Answer:
The turnover number is the maximum substrate quantitiy converted to product per enzyme and per second. It can be calculate as follows:
![k_c = \frac{V_{max}}{[E]}](https://tex.z-dn.net/?f=k_c%20%3D%20%5Cfrac%7BV_%7Bmax%7D%7D%7B%5BE%5D%7D)
with ![[E]](https://tex.z-dn.net/?f=%5BE%5D)
active enzyme concentration.
In this case we have Vmax an data to calculate
![[E]=\frac{3.38*10^{-14}}{0.01 L}=3.38*10^{-12} M](https://tex.z-dn.net/?f=%5BE%5D%3D%5Cfrac%7B3.38%2A10%5E%7B-14%7D%7D%7B0.01%20L%7D%3D3.38%2A10%5E%7B-12%7D%20M%20)
Now 
it is not like any option.
If we assume that 
have the non usual units of 
and it is 
So we need divide by the moles of E (in place of [E])
Now 
(pass from 
to 
dividing by 60)
The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
There are a number of
ways to express concentration of a solution. This includes molarity. Molarity
is expressed as the number of moles of solute per volume of the solution. The
concentration of the solution is calculated as follows:
<span> </span><span>Molarity = 15.5 g NaOH (1 mol NaOH / 40 g NaOH) / .250 L
solution</span>
<span>Molarity = 1.55 M</span>
