6 which is +2
so 2 metallic ion must be -2, meaning each metallic ion -1 in oxidation number. and originally it is +3 so 3-1=2 answer is B

 

[queen of the legend]

7 Flask X contains 5 dm3 of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa
pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9 kPa C 10 kPa

ans is A ...how ?

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf q2,3,4,6,11 actuallly most of the questions of the ppr bcz this ppr is kinda hard for me

for question 2 u will need to draw the orbitals with the electrons inside it in the form of arrows? uk that? if u do then see the first element has a proton no of 5 so 1s2 2s2 2p1 so it is not A because the p orbial has one unpaired electrons. when the proton no is 13 then u will have 1s2 2s2 2p6 3s2 3p1 which has one unpaired electron only in the p orbital also. now when the proton no is 15 u will have 1s2 2s2 2p6 3s2 3p3..now here we got 3 unpaired electrons in the 3p orbital so this will be our answer did u get it ?

ANSWER: C

for question 3, ok uk that as u cross the period the nuclear charge increases so the outermost electron is more attracted to the nucleus which will need more energy in order to remove that electron ( higher ionization energy across the period) and as u go down the group the element gains an extra outershell which means the outermost electron is further away from the nucleus and less attracted so easier to remove that electron ( lower ionization energy as u go down the group) you know that ionization energy is always endothermic because u r giving energy to the atom to remove its electron and form an ion so Rn is in group 8 and period 6 while the other r in period 7 so obviously Rn will have the most endothermic reaction. for Fr, it is in group 1 and Ra is in group 2 so Ra outermost electron is more attracted to the nuclues and harder to be removed so it is more endothermic than Fr

least endothermic will be Fr then Ra and the most endothermic will be Rn

ANSWER : A

 

[6Astarstudent]

7 Flask X contains 5 dm3 of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa
pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9 kPa C 10 kPa

ans is A ...how ?

Here is a quick equation
(V1P1 + V2P2)/(V1+V2) = new P
so (5x12 + 10x6)/(5+10) = 120/15 = 8

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf q2,3,4,6,11 actuallly most of the questions of the ppr bcz this ppr is kinda hard for me

for question 4, you know that bond breaking is always endothermic (+) and bond forming is exothermic (-).. P is bond breaking and will have a value of +193..Q is bond forming so will have a value of -244. in R we formed one C-Cl bond so we will have a value of -340. In S we broke one C-H bond so we will have a value of +410..the most negative will be R and the most positive will be S so it is R-->Q--->P---->S

ANSWER : C

for question 6, u have NH4NO3 -----> NO2. the oxidation no of nitrogen in NH4+ is -3 and the oxidation no of N in NO3- is +5 and the oxidation no of N in N2O is +1 so the answer is it changed from -3 to +1 so thats +4 and when it changed from +5 to +1 it is -4 so the answer will be +4, -4

ANSWER: D

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf q2,3,4,6,11 actuallly most of the questions of the ppr bcz this ppr is kinda hard for me

question 11, uk the haber process is an exothermic reaction so an increase in the temperature will shift the equilibrium position to the left and the yield of ammonia will be decreased

N2 + 3H2 --> 2NH3 + heat

u have less moles of products than reactants so an increase in the pressure will shift the equilibrium position to the right and therefore the yield of ammonia increases. now use that info to find the answer!

satisfied with my help or there is somethin u dont get still?

 

[iKhaled]

first you find the mol ratio of the metal ion and sulphite ion which is 2:1 because same concentration and 50:25 volume so you know 2 metallic ions are required to oxidise 1 sulphite ion.
then you find the change in oxidation number of sulphite ion to sulphate +4->+6 which is +2
so 2 metallic ion must be -2, meaning each metallic ion -1 in oxidation number. and originally it is +3 so 3-1=2 answer is B

thanks i got it now!!

 

[stella jazz]

can anyone help me with q 6 and 12 plz
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
both answers are C but why??

 

[Suno Yaar]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf question 10 plzzz quick

 

[Suno Yaar]

thanks i got it now!!

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf question 10 plzzz quick