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Practical p33 physics nd chemistry !!!!
- Thread starter QWERTY1
- Start date
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!
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!
- Messages
- 380
- Reaction score
- 16
- Points
- 28
hydrated ammonim salt titrated with KMnO4 THIS REACTION RESULTS IN AN EXPLOSION HOW MANY TIMES DO I WRITE THIS OVER HERE..................
- Messages
- 380
- Reaction score
- 16
- Points
- 28
I dont wil u solve it n paste answers if u dont mind....
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!!!!
- Messages
- 380
- Reaction score
- 16
- Points
- 28
ohhh sorrieee i did it bymistake......................
http://notezone.net/cambridgechem/chemi ... sis%5D.pdf below answer go page 11
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
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