Desperate help in Chemistry 9701 paper 5

[EmptyAxis]

what a coincedence i was doin this paper too today!!!

im still not sure if this is the correct way

x axis - mass CuO residue
y axis - mass of basic carbonate

so gradient = use value of student 1 = 0.6/0.87 = 20/29

so 20/29 = 2(63.5+16)/(221+18x)

solve for x = 0.53055555 approx = 0.5

 

[Youme&I]

what a coincedence i was doin this paper too today!!!

im still not sure if this is the correct way

x axis - mass CuO residue
y axis - mass of basic carbonate

so gradient = use value of student 1 = 0.6/0.87 = 20/29

so 20/29 = 2(63.5+16)/(221+18x)

solve for x = 0.53055555 approx = 0.5

But dont u have to draw lines in the graph and takes the gradient from the graph itself. Not from the table?

 

[Youme&I]

what a coincedence i was doin this paper too today!!!

im still not sure if this is the correct way

x axis - mass CuO residue
y axis - mass of basic carbonate

so gradient = use value of student 1 = 0.6/0.87 = 20/29

so 20/29 = 2(63.5+16)/(221+18x)

solve for x = 0.53055555 approx = 0.5

But dont u have to draw lines in the graph and takes the gradient from the graph itself. Not from the table?

According to the marking scheme ' Construction (lines) on graph and values read from the graph (with units) ' So I m not sure too whether is correct or not :S
Can anyone help us ?

 

[EmptyAxis]

if u construct a straight line passing through the origin that point would be on the line!!!!!

 

[diwash]

look u have plotted mass loss on y-axis and the mass of anhydrous oxide on x-axis
then gradient gives ....mass loss/ Mr
mass loss/ Mr = (Massof Co2 + mass of water vapour)/ Mr .......as solid residue only has mass
gradient= ((12+16*2)+(m+1) *18)/(Mr)
0.2719 = (44+ (m+1)18) /( 221+18x)
i think the gradient u have used is inappropriate i have done it and found the gradient 0.3125
therefore
0.3125 *(221+18x)= 44+18(x+1)
69.06+5063x = 62 +18x
12.37x = 7.06
X = 0.57....
approximately 0.5...

 

[Youme&I]

if u construct a straight line passing through the origin that point would be on the line!!!!!

Hahaha ok ok ... we dont usually do that wayy at least now I know !

 

[Youme&I]

look u have plotted mass loss on y-axis and the mass of anhydrous oxide on x-axis
then gradient gives ....mass loss/ Mr
mass loss/ Mr = (Massof Co2 + mass of water vapour)/ Mr .......as solid residue only has mass
gradient= ((12+16*2)+(m+1) *18)/(Mr)
0.2719 = (44+ (m+1)18) /( 221+18x)
i think the gradient u have used is inappropriate i have done it and found the gradient 0.3125
therefore
0.3125 *(221+18x)= 44+18(x+1)
69.06+5063x = 62 +18x
12.37x = 7.06
X = 0.57....
approximately 0.5...

Thanks diwash, I realize that my graph I plotted was slightly wrong, i got my gradient 0.31627 around 0.6 approx 0.5 so should be correct. Thanks!

 

[Youme&I]

diwash pm me your yahoo address

 

[Boss.Mentalist.Awesome]

i don't understand Oct nove 2009 P 52 Q 1 (d)....can some one please explain it to me...i am really confused... n my teacher isn't that amazing...so please answer me ASAP!!!​

 

[xxfarhaxx]

i don't understand Oct nove 2009 P 52 Q 1 (d)....can some one please explain it to me...i am really confused... n my teacher isn't that amazing...so please answer me ASAP!!!​

i have the same doubt too..it just dsnt make senss!! well i thought since the weak acids dont dissociate fully their cocn is suppose to b double n fr the strong acis its half!! but its not..the mark scheme is not amazing at all!!!

 

[Balkhyyy]

Re: Desperate help in Chemistry 9701 paper 5 (still no repli

u may have plotted
mass of Cuo on y-axis
and mass of anhydrous oxide on x-axis

the
gradient= (mass of Cuo)*2 / (Mr) .....according to the chemical eqn given
and Mr = 221+ 18x
suppose gradient is 20/29
...20/29 = mass of Cuo *2/Mr
20(221+18x) = 29((63.5+16)*2)
..X= 0.5

If u have plotted mass of cuo(residue) on x- axis and mass of anhydrous oxide on y-axis..reverse the ratio..
hope this helped..

Hey I hope you do reply.. Why do we find the gradient? I mean what does the gradient show?

 

[maryam fatima]

How to calculatethe relative formula mass of the basic copper(II) carbonate and the value of x in CuCO3.Cu(OH)2.xH2O by using data from the graph.

It is from Chemistry A level 9701 June 08 paper 5, Question no 2(f).

Please help me as soon as possible , I just dont understand how to find it by using the data from the graph.

i've spent half of my alevel duration in understanding this question