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PLEASE CONFIRM CHEM P33 DETAILS

rickyponting

rickyponting

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Hello people. The exam is just under 14 hours away. Can someone reliable PLEASE confirm the details of CHEM P33. So many people are saying so many different things that its really confusing. Anyone care to shed some light?! PLEASE!
 
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Anonymousx3

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zahraahmed said:
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!!!!
Click to expand...

I've tried to work out some of the questions and this is what I got:

1 d. (i) 1.26g
1 d. (ii) 1.04g
1 e. 0.058 mol
1 f. 8.29 x 10 ^ -3 mol
1 g. 152
1 h. 56 (Fe - iron)

I'm not sure if all of these are right so don't rely on them too much.
If you don't understand anything, let me know, I'll try and help you!
(If anyone thinks my answers are wrong, please correct them!)
 
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filza94

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n telme how did u solve for each question wat values u took plzz help me if u dont mind...:(
 
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filza94

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yes cox i hate ppl who copyng me so i get angry n abuse get it punch...........:p
 
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Anonymousx3

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filza94 said:
n telme how did u solve for each question wat values u took plzz help me if u dont mind...:(
Click to expand...

I don't mind. :)

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 gives 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

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 - 96 = 56

There you go! I hope this helped you! :)
 
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Anonymousx3

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hassam said:
dont use 152.......use the correct value 152.7..calculator anser
Click to expand...

Okay, thanks for letting me know - I have problems with using the right amount of significant figures.
 
rickyponting

rickyponting

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thanks =) but are you sure that the volumetric analysis question on pg11 is a legitimate guess of a probable question? As of yet, no one actually answered my question, what are the probable questions for tomorrow's exam?
 
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leeadam1234

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zahraahmed said:
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!!!!
Click to expand...


Do u have the complete answer .. ??
pls attach it .. thx =)
 
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filza94

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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
 
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