paper 32

[plastation4]

help please whats comming for chemistry paper 32 physics paper 32 and biology paper 32.

 

[aloona994]

i asked about what might come up in chemistry p32 and physics p32 couple of times but no one answered
is the 32 the same as 33???

 

[zahraahmed]

hydrated ammonim salt titrated with KMnO4 THIS REACTION RESULTS IN AN EXPLOSION HOW MANY TIMES DO I WRITE THIS OVER HERE..................

 

[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

 

[plastation4]

hydrated ammonim salt titrated with KMnO4 THIS REACTION RESULTS IN AN EXPLOSION HOW MANY TIMES DO I WRITE THIS OVER HERE..................

is this whats coming more info please.

 

[plastation4]

http://notezone.net/cambridgechem/chemical%20calc%5B1%5D.Problems%5Bvolumetric%20analysis%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
which paper is this

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

 

[filza94]

why u copied dix huh....!!!!