Answer:
is the maximum velocity of this reaction.
Explanation:
Michaelis–Menten 's equation:
![v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7BK_m%2B%5BS%5D%7D%3Dk_%7Bcat%7D%5BE_o%5D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7BK_m%2B%5BS%5D%7D)
![V_{max}=k_{cat}[E_o]](https://tex.z-dn.net/?f=V_%7Bmax%7D%3Dk_%7Bcat%7D%5BE_o%5D)
v = rate of formation of products =
[S] = Concatenation of substrate
![[K_m]](https://tex.z-dn.net/?f=%5BK_m%5D)
= Michaelis constant
= Maximum rate achieved
= Catalytic rate of the system
![[E_o]](https://tex.z-dn.net/?f=%5BE_o%5D)
= Initial concentration of enzyme
We have :
![[S]=0.110 mol/dm^3](https://tex.z-dn.net/?f=%5BS%5D%3D0.110%20mol%2Fdm%5E3)
![v=V_{max}\times \frac{[S]}{K_m+[S]}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7BK_m%2B%5BS%5D%7D)
![1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}](https://tex.z-dn.net/?f=1.15%5Ctimes%2010%5E%7B-3%7D%20mol%2Fdm%5E3%20s%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B0.110%20mol%2Fdm%5E3%7D%7B%5B%280.045%20mol%2Fdm%5E3%29%2B%280.110%20mol%2Fdm%5E3%29%5D%7D)
![V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D%5Cfrac%7B1.15%5Ctimes%2010%5E%7B-3%7D%20mol%2Fdm%5E3%20s%5Ctimes%20%5B%280.045%20mol%2Fdm%5E3%29%2B%280.110%20mol%2Fdm%5E3%29%5D%7D%7B0.110%20mol%2Fdm%5E3%7D%3D1.620%5Ctimes%2010%5E%7B-3%7D%20mol%2Fdm%5E3%20s)
is the maximum velocity of this reaction.
It's a weak base bacause H C N is weak
Answer:
There are 188.96 gm of moles in NaF
The solution that will have the lowest freezing point is 5.0 SODIUM CHLORIDE.
Adding solute to solvents usually result in the depression of the freezing point. The higher the quantity of the solute that is added, the lower the freezing point of the solution.
Option C: elements produce spectra with only few distinct lines.
The spectra are not continuos and are different for every element. This permits to identify elements.
