CHEMISTRY PAPER 3 DOUBTS HERE ....

[Violettamee]

:Bravo: :Bravo: :Bravo: :Bravo: Does anyone here know how to solve the following questions (they're from the 0620/32/M/J/09 past paper)

The first one is :- :Search: :unknown:
question(5) part (b) on page 7

the second ones is :- :Search: :unknown:
question (9) part (b) AND part (c)(i) and (c)(ii)


thnx alot for ur help guys.... :wink: :wink:

 

[gary221]

:Bravo: :Bravo: :Bravo: :Bravo: Does anyone here know how to solve the following questions (they're from the 0620/32/M/J/09 past paper)

The first one is :- :Search: :unknown:
question(5) part (b) on page 7

the second ones is :- :Search: :unknown:
question (9) part (b) AND part (c)(i) and (c)(ii)


thnx alot for ur help guys.... :wink: :wink:

to find the formula we need to see how much of sodium phospate reacts completely wth the mateal nitrate
according to d grph 6 cm3 of the phospate reacts completely wth 18 cm3 of the metal nitrate
so 6:18
=1:3
so d formula 4 d metal phuspate will be 3 moles of metal T for 1 mole of the phospate

thus formula is T3PO4

hope u got it!

 

[Violettamee]

thnx very much qary221..
i DID get it actually...

but do u also know how to solve question 9 (page 14 of the same past paper) part (b) and part (c)(i)and(ii) ???

thnx for ur help .. =D

 

[gary221]

for ur second ques first deduce the formula for the compound
valency of Al = 3 and C = 4
so compound = Al4C3

frame d equation,
Al4C3 + H2O = Al(OH)3 + CH4

now balance it
4 Al on the LHS, so add a 4 in front of the Al(OH)3
3 C on the LHS so add a 3 in front of the CH4


24 H on the RHS, so add a 12 b4 H2O to balance equation
Al4C3 + 12H2O = 4Al(OH)3 + 3CH4

 

[gary221]

for ur thrd ques its stoichiometry
calculate no of moles of F needed
mass/Mr=7.2/(19*2) = 0.189

so 0.189 moles of F combine wth 2*0.08 = 0.16 moles of Si

since 0.16 is less thn 0.189
Si is d limiting reagent!!


dude if i helped dont forget to click the thanks button!!!!

 

[gary221]

same stoiciometry needed
moles = mass of compound/RFM

calculate the mass of Si, mass = moles *Mr = 0.08*28 = 2.24 g
mass of F = 7.2g so total mass = 7.2+2.24 = 9.44g

RFM = 28+ (4*19) = 104

moles = 9.44/104
= 0.08 moles formed of SiF4

 

[Violettamee]

for ur thrd ques its stoichiometry
calculate no of moles of F needed
mass/Mr=7.2/(19*2) = 0.189

so 0.189 moles of F combine wth 2*0.08 = 0.16 moles of Si

since 0.16 is less thn 0.189
Si is d limiting reagent!!


dude if i helped dont forget to click the thanks button!!!!

wait wait wait a second .....
the formula says : Si + 2F2 --> SiF4
the ratio of the reactants is :

Si : 2F2
= 1 : 2

so, why do u multiply TWO * 0.08 = 0.16 moles of Si ????????????????
where did the "2" come from ??????? :no: :unknown: :unknown:

 

[Outofideaz]

hey violettamee ... i knw u

 

[Violettamee]

hey violettamee ... i knw u

heyyyy !!!!!! i know u tooo REALLY WEELLLL
!!!!!!!!
luv ya :* BBF

=D =D =D

 

[Violettamee]

hey violettamee ... i knw u

so can YOUUUU answer my other question ?????

 

[LimeReem]

http://www.xtremepapers.me/CIE/Cambridg ... _qp_31.pdf
Q4c can someone help me?

 

[Violettamee]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w09_qp_31.pdf
Q4c can someone help me?

Here it is :-
Click on the attachment below :-

hope i helped ya ! hehe

 

[breakheart]

i need help in this paper http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_3.pdf
question 4. part b. ii.
can u please explain how we get the graph? i mean the theory behind it etc. how we get the answer :/