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
Answer:
<u><em>Pentane </em></u>
Explanation:
since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it 
In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m .
What is rate of a reaction?
The speed at which a chemical reaction takes place is the rate of the reaction. It is the concentration change per unit time of a reactant in a reaction.
Since the concentration of NO reduces to half its initial concentration in 85 seconds that is from 1.12m to 0.520m, it can be said that 85 seconds is the half life interval for the reaction, <u>Hence on average, </u><u>half reaction</u><u> is completed in the time interval of </u><u>85 seconds</u><u>.</u>
To learn more about rate of a reaction from the given link below,
brainly.com/question/12172706
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Answer:thermal energy
Explanation:Thermal radiation is the emission of electromagnetic rays from all matter that’s greater then zero
Answer: 281 hours
Explanation:-
1 electron carry charge=
1 mole of electrons contain=
electrons
Thus 1 mole of electrons carry charge=

of electricity deposits 1 mole or 63.5 g of copper
0.0635 kg of copper is deposited by 193000 Coloumb
11.5 kg of copper is deposited by=
Coloumb

where Q= quantity of electricity in coloumbs = 34952756 C
I = current in amperes = 34.5 A
t= time in seconds = ?


Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.