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- Two different aqueous solutions are represented in the table above. (a) The Ka value for HF is equal to 6.8 x 10–4. Calculate t
- Using the equations 2 Fe (s) 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol Si(s) 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol
- 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
