2) 2HF + C₂O₄²⁻ ⇔ 2F⁻ + H₂C₂O₄ K= 0.24
Chemistry, 10.02.2021 19:50 CamBroOkaii2590
Given the reactions:
1) HF ⇔ H⁺ + F⁻ K= 6.8 x 10⁻⁴
2) 2HF + C₂O₄²⁻ ⇔ 2F⁻ + H₂C₂O₄ K= 0.24
Determine the value of the equilibrium constant for the reaction:
3) H₂C₂O₄ ⇔ 2H⁺ + C₂O₄²⁻
Answers: 2
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Given the reactions:
1) HF ⇔ H⁺ + F⁻ K= 6.8 x 10⁻⁴
2) 2HF + C₂O₄²⁻ ⇔ 2F⁻ + H₂C₂O₄ K= 0.24
2) 2HF + C₂O₄²⁻ ⇔ 2F⁻ + H₂C₂O₄ K= 0.24
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