It equals 2000 but in this case it would be 20,000mm3
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.
Answer:
The closest shell to the nucleus is called the "1 shell" (also called the "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus.
Answer: Option (2) is the correct answer.
Explanation:
Atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. So, it contains only 2 orbitals which are closer to the nucleus of the atom.
As a result, the valence electrons are pulled closer by the nucleus of oxygen atom due to which there occurs a decrease in atomic size of the atom.
Whereas atomic number of sulfur is 16 and its electronic distribution is 2, 8, 6. As there are more number of orbitals present in a sulfur atom so, the valence electrons are away from the nucleus of the atom.
Hence, there is less force of attraction between nucleus of sulfur atom and its valence electrons due to which size of sulfur atom is larger than the size of oxygen atom.
Thus, we can conclude that the oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
