Chemistry p33

[Feelguud77]

There are two methods specified in Q.1 of the paper. If you read it youll know!
And yeah, Hassam is right, Mohr's salt is coming, and you probably have a little dilution in the mix as well!!! This is kinda from the confidentials
And the salts MAY include Br^- , NH^4,

 

[umerkhadija]

hmm thanks 4 telling me..

r u sure abt thss....

 

[filza94]

maybe...

 

[ahmedrik1501]

http://notezone.net/cambridgechem/chemical%20calc%5B1%5D.Problems%5Bvolumetric%20analysis%5D.pdf can sum1 go in dix page 11 n do solve n gimme answers plzzzz i cant able to slove it n it might cum for exam help mee!!!!

i think you would have got till mass of water.... and mols of water...

if not then mols of water are 1.04/18 =0.057

(f) i hop it was this part troubling you.....
so if 7 mols of H2O are their in 1 mol of XSO4
then using ratio method

0.057/7 =8.14x10^-3

part G: Mr= mass/mol
hence 1.26/8.14x10^-3
=154.79
(h) : 155-(32+16*4)
=59.... hence it is close to Fe but right now u can say its Cu(II)SO4 but the best part its not ask to name the compound...

and for your info this question is from o/n/02 :O:

 

[filza94]

http://notezone.net/cambridgechem/chemi ... sis%5D.pdf plz go dix link pg 11 n c answers below of it

First of all you need the mass of FA1. In the first table you have the mass of the empty test tube and the

mass of the test tube + FA1. All you do is subtract the values 12.80 - 10.50 giving you 2.30g of FA1.

In the second table you need the mass of FA1 after heating. Again, you just subtract the values 11.76 -

10.50 = 1.26g FA1 after heating.

1 d. (i) The mass of anhydrous XS04 present in the crystals is what is left over after heating = 1.26g

1 d. (ii) The mass of water driven from the crystals is the loss in mass that has occurred when the FA1 was

heated. Before heating there were 2.30g of FA1 and after heating there were 1.26g. So the mass of water

lost is 2.30 - 1.26 = 1.04g

1 e. To calculate the number of moles of water in the sample of FA1 you use the formula no. of moles =

mass/mass of 1 mole which gives you 1.04/18 = 0.058 mol

1 f. The number of moles in XSO4
7 moles of H20 gives 1 mole XSO4 (from the formula XSO4.7H20)
So 0.058 moles of water give 0.058/7 = 8.29 x 10 ^ -3 moles of XSO4

1 g. To calculate the relative formula mass of XSO4 you use the formula mass of 1 mole = mass/no. of

moles which gives you 1.26/8.29 x 10 ^ -3 = 152.7

1 h. To calculate the Ar of element X you first find the Mr of SO4 which is 32 + (4 x 16) = 96
You subtract this from the RFM from part g giving you 152.7 - 96 = 56.7

 

[honeyDew]

@filza94
IS ths the complete experiment coming tomorrow? or just a guesssss ? :s

 

[filza94]

dis type of mean similiar question can come...

 

[filza94]

by masterboss365 » Wed May 11, 2011 5:08 pm
Tritration with KMnO4, sodium thiosulfate, potassium iodide
J07/32, J08/31, N08/32,

Heating related question. Find water of crystallisation. Heating (may be CuSO4 hydrated ) and then finding mass of water lost. Check for percentage of water present. J08/32, J09/32,

Salt anylsis
NH4+, Br-, NO3-, CO32-,
ammonium sulfate, zinc nitrate, ammonium bromide


Thanks,
Not sure 100%.