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Paper 5 Tips !! :)

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no its alrite...i am just panicked you know a day before the exam...u don't need to be sorry...can u plzz elaborate what the difference is in serial dilution?

okay i'll try to clear out the difference thing is you can use c1v1=c2v2 in both
the difference being though, is that.. lets say you start out with a solution of say HCl of 10mol/dm^3
in serial dilution you take a volume of that solution say 10 cm^3 then you add 10cm^3 of water you have a solution now that is 5mol/dm^3 HCl with a total volume of 20 cm^3
now take 10cm^3 of the solution you just made ( 5mol/dm^3 ) and add 10 cm^3 of water again.. now you have a 20 cm^3 of 2.5 mol/dm^3 HCl.. basically in serial dilution you use the solution last you made and use some of it to make a new concentration..

while in normal dilution you take 2 beakers containing 10 cm^3 each, of 10 mol/dm^3 of HCl and in one you add 10 cm^3 of water and in the other you add 30 cm^3 of water..
the result is the same.. just the difference that in normal you use the original concentration
hope it helped
 
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Can someone please have a look at qn 3 a of this paper?
I don't get how the qn a iii is asking for mass of Cu with 1 g of O, and then suddenly, it's like Cu:O ratio?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_52.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_52.pdf

Thanx a lot

MaxStudentALevel xhizors knowitall10
@ all other p5 candidates!
I have the same problem.. Someone help ?
 
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Can anyone please tell me about how to calculate soulbility when we have the mass of water and mass of solid in grams? And how do we calculate molar concentration/molarity?
 
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me too stuck at this point :(
help required here

xhizors
Gémeaux
knowitall10
MaxStudentALevel

or any one?

JazakAllah :)
I didn't draw the graph myself so I'd just put a dash where values from graph are required.
mass of basic carboate is 221 (of CuCo3 and Cu(OH)2 )+ 18x (of H2O)
loss in mass = mass of CO2 + (x+1) H2O = 44 + 18(x+1)

Take the ratio,
mass of basic carbonate : loss in mass
__(value from graph)___ : __(corresponding value from graph)___
221+18x : 44+ 18 (x+1)

Cross multiply and solve for x. :)
 
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Asslamu Alykum!
I'm going to explain it generally; try doing the question yourself then:)
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?
Can we use this in the may/june 2012 paper ?
 
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no its alrite...i am just panicked you know a day before the exam...u don't need to be sorry...can u plzz elaborate what the difference is in serial dilution?
It's perfectly okay :) I'm in ruins myself at the moment :p
This is in simplest of words the difference in them written on an other site.
In serial dilution you produce a series of dilutions, usually based on a dilution by 10. So, for example you might product 10% 1% 0.1% 0.01% and 0.001% solutions - each of which is 10 times weaker than the one before it.
You could produce such a dilution by taking 10 ml of the first solution and adding it to 90 ml of water. After mixing you take 10 ml of the second dilution and add it to 90 ml of water and so on....
Simple dilution is just that - a dilution of one solution to make a weaker one.
 
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I'm unable to find it D=
Dude..we've explained much of all these questions there... you might find it somewhere.. if not then please gimme a moment, i'll help you out.. sooryto bother u, but can u post ur doubts agian please?
Sorry for bothering you dude...
 
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