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
Make sure that there an exit and no over head power lines and wate for a backup
Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
Answer:
9.5X10^23 molecules N2 X (2 molecules NH3 / 1 molecule N2) = 1.9X10^24 molecules NH3
Explanation:
.