The enthalpy change for the formation step of this intermediate is 136 kJ/mol.
<h3>What is the enthalpy?</h3>
The enthalpy of reaction can be obtained from the enthalpies of formation of the species involved. The enthalpy of reaction is the heat evolved or absorbed in a reaction.
Thus, the enthalpy change for the formation step of this intermediate is 4 * 34 kJ/mol = 136 kJ/mol.
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Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] ·
[H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
Energy which a body possesses by virtue of being in motion.
Answer:
• long time lag between planning and operation.
• cost.
• weapons proliferation risk.
• meltdown risk.
• mining lung cancer risk.
• carbon-equivalent emissions and air pollution.
• waste risk.
Explanation:
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