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chem p33
- Thread starter aashish
- Start date
hey guyz recent info i got for u ppl for chemistry ppr 33
my class fellow get information that FA1 is ammonium iron salt for heating, Q2 is reaction KMnO4 with KI in the presence of H2SO4, then titration with Na2S2O3...and Q3 contain Ammonium Bromide and zincsulfate salts.....
my class fellow get information that FA1 is ammonium iron salt for heating, Q2 is reaction KMnO4 with KI in the presence of H2SO4, then titration with Na2S2O3...and Q3 contain Ammonium Bromide and zincsulfate salts.....
- Messages
- 48
- Reaction score
- 0
- Points
- 16
filza94 said:
filza94
You just uploaded my informations......
plz dnt confuse ppl...i told you that these are not sure.....
- Messages
- 380
- Reaction score
- 16
- Points
- 28
this is what the examiners say abt titrations:
Titrations were generally performed well. Burette readings for “accurate” titrations were recorded
to 2 decimal places (nearest 0.05 cm as required by the syllabus). The Examiners were pleased
to see that few candidates recorded “impossible” burette readings such as 27.43 cm
The majority of candidates produced consistent titres as described in the syllabus (2 titres within 0.10 cm3
. Many candidates, having obtained two titres within 0.10 cm wasted time by performing
further titrations: 3, or even 4, identical titres or titres within 0.10 cm3 was not unusual
The selection of titres for calculation of the “average” was less successfully performed. Many
candidates ticked only one titre. In this case Examiners accepted the candidate’s chosen value
when assessing accuracy. The difference between the chosen value and the next nearest was
used to calculate spread, and a penalty applied if necessary.
Titrations were generally performed well. Burette readings for “accurate” titrations were recorded
to 2 decimal places (nearest 0.05 cm as required by the syllabus). The Examiners were pleased
to see that few candidates recorded “impossible” burette readings such as 27.43 cm
The majority of candidates produced consistent titres as described in the syllabus (2 titres within 0.10 cm3
. Many candidates, having obtained two titres within 0.10 cm wasted time by performing
further titrations: 3, or even 4, identical titres or titres within 0.10 cm3 was not unusual
The selection of titres for calculation of the “average” was less successfully performed. Many
candidates ticked only one titre. In this case Examiners accepted the candidate’s chosen value
when assessing accuracy. The difference between the chosen value and the next nearest was
used to calculate spread, and a penalty applied if necessary.
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
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