The electron has a higher frequency compared to the neutron. It can be explained by the way an electron orbits the nucleus of an atom.
According to Quantum Mechanics, electrons do not really orbit the nucleus of an atom. In fact, the most tightly bound state, the 1s orbital, has no angular momentum at all. This would be the state with the most "kinetic energy" and yet there is no "orbital" motion at all in this state.
<span>However, there are frequencies associated with each orbital.</span>
Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of
be x
Initial concentration of
, c = 0.230 M

at eq'm c-x x x
Expression of
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so
.

Solving for x, we get:

Given, HCl with 0.230 M , it dissociates fully in water which means
= 0.230 M
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining
after neutralizing
ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
Answer:
A variable is anything that can change or be changed. In other words, it is any factor that can be manipulated, controlled for, or measured in an experiment.
Answer:
Explanation:
From the information given:
Feed F = 150.0 kmol/hr
The saturated liquid mixture of the distillation column is
= 30%
Reflux ration = 2.0%
methanol distillate mole fraction
= 0.990
recovery of methanol in the distillate = 97.0%
The distillate flow rate D can be determined by using the formula;

D = 0.97 × 150 × 0.3
D = 43.65 kmol/h
The bottom flow rate Balance B on the column is:
F = D + B
150 = 43.65 + B
B = ( 150 - 43.65 )kmol/h
B = 106.35 kmol/h
The methanol mole fraction in the bottom
can be computed by using the formula:

150(0.3) = 43.65(0.999) + 106.3(
)
45 = 43.60635 + 106.3(
)
45 - 43.60635 = 106.3(
)
1.39365 = 106.3(
)
= 1.39365 / 106.3
= 0.013
the fractional recovery of water in the bottoms f is calculated as:



f = 0.99969