Answer:
5 mg
Explanation:
If one half life is 4 hours, then 3 half lives is 12 hours.
This means that the sample will decay to 1/8 of its original amount.
So, the answer is 40(1/8) = 5 mg.
Answer:
1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. 6 moles of Cl2
Explanation:
1. The balanced equation for the reaction. This is illustrated below:
4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.
This is illustrated below:
From the balanced equation above,
4 moles of FeCl3 reacted with 3 moles of O2.
Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.
Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:
Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.
From the balanced equation above,
4 moles of FeCl3 will react to produced 6 moles of Cl2.
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I think you divide something
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
