Answer:30 L
Explanation:
Initial Volume
=
V
1
=
60
l
i
t
e
r
Initial Temperature
=
T
1
=
546
K
Final Temperature
=
T
2
=
273
K
Final Vloume
=
V
2
=
?
?
Sol:-
Since the pressure is constant and the question is asking about temperature and volume, i.e,
V
1
T
1
=
V
2
T
2
⇒
V
2
=
V
1
⋅
T
2
T
1
=
60
⋅
273
546
=
60
2
=
30
l
i
t
e
r
⇒
V
2
=
30
l
i
t
e
r
Hence the new volume of the gas is
30
l
i
t
e
r
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!
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