Troubles in chem p1

[Jinkglex]

AS

ohh, then im guessing you probably haven't studient polarizing power yet? in that case dont worry about that, that MCQ is out of syllabus for you as well

 

[cute97]

ohh, then im guessing you probably haven't studient polarizing power yet? in that case dont worry about that, that MCQ is out of syllabus for you as well

thankss but are sure cz the same question was repeated in may june 2012 varient 11??
yhh didnt take polarizing power

 

[Jinkglex]

Hmm, in that case it might have been referring to some other property..as you already know though, eff. refers to CO2 usually, and down group II reactivity decreases so we have to use harsher methods...maybe thats all they were relating to?

thankss but are sure cz the same question was repeated in may june 2012 varient 11??
yhh didnt take polarizing power

 

[cute97]

Hmm, in that case it might have been referring to some other property..as you already know though, eff. refers to CO2 usually, and down group II reactivity decreases so we have to use harsher methods...maybe thats all they were relating to?

yhhh maybe thanks for your time

 

[cute97]

Anyone pls
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf
Q1 and Q2
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q10
thanks in advance
mark scheme
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_ms_1.pdf
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_ms_1.pdf

 

[1357913579]

Use of the Data Booklet is relevant to this question.
When 3.00 g of an anhydrous nitrate of a Group II metal is decomposed, 1.53 g of gas is
produced.
What is the nitrate compound?
A beryllium nitrate
B calcium nitrate
C magnesium nitrate
D strontium nitrate help plz

can u please tell what is the answer for this

 

[cute97]

I know this sounds long and tedious, but the sure-shot way would be for you to calculate the MR of all the mentioned nitrates, divide 3 grams on each to see the number of moles it would represent, and then see if said nitrates decomposition product [lets call it XO] fits the bill, as it will have the same number of moles, and if 1.53 is the answer you get [by multiplying the number of moles mentioned previously with the MR of the decomposition product], there's your answer.

LOL i knw you wanna kill me at the moment but i didnt understand it i told u am bad in chem hope ur not regretting that u post an answer

 

[queen of the legend]

can u please tell what is the answer for this

for this one first write the elements given with their Ar:

now that you have written their masses write an equation for any of them
Be : 9
Ca: 40
Mg: 24
Sr: 87

Ca(NO3)2 ______> CaO + 2NO2 + 1/2 O2

the mass of nitrate of any of these elements is = (14*2 +6*16) =124
the total mass of gases produced is = (2*14 +4*16 + 16)= 108

now just cross multiply for each element take the mass as follows :

Ca(NO3)2 ______> CaO + 2NO2 + 1/2 O2
40 + 124 g yields 108 g gases
so
3 g will yield

(108 *3 ) / 164 = 1.97 you can eliminate calcium now

same procedure using other metals

 

[1357913579]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
number-28,21,39
thnks

 

[Jinkglex]

LOL i knw you wanna kill me at the moment but i didnt understand it i told u am bad in chem hope ur not regretting that u post an answer

ahaha, no, its no problem, but as you can see queen has already answered the question in full now

 

[cute97]

ahaha, no, its no problem, but as you can see queen has already answered the question in full now

oohh this is the same Q okaay thanks

 

[HorsePower]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Q1
plz help

 

[cute97]

HELP PLS
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q11 and Q14
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9
Thanks is advance

 

[cute97]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Q1
plz help

same doubt here ANYONE PLEASE

 

[Jinkglex]

once more guys please post marking scheme as well!

same doubt here ANYONE PLEASE

as the question says, PbCl4 turns into PbCl2, meaning 2 chlorine are lost, thus 2 moles of bromine are released.

so if you form equation, it becomes:

PbCl4 + 2NaBr -> PbCl2 + 2NaCl + Br2.

As we have 6.98grams we divide by Mr of PbCl4 and get 0.02 moles

multiply by 2, thats 0.04 moles of NaBr [as the mole ratio is 1:2 between the two reactants]

0.04 moles of NaBr reacting means 0.04 moles of Bromine gas being released as seen in the above equation.

0.04 x 79.9 = 3.196 = Option C= +1 mark

 

[HorsePower]

once more guys please post marking scheme as well!



as the question says, PbCl4 turns into PbCl2, meaning 2 chlorine are lost, thus 2 moles of bromine are released.

so if you form equation, it becomes:

PbCl4 + 2NaBr -> PbCl2 + 2NaCl + Br2.

As we have 6.98grams we divide by Mr of PbCl4 and get 0.02 moles

multiply by 2, thats 0.04 moles of NaBr [as the mole ratio is 1:2 between the two reactants]

0.04 moles of NaBr reacting means 0.04 moles of Bromine gas being released as seen in the above equation.

0.04 x 79.9 = 3.196 = Option C= +1 mark

thnks... i cudnt form the equation, the rest is fine

 

[cute97]

once more guys please post marking scheme as well!



as the question says, PbCl4 turns into PbCl2, meaning 2 chlorine are lost, thus 2 moles of bromine are released.

so if you form equation, it becomes:

PbCl4 + 2NaBr -> PbCl2 + 2NaCl + Br2.

As we have 6.98grams we divide by Mr of PbCl4 and get 0.02 moles

multiply by 2, thats 0.04 moles of NaBr [as the mole ratio is 1:2 between the two reactants]

0.04 moles of NaBr reacting means 0.04 moles of Bromine gas being released as seen in the above equation.

0.04 x 79.9 = 3.196 = Option C= +1 mark

Thanks

 

[cute97]

thnks... i cudnt form the equation, the rest is fine

1)They said in the question 6.980 g of lead(IV)chloride so this is PbCl4------ agree
2)then they said this PbCl4 is added( reacted) to an excess of NaBr so this is Pbcl4+NaBr---- agree
3)Now look at the first part of the question where they says the PbCl4 oxidise the bromide to bromine so a bromine atom is formed Br2 so this is one of the products. Completing the equation PbCl4+NaBr-->Br2---- agree
4)they also said that"" Pb 4+"" that is when it is in the compound ""PbCl4"" this is reduced to ""Pb 2+""so another product formed for this compud which is PbCl2--- agree
5) Now the equation is PbCl4+NaBr--->PbCl2+Br2 no check the equation the reactants has Na which is not there in products and there is Cl2 remaining from Cl4 so forming NaCl
6) PbCl4+NaBr--->PbCl2+Br2+NaCL --- agree
now balance it
PbCl4+2NaBr--->PbCl2+Br2+2NaCL
Hope you understood am not really good in explaining

 

[Rajvinder]

Hmm, in that case it might have been referring to some other property..as you already know though, eff. refers to CO2 usually, and down group II reactivity decreases so we have to use harsher methods...maybe thats all they were relating to?

doesn't reactivity increase down group 2 ? i thought reactivity decreases down group 7 (halogens)

 

[cute97]

doesn't reactivity increase down group 2 ? i thought reactivity decreases down group 7 (halogens)

yhh the reactivity of the halogens decreases on descending the group AND only the halogen halides increases down the group from HF to HI this is wat i knw