Chemistry Paper 4-theory- doubts =D

[Soldier313]

Assalamu Alykum wa Ra7matullahi wa Barakatuhu!!!
Al7amdulillah, we're done with the pathetic Paper 5 finally
So now we can start our Paper 4 practice... and we need to work extremely hard if we're really hoping for an A* inshAllah..
and the rest of it will be left to Allah (SWT), the One Who's Worthy of all Praise and Worship. If He Wills, nothing can ever stop us from succeeding!


Alright.. i don't know all the A-level Candidates on this forum, so i'm only tagging a few, but please brothers and sisters, tag fellow brothers and sisters if they do not know about this thread. Thank you, Jazakum Allahu Khair

Soldier313
Zari
Jiyad Ahsan
fadilah
xhizors
PhyZac
MaxStudentALevel
pearl angel
@everyone else...wow.. long list MashAllah

Wa alaykum salaam wr wb!
That's a great idea knowitall10 JazakAllah khair
So we need to start working asap guys.
If anyone's got short notes for any of the tough topics (like NMR, group trends as suggested above), i think we should collect them here.
And we could discuss all the hard qns inshaAllah
Hoping this thread will help us get amazing grades inshaAllah

 

[knowitall10]

hey guys i will join you after i am done with my mechanics paper on monday

inshAllah Good Luck!!!

 

[knowitall10]

how to find orders of reagents ?

Assalamu Alaykum
The electrode potential values... the more positive, the easier it is to reduce the ions on the left.. so the ion on the right is relatively unreactive and a relatively poor reducing agent.

 

[knowitall10]

Assalamu Alaykum
The electrode potential values... the more positive, the easier it is to reduce the ions on the left.. so the ion on the right is relatively unreactive and a relatively poor reducing agent.

I doubt i answered ur question

 

[Ariel Robert]

No it depends on their concentrations, comparing their concentration and rate of reaction you can find out if they are first second or third order...

 

[Ariel Robert]

No it depends on their concentrations, comparing their concentration and rate of reaction you can find out if they are first second or third order...

First order is when the concentration increase and the rate of reaction increases to a ratio of 1:1
2nd order 1:2
3rd order 1:3

 

[xhizors]

Assalamu Alaykum
The electrode potential values... the more positive, the easier it is to reduce the ions on the left.. so the ion on the right is relatively unreactive and a relatively poor reducing agent.

very bad u mixed electrochemistry with reaction kinetics

 

[xhizors]

You can find the order by either drawing a graph of rate against concentration and check ita proptionality initial rates method
or by rate determining facts
slow reactions reagants

 

[knowitall10]

very bad u mixed electrochemistry with reaction kinetics

yo man! u said order of reagent! not reaction!

 

[knowitall10]

What

You can find the order by either drawing a graph of rate against concentration and check ita proptionality initial rates method
or by rate determining facts
slow reactions reagants

What do u mean "slow reactions reagents"?

 

[knowitall10]

And guys/gals, this is not a test or anything, but can someone please explain what's the half equation of the half cell on the left (the one with the solid sulfer deposit) and please provide me with an explanation
Thank you very much!
Jazakum Allahu Khair

 

[knowitall10]

Another question: When we change the direction of a half equation, does the electrode potential value become negative if it were positive before?

 

[knowitall10]

no test... i swear...

 

[knowitall10]

LOL! Look at this

 

[Gémeaux]

Another question: When we change the direction of a half equation, does the electrode potential value become negative if it were positive before?

No it stays the same.

 

[knowitall10]

No it stays the same.

Ahhh.. thank God u answered!!! JazakAllah! thank u! Phew! I was dying already.. there's so much of the crappy syll to finish in 3 days!!!
But thanks man(or girl?)!! I really appreciate ur reply.. at least some1's awake at ths time

 

[Gémeaux]

And guys/gals, this is not a test or anything, but can someone please explain what's the half equation of the half cell on the left (the one with the solid sulfer deposit) and please provide me with an explanation
Thank you very much!
Jazakum Allahu Khair

The voltmeter shows that sulphur is on the +ve side, which means that it undergoes reduction (and hydrogen ions get oxidized).
Reduction is gain of electrons therefore the equation linking solid sulfur and sulfur ions becomes: S + 2e- ---> S^-2

 

[Gémeaux]

Ahhh.. thank God u answered!!! JazakAllah! thank u! Phew! I was dying already.. there's so much of the crappy syll to finish in 3 days!!!
But thanks man(or girl?)!! I really appreciate ur reply.. at least some1's awake at ths time

Ahaha you're welcome but 3 days??! I thought we still had seven days.

 

[knowitall10]

The voltmeter shows that sulphur is on the +ve side, which means that it undergoes reduction (and hydrogen ions get oxidized).
Reduction is gain of electrons therefore the equation linking solid sulfur and sulfur ions becomes: S + 2e- ---> S^-2

Ahaha you're welcome but 3 days??! I thought we still had seven days.

Hey man! Thanks a LOT!!!!! Jazak Allahu Khair!!!
Oh yeah, i planned on finishing the course off in 3 days thoroughly and then do some P4 butt kicking... although that snap shot i posted kinda discouraged me, lol

 

[knowitall10]

Hey guys/gals, i just figured out that the cell potential of a viable reaction must be positive... did u know that? I didnt.. that's news to me