Answer:
1 : 1.5
Explanation:
First Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 35 = 35g
35g : 35g
1 : 1
Second Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 28 = 42g
28g : 42g
1 : 1.5
Further reducing it to make oxygen 1;
0.6667 : 1
ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds;
0.6667 : 1
1 : 1.5
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
They connect at the joint
That is FALSE because neutrons have no charge and as their name suggest are neutral particles.