Answer: the answer is C oxidizing
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!
Paraphrasing and summarizing
Answer:
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
Step-by-step explanation:
The unbalanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃
Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.
One way to simplify the balancing is to replace them with a single letter.
(a) For example, let <em>X = NO₃</em> and <em>Y =CrO₄</em>. Then, the equation becomes
PbX₂ + K₂Y ⟶ PbY + KX
(b) You need 2X on the right, so put a 2 in front of KX.
PbX₂ + K₂Y ⟶ PbY + 2KX
(c) Everything is balanced. Now, replace X and Y with their original meanings. The balanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
