What is the second most abundant isotope of oxygen
1 answer:
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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
(a) 33.6 L of oxygen would be produced.
(b) 106 grams of would be needed
1 mole of gas = 22.4 L
(a) From the equation, 2 moles of produce 3 moles of
. 1 mole of
will, therefore, produce 1.5 moles of
.
1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L
(b) 22.4 L of is produced at STP. This means that 1 mole of the gas is produced.
From the equation, 1 mole of requires 1 mole of
.
Molar mass of = (23x2)+ (12)+(16x3) = 106 g/mol
Mass of 1 mole = 1 x 106 = 106 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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