Answer:
(2) Half of the active sites are occupied by substrate.
Explanation:
The Michaelis–Menten equation is the rate equation for a one-substrate enzyme-catalyzed reaction. It is an expression of the relationship between the initial velocity V₀ of an enzymatic reaction, the maximum velocity Vmax, and substrate concentration [S] which are all related through the Michaelis constant, Km.
Mathematically, the Michaelis–Menten equation is given as:
V₀ = Vmax[S]/Km + [S]
A special relationship exists between the Michaelis constant and substrate concentration when the enzyme is operating at half its maximum velocity, i.e. at V₀ = Vmax/2
substituting, Vmax/2 = V₀ in the Michaelis–Menten equation
Vmax/2 = Vmax[S]/Km + [S]
dividing through with Vmax
1/2 = [S]/Km + [S]
2[S] = Km + [S]
2[S] - [S] = Km
[S] = Km
Therefore, when the enzyme is operating at half its maximum velocity, i.e. when half of the active sites are occupied by substrate, [S] = Km
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
Uranus's atmosphere is mostly composed of hydrogen and helium
Answer:
pH = 2.462.
Explanation:
Hello there!
In this case, according to the reaction between nitrous acid and potassium hydroxide:
It is possible to compute the moles of each reactant given their concentrations and volumes:
Thus, the resulting moles of nitrous acid after the reaction are:
So the resulting concentration considering the final volume (20.00mL+13.27mL) is:
In such a way, we can write the ionization of this weak acid to obtain:
So we can set up its equilibrium expression to obtain x as the concentration of H3O+:
Next, by solving for the two roots of x, we get:
Whereas the correct value is 0.003451 M. Finally, we compute the resulting pH:
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