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
do you have anymore numbers to add to this problem??
Ethanoic acid ionizes in aqueous solutions to form two ions which are
and 
<h3>Ionization of ethanoic acid</h3>
Ethanoic acid goes by the chemical formula
.
In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single 
More on ethanoic acid can be found here: brainly.com/question/9991017
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1. The answer is option E, that is None of the above is correct.
As a polymer becomes more crystalline,
its melting point doesn't decreases, its density doesn't decreases, its stiffness doesn't decreases and its yield stress doesn't decreases.
2. The answer is option B, that is the molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
In the smectic A liquid-crystalline phase, molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
3. For a substitutional alloy to form, the two metals combined must have similar atomic radii and chemical bonding properties.