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Paper 5 Tips !! :)

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ohh since ur axis is diff frm mine, im nt rely sure hw to do it. bt fr my graph ths is hw i gt it...my x axis ws mass of carbonate n y- mass of CuO
so u gt a point in da graph n read its values ... mine were x-4.4 y-3.05
frst find the moles of CuO. (mr=79.5) do divide 3.05/79.5= 0.038
consider ratio of CuO : CuCo3
2 :1
so 0.038---> x
find x. i.e 0.038/2=0.0192 of mols of carbonate. usin ths fnd the mr of the crbonate i.e. mols=mass/mr, calculated value will b 229.4
nw to fnd da mr of the CuCO3.Cu(OH)2.xH2O, mr of ths= 221+18x
221+18x=229.4
x=0.46 whch is appro=0.5

i knw its a mess, bt hope u gt it sumhw..i rely dnt knw hw to do fr ur graph!
Okay heyy.. another question from o/n 08... i always get stuck in these calculations in q2 :/
its the part (e) this time.
here is the link.
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_5.pdf
 
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Wallah guys..im seriously scared after reading some of the posts... ..
im givin it for the 2nd time...please pray.i really wana score good....
 
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Can somebody pls explain me serial dilution making and concentration related question. Vl b very thankful. Plzz
Asslamu Alykum Sis.
You use the formula C1V1=C2V2.
The total volume of your solution is to be kept constant, say 50cm3. And you are to make dilutions of a given concentration, say 2.00 moldm3.
So you first decide which concentrations you want to make. marking schemes usually prefer a minimum of five concentrations, so:
1.00, 1.20, 1.40, 1.60, 1.80, and 2.00.
If you're going to make the 1.00moldm3 conc, look at the following steps:
C1V1=C2V2
C1= 2.00 (the concentraiton you're given.)
V1= x (what the volume of the new concentration will be)
C2= 1.00 (the conc you're planning to make)
V2= 50 (the total volume)
So, 2x= 1 x 50
x= 25
So now that your V2 is 25, to know what volume of water you must add to make it up to that conc, 5o-25= 25.
So your procedure to make 1.00 moldm3 from the given 2.00moldm3 would be to add 25cm3 of water into 25cm3 of acid.
Got it?
 
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i just want to be sure if i am doing the serial dilutions correctly
Lets say i have 2 moldm^-3 of and acid...
and i am making my total volume constant to 100 cm^3
I want a solution of acid with concentration 1 moldm^-3
So in this case will case i will add 50 cm^3 of distilled water and 50cam^3 of acid
now lets say i want to make a concentration of 0.5 moldm^-3
so i will add 25 cm^3 of acid n 75 cm^3 of distilled water
(Please tell me ASAP...if i am correct or not)
JazakAllah!
see use the equation CV = C1V1 suppose now u want to prepare a solution of 1M from ur 2M u will use it that way

2V = 1(100)
v = 50 cm^3 so u will need a volume of 50cm^3 acid and 50cm^3 of water. now suppose that u need to prepare a solution of 0.5M from ur 2M solution u will say

2V = 0.5(100)
V = 0.5(100)/2
V= 25 cm^3 of acid..now fill the rest with water so u will need 75cm^3
 
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Asslamu Alykum Sis.
You use the formula C1V1=C2V2.
The total volume of your solution is to be kept constant, say 50cm3. And you are to make dilutions of a given concentration, say 2.00 moldm3.
So you first decide which concentrations you want to make. marking schemes usually prefer a minimum of five concentrations, so:
1.00, 1.20, 1.40, 1.60, 1.80, and 2.00.
If you're going to make the 1.00moldm3 conc, look at the following steps:
C1V1=C2V2
C1= 2.00 (the concentraiton you're given.)
V1= x (what the volume of the new concentration will be)
C2= 1.00 (the conc you're planning to make)
V2= 50 (the total volume)
So, 2x= 1 x 50
x= 25
So now that your V2 is 25, to know what volume of water you must add to make it up to that conc, 5o-25= 25.
So your procedure to make 1.00 moldm3 from the given 2.00moldm3 would be to add 25cm3 of water into 25cm3 of acid.
Got it?
damn i guess i was late in explaining the equation..ehmm someone posted it faster than me :p
 
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i just want to be sure if i am doing the serial dilutions correctly
Lets say i have 2 moldm^-3 of and acid...
and i am making my total volume constant to 100 cm^3
I want a solution of acid with concentration 1 moldm^-3
So in this case will case i will add 50 cm^3 of distilled water and 50cam^3 of acid
now lets say i want to make a concentration of 0.5 moldm^-3
so i will add 25 cm^3 of acid n 75 cm^3 of distilled water
(Please tell me ASAP...if i am correct or not)
JazakAllah!
I think that would be normal dilution. For serial dilution it must be that for 0.5mol/dm^3, you take 50cm^3 distilled water and 50cm^3 of the previously diluted solution (i.e. 1 mol/dm^3).
 
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I think that would be normal dilution. For serial dilution it must be that for 0.5mol/dm^3, you take 50cm^3 distilled water and 50cm^3 of the previously diluted solution (i.e. 1 mol/dm^3).
waittt...u got me confused now...!!!...what do u mean?...that my method is wrong??...
 
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Can some one list down examples of the reasons when the anomalous point is befor or after what reasons do we give??
 
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just for the note... where will the water be collected??.. will it stay in the tube???
that's what i assumed :/ coz the ms didn't specify :/ i dunno if anyone else has a different way to collect the water, please do suggest....
 
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Yeah exactly,i did get two curves and when i checked with the marking scheme, it says that the first line (at lower temp) has a steeper gradient and vice versa while i got the oppposite. Is the transition temp around 49C?? Plus all my solubilities were above 100! Is that corrct??
solubility cant be above 100g because here we calculating solubility in 100g of water u might have did something wrong
what are your column headings??
 
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