Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
Answer: 0.4 moles
Explanation:
Given that:
Volume of gas V = 11L
(since 1 liter = 1dm3
11L = 11dm3)
Temperature T = 25°C
Convert Celsius to Kelvin
(25°C + 273 = 298K)
Pressure P = 0.868 atm
Number of moles N = ?
Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)
9.548 atm dm3 = n x 24.47atm dm3mol-1
n = (9.548 atm dm3 / 24.47atm dm3 mol-1)
n = 0.4 moles
Thus, there are 0.4 moles of the gas.
This combination in non polar.
It allows you to determine the relation between the reactants and the products.
Solid should be the anwser
