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chemistry p3 help

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is the correct way of making a table to record titre values of a titration. are we supposed to have the concurrent reading table?????
 
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No, I don't think you need the concurrent reading column.
I haven't read anything about it in the mark schemes nor have I been told about it in school.
The first four columns are right, though. You don't need the last column.
 
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Oh one more thing, shouldn't you be taking 3 titrations? The chart above is for two only.
 
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We were told to take three titrations. One is more than enough. And about the rest two increase the reading by about 0.1 to 0.2.
 
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my teacher told me take 1st reading as rough n then make 2 acurrate for titrationss
 
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hey can any of u help me to how tro represent data in chem oct nov p34 2010 ......Q2..............
screenshot will be a better option.............
 
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http://notezone.net/cambridgechem/chemi ... nalysis%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
 
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filza94 said:
my teacher told me take 1st reading as rough n then make 2 acurrate for titrationss

But it will take a lot of time to complete 3 titrations. I'd suggest to do one accurate one. Time is quite less. I've appeared before too.
 
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