Chemistry: Post your doubts here!

[MaxStudentALevel]

where did u disappear ?

Sorry! Had to go out for something! I labelled X (mass of basic carbonate) and Y (mass of CuO)

 

[iKhaled]

Sorry! Had to go out for something! I labelled X (mass of basic carbonate) and Y (mass of CuO)

alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can

 

[MaxStudentALevel]

alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can

OHHHH GOT IT!!! Thanks! I just wasn't understanding how to use the gradient in it!!! Thank you!!!

 

[iKhaled]

OHHHH GOT IT!!! Thanks! I just wasn't understanding how to use the gradient in it!!! Thank you!!!

no problem

 

[MaxStudentALevel]

alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can

Could you help me with ON 2007 Q1 f also?? I don't understand the equation!!

 

[unseen95]

how to know if a substance is a reducing or oxidising agent. For example: how is CO a reducing agent.

 

[bital]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_22.pdf
please someone help in 1(c)(i)

 

[iKhaled]

how to know if a substance is a reducing or oxidising agent. For example: how is CO a reducing agent.

if a substance is oxidized then it is a reducing agent and if it is reduced then it is an oxidising agent

 

[iKhaled]

Could you help me with ON 2007 Q1 f also?? I don't understand the equation!!

i will need to read the whole question and do it and i am kind of busy now, i am sorry :/

 

[Alice123]

it is supposed to be +226

Thanks..... Chem exam tomoro so i might disturb u the whole day

 

[Shai_n]

to get the

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_22.pdf
please someone help in 1(c)(i)

to get the equlibrium for CH3CO2H, you will substract the initial amount whinc is 0.10 mol from the answre you had for b(iv) which is 0.04 giving you: 0.10-0.04=0.06

 

[MaxStudentALevel]

i will need to read the whole question and do it and i am kind of busy now, i am sorry :/

Aww that's fine. Could you help me before Friday?

 

[xhizors]

Reducing

how to know if a substance is a reducing or oxidising agent. For example: how is CO a reducing agent.

reducing agents are those ppl who like to reduce oxygens (Oxidation Guys)

 

[iKhaled]

Aww that's fine. Could you help me before Friday?

ofc!! because the next coming days i will be doing paper 5 mostly and i will need u and all the other paper 5 candidates to discuss stuff about it and doubts like how yesterday we kept discussing. i guess this will help us to score a good grade, right ?

 

[iKhaled]

Thanks..... Chem exam tomoro so i might disturb u the whole day

if its doubts about paper 2 you can disturb me the whole day !! its okay will do my best to answer your doubts

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_22.pdf

please someone help in 1(c)(i)

if you made your calculations above correctly then you will have 0.04 moles of NaOH that reacted with ethanoic acid which means at equilibrium we will have 0.1-0.04= 0.06 moles of ethanoic acid at equilibrium. acid to alcohol ratio is 1:1 so also 0.06 moles of alcohol left at equilibrium and 0+0.04 = 0.04 moles of water and sodium ethanoate al equil.

THIS IS THE LOGICAL WAY. i don't know why the mark scheme made the opposite anyone out here can explain ??

 

[Shai_n]

if its doubts about paper 2 you can disturb me the whole day !! its okay will do my best to answer your doubts

someone help me...please? chemisr=try is hard for me especially organic chemisrty

 

[Shai_n]

if you made your calculations above correctly then you will have 0.04 moles of NaOH that reacted with ethanoic acid which means at equilibrium we will have 0.1-0.04= 0.06 moles of ethanoic acid at equilibrium. acid to alcohol ratio is 1:1 so also 0.06 moles of alcohol left at equilibrium and 0+0.04 = 0.04 moles of water and sodium ethanoate al equil.

THIS IS THE LOGICAL WAY. i don't know why the mark scheme made the opposite anyone out here can explain ??

you mean what i said was wrong?

 

[iKhaled]

you mean what i said was wrong?

oh what did u say ?

 

[Shai_n]

i mean the calculation to number i c(i) ?
anyways any help with organic chemitry?