Here we have to write a simple equation which describes the action of the enzyme catalase.
The equation is: The concentration of the complex [ES] = ![\frac{[E]0}{1+\frac{Km}{[S]} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BE%5D0%7D%7B1%2B%5Cfrac%7BKm%7D%7B%5BS%5D%7D%20%7D)
Let us consider an enzyme catalyses reaction E + S ⇄ ES → E + P
Where E, S, ES and P are enzyme, substrate, complex and product respectively.
The concentration of the complex [ES] = , where 
is the Michaelis constant.
[E]₀ and [S] is the initial concentration of enzyme and concentration of substrate respectively.
9 x 3 = 27
27 moles of O reacted
27 / 2 = 13.5 O2 reacted
round up to 14 moles of O2
C. 343K
70 degrees C + 273 = 343
Because it is made up of sodium, a metal, and carbon, a nonmetal.
Hope I helped ♡
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
