http://www.youtube.com/watch?v=vKaIZIHaenU
enjoy guyx....mohr salt.......and other is about XSO4.7H2O....WE have to find X
enjoy guyx....mohr salt.......and other is about XSO4.7H2O....WE have to find X
-
Prepare by Xtremepapers!
🚀 NEW from the xtremepape.rs team: AI exam prep — 120,000+ worked solutions, and it marks your handwritten working from a photo. Sign in with your forum account & try it free → prepare.xtremepape.rs
video for q1 p33 chemistry
- Thread starter hassam
- Start date
- Messages
- 380
- Reaction score
- 16
- Points
- 28
THANKS..
http://notezone.net/cambridgechem/chemi ... sis%5D.pdf go page 11 below answers
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