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:
2C3H6 + 9 O2 ---> 6 CO2 + 6 H2O
Explanation:
Answer:
V₂ = 2.96 L
Explanation:
Given data:
Initial volume = 2.00 L
Initial temperature = 250°C
Final volume = ?
Final temperature = 500°C
Solution:
First of all we will convert the temperature into kelvin.
250+273 = 523 k
500+273= 773 k
According to Charles's law,
V∝ T
V = KT
V₁/T₁ = V₂/T₂
V₂ = T₂V₁/T₁
V₂ = 2 L × 773 K / 523 k
V₂ = 1546 L.K / 523 k
V₂ = 2.96 L
Answer: heat
Explanation:
Is that heat boils the liquid and the liquid eventually evaporated into gas
