Answer:
-586.56 kJ/mol is the standard enthalpy of the 3rd reaction.
Explanation:
...[1]
...[2]
..[3]
The unknown standard enthalpy of third reaction can be calculated by using Hess's law:
The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.
[2] - 2 × [1] = [3]
The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.
Answer:
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Explanation:
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Answer:
- 416 kJ/mol
Explanation:
The standard enthalpy of the reaction (Δ
H
∘
rxn) is independent of the pathway, so it can be calculated by the enthalpy of formation of the reactants and the products:
Δ
H
∘
rxn = ∑n*Δ
H
∘f products - ∑n*Δ
H
∘f reactants
Where n is the number of moles in the balanced reaction. So, for the reaction given:
Na₂O(s) + 1/2O₂(g) → Na₂O₂(s)
Because O₂ is formed by only one elements, its Δ
H
∘f is 0 kJ/mol:
-89.0 = (1*(-505) - (1*Δ
H
∘fNa₂O)
Δ
H
∘fNa₂O = -505 + 89
Δ
H
∘fNa₂O = - 416 kJ/mol
Answer:
Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy.
Explanation:
