Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Actually, I strongly believe it is a switch.
Answer:
The answer to your question is: 2.20 x 10 ²³ molecules
Explanation:
Data
mass = 45.7 g
molecules of CF₂Cl₂ = ?
Process
1.- Calculate the mass number of CF₂Cl₂
C = 12 F =2 x 19 Cl = 2 x 35.5
total = 12 + 38 + 71
total = 121 g
2.- Use the Avogradro's number to solve the problem
121 g ------------------- 6.023 x 10²³ molecules of CF₂Cl₂
45.7 g --------------- x
x = (45.7 x 6.023 x 10²³) / 121
x = 2.75 x 10²⁵ / 121
x = 2.20 x 10 ²³ molecules
Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2
Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams