The below is about the energy exchanges in earth systems.
<u>Explanation</u>:
- Energy exchanges in earth systems are of many types. The earth systems are atmosphere, geosphere, stratosphere, hydrosphere, and biosphere. All these earth systems exchange energy with each other.
- The earth gains energy reflected from the sky. It converts that energy back to space. That energy is equally given to all the planets in the sky.
- Each planet will absorb that energy and radiate heat. This heat is absorbed by all the places on the earth. So this is the energy exchange in the earth systems.
<u>Answer:</u>
<u>For A:</u> The 
for the given reaction is 
<u>For B:</u> The 
for the given reaction is 1642.
<u>Explanation:</u>
The given chemical reaction follows:
The expression of for the above reaction follows:
We are given:
Putting values in above equation, we get:
Hence, the for the given reaction is
Relation of with is given by the formula:
where,
= equilibrium constant in terms of partial pressure =
= equilibrium constant in terms of concentration = ?
R = Gas constant = 
T = temperature = 500 K
= change in number of moles of gas particles = 
Putting values in above equation, we get:
Hence, the for the given reaction is 1642.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
