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chemistry p5
- Thread starter zeebujha
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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 ... 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
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
Topic To Stress On Biology Paper 33 AS
Gaseous exchange
Infectious diseases
Immunity
Transport: In plants- all the pathways in the cell (apoplast n symplast)
in humans- cardiac cycle, pressure graph, etc...
Nitrogen cycle
Itx surely cuming as my Biology Teacher predicted them every year she does n it cumss last year 4 my O level also she predict so i got A* n dat all camee now she precdicted for AS Paper 33 so evryone study all topics anyway so focusing on these shouldn't make u forget the others!!!!
Gaseous exchange
Infectious diseases
Immunity
Transport: In plants- all the pathways in the cell (apoplast n symplast)
in humans- cardiac cycle, pressure graph, etc...
Nitrogen cycle
Itx surely cuming as my Biology Teacher predicted them every year she does n it cumss last year 4 my O level also she predict so i got A* n dat all camee now she precdicted for AS Paper 33 so evryone study all topics anyway so focusing on these shouldn't make u forget the others!!!!
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How to draw construction lines on the graph to find the no, of moles??may june 2007 question???/ plz helpp!)
depends on what you plotted in your graph.olenka1611 said:
I, for one, plotted mass of HgClx on X-axis and mass of mercury on the Y-axis. The slope of the graph gives nothing more than a more accurate ratio of the mass of Hg to HgClx.
The construct lines simply refer to a triangle that you would draw to find the slope of line in say physics P5. Draw two lines: one horizontal from one of the two points considered for the slope and a vertical line from the other point being used for the slope. They should be as far apart as possible (and yeah they should lie on the line as well).
You should find the two lines intersect to form a close right angled triangle with the line of best fit forming the hypotenuse.
Now, find the slope as you would normally do. Suppose you got the value of slope to be : 4.75/6.50 (It is BEST not to round off the value of slope when you need to use it for further calculations!). Now all that means for my graph is a mass of 6.50 g of HgClx gives 4.75 g of Hg on reaction.
Now:
Find the moles of Hg corresponding to 4.75 g of Hg ( 4.75/ 201). Again do NOT round off.
So, as 1 mole of HgClx should obviously give one mole of Hg, the answer is also the number of moles of HgClx.
But we already now that the mass of corresponding HgClx is 6.50 g
SO: (4.75/201) moles of HgClx weighs 6.50g. So, 1 mole of HgClx weighs (6.50x201/4.75)g = (1306.5/4.75) g. (STILL NO ROUNDING OFF)
FInally:
we now that one mole of HgClx consists of x mole of Cl and 1 mol of Hg
Therefor: 1x Ar of Hg + x x Ar of Cl= 1306.5/4.75
If you do the necessary calculations to solve for x, you will get x= 2.085 (rounded off answer)
However, the value of x has to be an integer (we can't have 2.085 atoms of Cl in a molecule). So, round it off to the closest integer value (which in this case obviously 2)
so, the formula is HgCl2.
Hope this helped 8)
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there is a possibility ov electochemistry question as well..! but itz bttr 2 n0 evrytn frm AL portion..! 