Chemistry: Post your doubts here!

[My Name]

Hi,

Have you ever come across any question relating to it in P4? If you see any question, please do tell me its year.

Yep I have.
I dont think its like just one year though.

 

[Rizwan Javed]

Yep I have.
I dont think its like just one year though.

Can you tell me it's year. It's very difficult to go through every paper for finding it for me. :/

 

[My Name]

Can you tell me it's year. It's very difficult to go through every paper for finding it for me. :/

I will tell you.
But I might miss some. :3 I havent solved everything yet.

 

[qwertypoiu]

Hi,

Have you ever come across any question relating to it in P4? If you see any question, please do tell me its year.

Here's one:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_41.pdf

Go to question 6.

If you have troubles I've answered it here.

Good luck

 

[The Sarcastic Retard]

http://dynamicpapers.com/wp-content/uploads/2015/11/9701_w15_qp_43.pdf
Q2(d)(ii)(iii) Can anyone explain me this? THanks.

 

[ashcull14]

http://dynamicpapers.com/wp-content/uploads/2015/11/9701_w15_qp_43.pdf
Q2(d)(ii)(iii) Can anyone explain me this? THanks.

d (2) to find an oxidation no of any compund u have to add the valencies of all the compunds and equate them wid 0
so it will b:-
oxid no of y + oxid no of Ba...and so on = 0
keeping Cu as x
3 +2(2) + 3x + (2*-7) = 0
x= 7/3 = 2.3

 

[The Sarcastic Retard]

d (2) to find an oxidation no of any compund u have to add the valencies of all the compunds and equate them wid 0
so it will b:-
oxid no of y + oxid no of Ba...and so on = 0
keeping Cu as x
3 +2(2) + 3x + (2*-7) = 0
x= 7/3 = 2.3

(iii)?
Oh silly me!! LOL Thanks. ^_^

 

[holoholo]

can anyone please explain this ?

 

[holoholo]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_53.pdf

Can someone please draw question 1 c ?

 

[qwertypoiu]

View attachment 59927
can anyone please explain this ?

Excess ammonia with Co2+ forms [Co(H2O)2(NH3)4]2+, in a LIGAND EXCHANGE reaction, just as Cu2+ would.
OH- forms Co(H2O)4(OH)2 (s), in a PRECIPITATION reaction, just as Cu2+ would.
Reacting with S2O8 2-, which you know is a very good oxidizing agent, will oxidize Co2+ to Co3+, in a REDOX reaction. We actually wouldn't expect this from Cu2+, since (as far as I know) Cu3+ doesn't exist. The Data Booklet however will show you that Co3+ does exist.

 

[sfhn_128]

A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen.
After cooling to room temperature, the residual gas was passed through aqueous potassium
hydroxide.
All gas volumes were measured at the same temperature and pressure.
What volume of gas was absorbed by the alkali?
A 15 cm3 B 20 cm3 C 30 cm3 D 40 cm3
please can some one explain this question

 

[Saad the Paki]

A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen.
After cooling to room temperature, the residual gas was passed through aqueous potassium
hydroxide.
All gas volumes were measured at the same temperature and pressure.
What volume of gas was absorbed by the alkali?
A 15 cm3 B 20 cm3 C 30 cm3 D 40 cm3
please can some one explain this question

First write the equations of combustion of both methane and ethane.
CH4+202-->C02+2H20 ; so ratio of volumes is the same as the ratio of moles, ratio of CH4 to CO2 is 1:1 so 10cm^3 of CH4 will make 10cm^3 of CO2.
C2H6+3.502-->2CO2 + 3H20 ; ratio of C2H6 to CO2 is 1:2 so 10cm^3 ethane will make 20cm^3 of CO2. Now add the amounts of CO2 10+20=30cm^3 so C is the ans.
(Potassium hydroxide is an alkali so only the acidic gas will be absorbed by it, O2 and H20 are neutral and CO2 is the only acidic gas that's why we only calculate the volume of CO2)

 

[holoholo]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_53.pdf

Can someone please draw question 1 c ?

 

[ashcull14]

(iii)?
Oh silly me!! LOL Thanks. ^_^

if u look at the compound Cu3O7 it shows that Cu has an overall charge of +7
Thus, 7 is the sum of combination of oxidation no.s whose mean oxidation no. is 2.3 u can either factorise 7 to obtain combinations or either judge by the decimal .3 which is less than 0.5 thus closer to 2 which means we hav more +2 charges out of a total of 3 oxidation no.s
so 2+2 =4 ...7-4 = 3
so two +2 charges and one +3

 

[ashcull14]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_53.pdf

Can someone please draw question 1 c ?

 

[The Sarcastic Retard]

View attachment 59934

from where u got this pic?

 

[ashcull14]

from where u got this pic?

http://www.slideshare.net/KasturiNaidu/14-rate-of-reaction12dbiology
slide 26/74

 

[The Sarcastic Retard]

(

http://www.slideshare.net/KasturiNaidu/14-rate-of-reaction12dbiology
slide 26/74

 

[holoholo]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf
question 5 c(iv). Why can't we add a side chains ? Doesn't this increase van der waals force and so it becomes more rigid ?

 

[sfhn_128]

First write the equations of combustion of both methane and ethane.
CH4+202-->C02+2H20 ; so ratio of volumes is the same as the ratio of moles, ratio of CH4 to CO2 is 1:1 so 10cm^3 of CH4 will make 10cm^3 of CO2.
C2H6+3.502-->2CO2 + 3H20 ; ratio of C2H6 to CO2 is 1:2 so 10cm^3 ethane will make 20cm^3 of CO2. Now add the amounts of CO2 10+20=30cm^3 so C is the ans.
(Potassium hydroxide is an alkali so only the acidic gas will be absorbed by it, O2 and H20 are neutral and CO2 is the only acidic gas that's why we only calculate the volume of CO2)

thanks!