Attending an in state college
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.
CO2<span> is a linear molecule and the Oxygen (O) atoms on each end are symmetrical. Polarity results from an unequal sharing of valence electrons. Because of this symmetry there is no region of unequal sharing and </span>CO2<span> is a</span>nonpolar<span> molecule</span>
