ALL CIE CHEMISTRY DOUBTS HERE !!

[Deadly_king]

Ah, damn. I didnt think of it in that way. 

+rep.

It's alright dude 

[Deadly_king]

plz help in Nov.2009 ppr22 Q2(v) &

also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)

thanx in advanc.

Nov 09 P2 Q2(v)

In most cases, Hydrogen has an oxidation number of -1.

Therefore in the first 3 cases. take oxidation number of H to be -1. Since they are all neutral molecules, resultant charge should become to zero. This principle is applied to ionic compounds only.

MgH₂ :
Let oxidation number of Mg be x.
x  + 2(-1) = 0 ---> x = +2

Apply same principle for AlH₃ and you'll get +3.

As for the other two, they are covalently bonded such that oxidation state of hydrogen is +1

PH₃
Let oxidation no of P be x.
x + 3(+1) = 0
x = -3

same method for H₂S. 

[nid404]

Hehe......yeah I guess so 

Anyway it would be nice if you could confirm the method since I didn't obtain the exact answer 

Your method is not wrong really...

what I would have done instead

at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3

1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3

[Deadly_king]

Your method is not wrong really...

what I would have done instead

at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3

1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3

Oooh.........yeah.
I guess I chose the long way

It's far easier to understand like you described 

Thanks

[nid404]

Oooh.........yeah.
I guess I chose the long way

It's far easier to understand like you described 

Thanks

Considering it's an MCQ , you wouldn't want to do it the long way

My pleasure

[Deadly_king]

Nov 09 P4 Q8

c)(ii) If two chlorine atoms are present, the three possibilities are : Cl³⁵-Cl³⁵, Cl³⁵-Cl³⁷ and Cl³⁷-Cl³⁷

Cl³⁵ and Cl³⁷ exist in the ratio 3 : 1

Probability of obtaining Cl³⁵-Cl³⁵ = 3/4 x 3/4 = 9/16 ---> M peak
Probability of obtaining Cl³⁵-Cl³⁷ = 2(3/4 x 1/4) = 6/16 ---> M+2 peak
Probability of obtaining Cl³⁷-Cl³⁷ = 1/4 x 1/4 = 1/16 ----> M+4 peak

Therefore the ratio will be 9 : 6 : 1

NOTE : The probability for M+2 peak is multiplied by two since it can exist both as Cl³⁵-Cl³⁷ and Cl³⁷-Cl³⁵

d)(ii) PCBs.
This is because when you compare the graphs upstream and downstream, you will note that PCBs is already in very high concentration unlike the others which are present in minimal amount.

Therefore after passing the mill, the concentration of the other three chemicals increase in concentration showing that the mill releases these chemicals.

(iv) Four chlorine atoms will result in 5 possible peakss.
1. Cl³⁵-Cl³⁵-Cl³⁵-Cl³⁵
2. Cl³⁷-Cl³⁷-Cl³⁷-Cl³⁷
3. Cl³⁵-Cl³⁷-Cl³⁷-Cl³⁷
4. Cl³⁵-Cl³⁷-Cl³⁵-Cl³⁷
5. Cl³⁷-Cl³⁷-Cl³⁷-Cl³⁵

NOTE : The arrangement of the atoms in each case may differ but they will form the same peak.

Hope it helps

[Deadly_king]

Considering it's an MCQ , you wouldn't want to do it the long way

My pleasure

Absolutely right madam 

[ashish]

Good job pal........though it was a bit long 

+ rep

^_^ thank you !

[$!$RatJumper$!$]

For chemistry practical.. what is the best way to study for it other than actually doing it in school? We're on study leave so dont have access to lab equipment. How is the best way to do it at home?

[elemis]

For chemistry practical.. what is the best way to study for it other than actually doing it in school? We're on study leave so dont have access to lab equipment. How is the best way to do it at home?

Watch videos on youtube.

[$!$RatJumper$!$]

Watch videos on youtube.

Can you please provide me a link to get started?

[Freaked12]

Am i late  

What the heck

moles CO2 = 0.352 g / 44.009 g/mol=0.00800
moles H = 2 x 0.072 /18.02 g/mol=0.00800

[Freaked12]

MgH2 is Mg 2+ and 2 H- ions so oxidation No of Mg = +2 (Group 2 - loses 2 electrons to form Mg2+ ion 2 -2 empties outer shell)

AlH3 is Al3+ + 3 H- ions so oxidation number of Al = +3 (Group 3 - loses 3 electrons to form Al3+ ion 3 - 3 empties outer shell)

PH3 Is P3- and3 H+ so oxidation number of P = -3 (Group 5 - gains 3 electrons to form P3- ion : 5 + 3 = 8 fills up outer shell)

H2S is 2 H+ and S2- ions so oxidation number of S is -2 (Group 6 - gains 2 electrons to form S2- ion ; 6 + 2 = 8 fills up outer shell)



Sorry
Just wanted to get involved

[moon]

Nan Ari......the data provided is sufficient to find the answer.

C_xH_y + (x + y/4)O₂ ----> xCO₂ + (y/2) H₂O

Number of moles CO₂ = Mass / Mr = 0.352/44 = 0.008

Number of moles of H₂O = 0.072/18 = 0.004

1 mole of the hydrocarbon gives 0.008 moles of CO₂ and 0.004 moles of H₂O

Therefore from equation above we can note that x = 0.008 while y/2 = 0.004

Empirical formula will hence be C₈H₈

why didn't we find the mole ratio CO₂: H₂O
so that the formula can be C₂H₂...

[moon]

for nov 2009 ppr 2

The oxidation state of hydrogen in the metal hydrides is always -1 so why in PH₃ the oxidation state is +1 and hence x+3(-1)=0 so x=+3 really need help..Thanks n advance