Answer:
The percentage efficiency of the electrical element is approximately 82.186%
Explanation:
The given parameters are;
The thermal energy provided by the stove element, = 3.34 × 10³ J
The amount thermal energy gained by the kettle, = 5.95 × 10² J
The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;
Therefore, we get;
The percentage efficiency of the electrical element, η% ≈ 82.186%.
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