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- Using the equations H₂ (g) F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol C (s) 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol 2 C(s) 2 H₂
- For the chemical reaction H₂ (g) F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol. What is the molar enthalpy (in kJ/mol) for the reac
- Using the equations H₂ (g) F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol C (s) 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol Determine t
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