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
The reaction is
CH3OH(l)→CH3OH(g)
To calculate ΔG°
ΔG°=ΔG°f of CH3OH(g)-ΔG°f of CH3OH(l)
=162.3kJ/mol - (-166.6kJ/mol
=4.3kJ/mol.
Hey there!
Given the reaction:
B + H⁺ => HB⁺
At half-equivalence point : [B] = [HB⁺]
=> [B] / [HB⁺] = 1
Henderson-Hasselbalch equation :
pH = pKa + log ( [B] ) / ( HB⁺)]
pH = 14 - pKb + log ( 1 )
pH = 14 - 7.95 + 0
pH = 6.05
Answer C
Hope that helps!
Answer: The student should have worn splash goggles, an apron, and gloves during the investigation.
Explanation: