The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer: Increase the rate of surface water evaporation
Irrigation is a process, in which the crops are supplied with water, to ensure their proper growth. Increased in the irrigation supply can be because of the environmental conditions like hot summer season or may be because of the type of soil is semi arid or arid means the soil does not retains moisture efficiently. This will result in increase in the rate of evaporation of the surface water.
Answer:
The answer to your question is V2 = 1.82 l
Explanation:
Data
Volume 1 = 77 l
Pressure 1 = 18 mmHg
Volume 2 = ?
Pressure 2 = 760 mmHg
Process
Use Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (18 x 77) / 760
-Simplification
V2 = 1386 / 760
-Result
V2 = 1.82 l
Answer:
Moon has to be in-between the Earth and the Sun.
2. Moon's umbra should sweep your place.
3. Latitude and longitude of your place should be within the befitting limits.