Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;

If
&





E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
They function as catalysts, they transport and store other molecules such as oxygen, they provide mechanical support and immune protection, they generate movement, they transmit nerve impulses, and they control growth and differentiation
The mass of is K2CrO7 in the 250 ml of solution is mathematically given as
x=1.25mg of K2CrO7.
Hence, 250 ml of the solution will contain 1.25mg of K2CrO7
<h3>The mass of is K2CrO7</h3>
Question Parameters:
The total of potassium dichromate are there in a 250 mL bottle filled with 5 ppm solution
Generally the mass of k2CrO7 is mathematically given as

x=1.25mg of K2CrO7
Therefore,
250 ml of the solution will contain 1.25mg of K2CrO7
For more information on chemical Reaction
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