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!
a is the answer because all of the other answers are wrtong
1. Complete ionization in water.
2. Ionization constant.
3. A good hydrogen-ion acceptor.
4. Weak acid.
5. This base ionizes slightly in aqueous.
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<u>Answer:</u> The number of moles of gas present is 0.276 moles
<u>Explanation:</u>
To calculate the number of moles of gas, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of the gas = 725 mm Hg
V = Volume of the gas = 7.55 L
n = number of moles of gas = ?
R = Gas constant = 
T = Temperature of the gas = 
Putting values in above equation, we get:
Hence, the number of moles of gas present is 0.276 moles
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