Answer:
No, it is not appropriate to mix water and DMSO
Explanation:
We have to realize that DMSO is a highly polar solvent and water is a highly polar solvent. The question explicitly says that our target is to produce a solvent of<u><em> intermediate polarity</em></u>.
We can only do this by mixing a polar and a nonpolar solvent. We have been given the example of the mixture of acetone/hexane which is quite a perfect mixture.
Thus, it is inappropriate to mix DMSO and water.
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
The energy can be shown as:
Q = ms dT
Whereas, m is the mass of block
s is specific heat
dT is change in temperature.
Copper block having the lowest specific heat and thus having the higher change in temperature and therefore having the higher final temperature.
Answer:
A ground wire helps those positive charges get to the ground in a safe, direct and controlled way, where they can be discharged without the risk of electrical shock or fire. Shock Absorber Excess electrical charges are common in any home.
Explanation:
:3
Answer is B.
As the diaphragm contracts and flattens, it increases the volume of the thorax where the lungs are located. This results in a decrease in pressure (Boyle’s Law, if you know it) that creates a pressure gradient from outside to inside. This is what causes air to move into the lungs.
