PH of solution at 25ºC = 8.3
![[ H_3O^+] = 10 ^{-pH}](https://tex.z-dn.net/?f=%5B%20H_3O%5E%2B%5D%20%3D%2010%20%5E%7B-pH%7D%20)
![{H_3O^+] = 10 ^{-8.3}](https://tex.z-dn.net/?f=%7BH_3O%5E%2B%5D%20%3D%2010%20%5E%7B-8.3%7D)
![[H_3O^+] = 5.011*10^{-9} M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%205.011%2A10%5E%7B-9%7D%20%20M)
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1 answer:
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Explanation:
It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.
As the given values are as follows.
= 72.0 mm Hg,
= 404 K
= ? ,
= 249 K
As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.
(at constant volume)
or, = k
Therefore, the formula to calculate the value of is as follows.
=
=
= 44.37 mm Hg
As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.
As a result, no condensation will occur and only vapors of iodomethane will be present.
For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.
1) D 5 mols
2) A 0.55 mols
Answer:
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Given: Kc = 0.50
Temperature =
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(3+1) = -2
Thus, Kp is: