Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the
Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So,
the pH of this buffer solution is 10,04Have a nice day!
Answer:
option C= hydrolysis and break down
Explanation:
All other three pairs are correct coupling of each others.
Option A= dehydration synthesis and hydrolysis
Dehydration synthesis:
In dehydration synthesis monomers combine through the covalent bonds and form large molecules. The large molecules are called polymers. The water as a byproduct also released when monomers joints together.
Hydrolysis:
In hydrolysis the polymers are break down into monomers by using water molecules. The catalysts are also required in this process.
Option B= Catabolic and Anabolic
Anabolic:
In this process smaller molecules combine to gather to form large complex molecules by using energy.
For example simple glucose molecules join together to form large disaccharides.
Catabolic:
It is the break down of large complex molecules to the smaller molecules.
For example during cellular respiration sugar molecules break down and generate energy.
Option D= Break down and synthesis
The break down and synthesis are also reverse pair of each others. The synthesis involve the formation of molecules form smaller component while the break down involve destruction of molecules into smaller units.
Answer:
Kc = [CO2], that is to say it is equal to the concentration of CO2
Explanation:
It is a heterogeneous equilibrium since the substances that participate in the reaction are in different phases
In the heterogeneous limestone decomposition reaction:
CaCO3(s) --> CaO(s) + CO2(g)
The equilibrium constants are:
Kc = [CO2(g)]; Kp = PCO2(g); Kc = Kp (R T)^
−(1−0) = Kp (R T)^
−1
The equilibrium situation is not affected by the amount of solid or liquid, as long as these substances are present.
The equilibrium constant is independent of the amounts of solids and liquids in equilibrium.
