Chemistry: Post your doubts here!

[My Name]

oh thank u so much

Welcome.

 

[Eugene99]

Is Manganese(II) sulfate same as manganese sulfate? of course manganese II sulfate has 2+ ions but I have a salt labelled just 'manganese sulfate', it's pale pink, and I need manganese(II) sulfate, could it be the same as that?

 

[Rizwan Javed]

Is Manganese(II) sulfate same as manganese sulfate? of course manganese II sulfate has 2+ ions but I have a salt labelled just 'manganese sulfate', it's pale pink, and I need manganese(II) sulfate, could it be the same as that?

Mn +2 salts should be white. Pink salts of manganese contain Mn+7, i guess.

 

[Eugene99]

Mn +2 salts should be white. Pink salts of manganese contain Mn+7, i guess.

no, it's very pale pink, not really pink. I just googled and see the one in the image:
https://www.google.com.pk/#q=manganese+sulfate
mine is exactly the same color
but it's labelled magnesium sulfate only and with the formula MnSo2 which is probably wrong MnSO2 is no salt xD, these labels aren't very authentic
do you think I can use it where I need manganese II sulfate?

 

[Rizwan Javed]

no, it's very pale pink, not really pink. I just googled and see the one in the image:
https://www.google.com.pk/#q=manganese+sulfate
mine is exactly the same color
but it's labelled magnesium sulfate only and with the formula MnSo2 which is probably wrong MnSO2 is no salt xD, these labels aren't very authentic
do you think I can use it where I need manganese II sulfate?

Lol, give it a try. If the color matches with that, it may be MnSO4. And thanks for correcting me that MnSO4 is pale pink in colour.

 

[Rizwan Javed]

It is there for the AS part.
So if you're doing AS or AL then you need to know it.


http://www.cie.org.uk/images/164502-2016-2018-syllabus.pdf

A Level material moved to AS material
10.4 (c) deduce the presence of a CH3CH(OH)– group in an alcohol from its reaction with alkaline aqueous iodine to form tri-iodomethane 10.5 (e) describe the reaction of CH3CO– compounds with alkaline aqueous iodine to give tri-iodomethane

Hi,

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

 

[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