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:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
Answer:
0.3192 M
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5.32 mL Molarity of stock solution (M1) = 6 M
Volume of diluted solution (V2) = 100 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
6 × 5.32 = M2 ×100
31.92 = M2 × 100
Divide both side by 100
M2 = 31.92 / 100
M2 = 0.3192 M
Therefore, the molarity of the diluted solution is 0.3192 M.
Answer:
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