Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Li(s) (answer A)
Li is strongest reducing agent because of the lowest standard reduction potential. when something is oxidized, it reduces another substance, becoming a reducing.Hence Lithium is strongest reducing agent. Reducing agent is stronger when it has a more positive oxidation potential.
Water molecules move througout the solute
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