Answer:
Metre
Explanation:
the SI unit of length is <em>Me</em><em>tre</em>
Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
I think it would seeing as you typicaly have to use a match to light a gas burner
Answer:
The value of the equilibrium constant
at this temperature is 3.42.
Explanation:
Partial pressure of the sulfur dioxide =![p_1=0.564 atm](https://tex.z-dn.net/?f=p_1%3D0.564%20atm)
Partial pressure of the oxygen gas =![p_2=0.102 atm](https://tex.z-dn.net/?f=p_2%3D0.102%20atm%20)
Partial pressure of the sulfur trioxide =![p_3=0.333](https://tex.z-dn.net/?f=p_3%3D0.333)
![2 SO_2(g) + O_2(g)\rightleftharpoons 2 SO_3(g)](https://tex.z-dn.net/?f=2%20SO_2%28g%29%20%2B%20O_2%28g%29%5Crightleftharpoons%202%20SO_3%28g%29)
The expression of an equilibrium constant is given by :
![K_p=\frac{(p_3)^2}{(p_1)^2\times p_2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28p_3%29%5E2%7D%7B%28p_1%29%5E2%5Ctimes%20p_2%7D)
![K_p=\frac{(0.333 atm)^2}{(0.564 atm)\times (0.102 atm)}=3.42](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%280.333%20atm%29%5E2%7D%7B%280.564%20atm%29%5Ctimes%20%280.102%20atm%29%7D%3D3.42)
The value of the equilibrium constant
at this temperature is 3.42.
This means that the reaction is in dynamic equilibrium.
Hope this helps because I don't really know the question you asked :)))