Calculate the standard molar enthalpy of formation for na2o(s), given that the standard enthalpy of formation for na2o2(s) is -505 kj/mol and the enthalpy change for the following reaction is -89.0 kj/mol. na2o(s) + 1/2 o2(g) → na2o2(s)

- 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

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph