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:
A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.
A disulfide bond links the two 80 kDa subunits (possibly identical).
Explanation:
Given that:
A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.
A disulfide bond links the two 80 kDa subunits (possibly identical).
As a result of SDS and dithiothreitol analysis treatment, the molecular masses can not be 360 in total. They are 280, which implies that they are in short of 80 kDa. This means that there are possibilities that two groups with a molecular mass of 80 kDa which are joined by a disulfide bond.
The presence of SDS and dithiothreitol acts as a reducing agent, and they can break disulfide bonds whose pH is greater than 7, i.e. those in basic condition.
Answer:
7.32 g of F₂
Solution:
The equation is as follow,
2 LiI + F₂ → 2 LiF + I₂
According to equation,
51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂
Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g
X = 7.32 g of F₂
<h2>
Answer:</h2>
The magnesium ribbon, <u>D. It forms a material to cast the tool mark</u>.
<h2>
Explanation:</h2>
When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.
Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.
