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
Bromine attracts electrons more strongly. Cesium is In fact the least electro negative element.
Sodium is more likely to lose an electron because is is less electro negative. Strong electronegativity make the element want more electrons. Sodium has loose electrons with a lower electronegativity so it gives it up easier.
Given :
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
To Find :
The volume of the gas after it is heated.
Solution :
Since, their is no information about pressure in the question statement let us assume that pressure is constant.
Now, we know by ideal gas equation at constant pressure :

Hence, this is the required solution.
The answer is a. hemoglobin is the molecule that transports oxygen in red blood cells :))))
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