JUNE 07 CHEM P5 HELP!

[Youme&I]

I need help asap for question 2(e) in June 07 Chem p5, I have already drawn the graph however I just dont know how to find 'x' ...

http://www.xtremepapers.me/CIE/Internat ... 7_qp_5.pdf

 

[filza94]

FOR Chemistry PAPER 33!!!!!!!!!!

hydrated ammonim salt titrated with KMnO4
usual salt analysis
one thermal question

We have to titrate ammonium nitrate with KMn04 or would we need to oxidize potassium iodide in the

presence of ammonium nitrate and then titrate the resulting solution with sodium thiosulphate? :|

see question no.1 in june2008-31
a similar question is expected!

 

[zahraahmed]

ohhh sorrieee i did it bymistake......................

 

[filza94]

http://notezone.net/cambridgechem/chemi ... sis%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!!!!

 

[xarak]

I need help asap for question 2(e) in June 07 Chem p5, I have already drawn the graph however I just dont know how to find 'x' ...

http://www.xtremepapers.me/CIE/Internat ... 7_qp_5.pdf

ohkay i did a very very weird thing buh i guess i got the answer rite.first i found out the gradient.mass of chloride/mass of mercury. my answer came 0.40

thn i found number of moles of Hg which i calculated as 0.019 nd thn i did
0.019 x Mr divided by 0.019 x 201 equals 0.4 (the gradient)
my answer came out as 80.86 showin the total Mr is 80.86 so if one Cl is 35.5 so 80.86/35.5 which equals 2
i think this is it, it makes abosolutely no sense! if ny1 noes an easier way plz help

 

[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