Answer:
ΔH°r = -483.64 kJ
Explanation:
Let's consider the following balanced equation.
2 H₂(g) + O₂(g) ⇒ 2 H₂O(g)
We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.
ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr
where
ΔH°f: standard heat of formation
n: moles
p: products
r: reactants
ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol
ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol
ΔH°r = -483.64 kJ
Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as

Where
is the quantum number
now
We can look at the given angular momentum in this form as

comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
Explanation:
The given reaction will be as follows.

So, equilibrium constant for this equation will be as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
As it is given that concentration of all the species is 2.4. Therefore, calculate the value of equilibrium constant as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
= 
= 0.173
Thus, we can conclude that equilibrium constant for the given reaction is 0.173.
The correct answer would be C. Convection, which describes that heat rises.