Evaporation of the solution
Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
Answer:
.259 g
Explanation:
PV = nRT
n = PV / RT
= .986 x 0.144 / .082 x 293.6
= .005897 moles
= .005897 x 44 g
= .259 g
Answer:
![0.21\text{ mol}](https://tex.z-dn.net/?f=0.21%5Ctext%7B%20mol%7D)
Explanation:
Here, we want to calculate the number of moles of methane in the container
From the ideal gas law:
![\begin{gathered} PV\text{ = nRT} \\ n\text{ = }\frac{PV}{RT} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20PV%5Ctext%7B%20%3D%20nRT%7D%20%5C%5C%20n%5Ctext%7B%20%3D%20%7D%5Cfrac%7BPV%7D%7BRT%7D%20%5Cend%7Bgathered%7D)
where:
P is the pressure inside the container which is 3 atm
V is the volume of the container which is 2 L
R is the molar gas constant which is 0.0821 Latm/mol.k
T is the temperature in Kelvin (we convert the temperature in Celsius by adding 273 : 273 + 77 = 350 K)
n is the number of moles that we want to calculate
Substituting the values, we have it that: