The answer is D. have nausea and hair loss.
c. Electrolytes
b. Lead sulfate
d. Zinc oxide
Answer:
b
Explanation:
Answer:
-196 kJ
Explanation:
By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ
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2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)
Simplifing the compounds that are in both sides (bolded):
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ
The enthalpy of the reaction where sulfur dioxide is oxidized to sulfur trioxide is -395 kJ.
The calculation of the enthalpy change of the reaction in which sulfur dioxide is oxidized to sulfur trioxide involves Hess's Law, which states that the enthalpy change of a chemical reaction is the same whether it takes place in one step or several steps. This can be solved by comparing the enthalpy changes given in the two reactions presented.
First, consider the reactions given:
2S(s) + 3O₂(g) → 2SO₃(g), ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g), ΔH = -297 kJ
From these reactions, it is seen that the first reaction can be re-written as:
2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = -790 kJ
However, this reaction contains two moles of SO₂ whereas the reaction in question only requires one mole. Thus, the enthalpy change for the reaction becomes: ΔH = -790 KJ / 2 = -395 kJ.
#SPJ3
b. forming solutions.
c. exothermic reactions.
d. endothermic reactions.