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final final details abt practicals!
- Thread starter princesszahra
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Hey can som1 pm me abt bio ppr 3 plzz especially abt da slide !!
Hey all.. Yhe 4 2mrw We might have Urease... N Transverse Section of leaf.. Hopefully..
http://www.scienceinschool.org/print/607 a link realted to enzymes urease.....
Ok Duh we're getting urease & urea tomorrow in Bio 33, Now what wud be the Improvements, Sources of error & Limitations for the experiment below?
Enzyme Concentration
In this investigation, we will examine what happens to the rate of a reaction when the amount of enzyme is reduced. We will use urease, an enzyme that converts urea to ammonia. The ammonia causes the pH of the water to increase (it becomes more basic). You will be able to tell when a reaction occurs because the urea solution also contains a pH indicator that is becomes yellow in acid but turns red when the solution becomes basic.
The object of this experiment is to measure the amount of time it takes for the solution to turn red if less enzyme is used.
C1. Create a hypothesis regarding the the amount of urease and the rate of reaction of Urea.
C2. Obtain four test tubes and add 2 cm of urea to each.
C3. Label three of these tubes 1 through 3; the remaining tube will not be used; it will serve as a control.
C4. Have your lab partner start timing as you add 15 drops of urease to tube #1 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C5. Add 5 drops of urease to tube #2 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C6. Add 1 drop of urease to tube #3 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C7. Record your results in the answer sheet.
C8. Did using less enzyme produce a reaction?
C9. What was the effect of using less enzyme in your experiment? If your experiment did not work as expected, what should have happened?
C10. In general, what happens to the rate of reaction as the amount of enzyme is decreased?
C11. Do your results support your hypothesis? Explain.
http://www.scienceinschool.org/print/607 a link realted to enzymes urease.....
Ok Duh we're getting urease & urea tomorrow in Bio 33, Now what wud be the Improvements, Sources of error & Limitations for the experiment below?
Enzyme Concentration
In this investigation, we will examine what happens to the rate of a reaction when the amount of enzyme is reduced. We will use urease, an enzyme that converts urea to ammonia. The ammonia causes the pH of the water to increase (it becomes more basic). You will be able to tell when a reaction occurs because the urea solution also contains a pH indicator that is becomes yellow in acid but turns red when the solution becomes basic.
The object of this experiment is to measure the amount of time it takes for the solution to turn red if less enzyme is used.
C1. Create a hypothesis regarding the the amount of urease and the rate of reaction of Urea.
C2. Obtain four test tubes and add 2 cm of urea to each.
C3. Label three of these tubes 1 through 3; the remaining tube will not be used; it will serve as a control.
C4. Have your lab partner start timing as you add 15 drops of urease to tube #1 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C5. Add 5 drops of urease to tube #2 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C6. Add 1 drop of urease to tube #3 and then swirl the tube until it changes to a red color. Record the amount of time that it took for the urease to change to a red color.
C7. Record your results in the answer sheet.
C8. Did using less enzyme produce a reaction?
C9. What was the effect of using less enzyme in your experiment? If your experiment did not work as expected, what should have happened?
C10. In general, what happens to the rate of reaction as the amount of enzyme is decreased?
C11. Do your results support your hypothesis? Explain.
http://www.xtremepapers.me/CIE/Internat ... 3_qp_5.pdf realated to enzyme urease...
FOR Chemistry PAPER 33!!!!!!!!!!
hydrated ammonim salt titrated with KMnO4
usual salt analysis
one thermal question
We have to titrate ammonium nitrate with KMn04 or would we need to oxidize potassium iodide in the
presence of ammonium nitrate and then titrate the resulting solution with sodium thiosulphate? :|
see question no.1 in june2008-31
a similar question is expected!
hydrated ammonim salt titrated with KMnO4
usual salt analysis
one thermal question
We have to titrate ammonium nitrate with KMn04 or would we need to oxidize potassium iodide in the
presence of ammonium nitrate and then titrate the resulting solution with sodium thiosulphate? :|
see question no.1 in june2008-31
a similar question is expected!
- Messages
- 380
- Reaction score
- 16
- Points
- 28
ammonium nitrate+KMnO4 results in an explosion!!!!!!!!!!!!!!!!!!!!!!!!! confirm it before writing it everywhere!!!!
- Messages
- 380
- Reaction score
- 16
- Points
- 28
yes FeSO4 is coming!!! a hydrated salt ....................thts wht i heard!!
besides no electricity question in phys 33..........
besides no electricity question in phys 33..........
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
- 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