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!
Answer:
(c) 
Explanation:
The solubility product of a solid is the amount of solid dissociates into its respective ions in the solution. Thus more the value of the Ksp, the more is the salt soluble in the solvent.
So, Given that:-
The salt having highest value of Ksp is AgCN. So, it is most soluble.
The answer should be:
KOH (aq) + HCl (aq) --> KCl (aq) + H20 (l).
KOH is a base, because OH can accept a H+.
HCl is an acid because it can donate a H+.
In general, bases are : OH-, and acids are : H+.
It's made up of two different metals bonded together, which expand by different amounts as they heat up. As the temperature changes, the bimetallic strip curves more or less tightly (contracts or expands) and the pointer, attached to it, moves up or down the scale.
