Answer:
The
of a substrate will be "10 μM".
Explanation:
The given values are:

![[Substract] = 40 \ \mu M](https://tex.z-dn.net/?f=%5BSubstract%5D%20%3D%2040%20%5C%20%5Cmu%20M)

Reaction velocity, 
As we know,
⇒ ![Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}](https://tex.z-dn.net/?f=Vo%3D%5Cfrac%7BK_%7Bcat%7D%5BE_%7Bt%7D%5D%5BS%5D%7D%7BK_%7Bm%7D%2B%5BS%5D%7D)
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
On subtracting "40" from both sides, we get
⇒ 
⇒ 
Calcium fluoride: CaF₂
Ca(2+) >>> Ar (argon)
F(-) >>> Ne (neon)
Water's specific heat capacity is 4200 J/Kg°C
95-28=67
72.5grams in kg is 0.0725kg
Energy = 67×0.0725×4200
Energy = 20,401.5 J or 20.4015 kJ
Answer:
What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
A. 0.0625
How does Q compare to Keq?
B. Q < Keq
Explanation:
they are made by of two types of elementary particles
electrons and quarks
quetion answered by
(jacemorris04)