Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:

Rate factor in the presence of catalyst:

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:

![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;

Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;



The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Hey there!
The correct answer is C
Hope this helps you!
Always remember, you are a Work Of Art!
- Nicole :) <3
Answer:
Its atomic mass increases by 1
An isotope of that element is fotmed with mass differences by 1
Explanation:
Answer:
Explanation:
This is a direct application of the equation for ideal gases.
Where:
- P = pressure = 1.25 atm
- V = volume = 25.2 liter
- R = Universal constant of gases = 0.08206 atm-liter/K-mol
- T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
- n = number of moles
Solving for n:
Substituting:

Answer:
option 2 is correct answer. its nitrogen.