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  • 0732. Решение задачки по химии
2. Вычислить ЭДС ячейки Pt(т) | Hg(ж)| Hg2Cl2(т)| HCl(aq)| H2(г)| Pt(т) при активности HCl(aq) равной 1.256 и фугитивности водорода 1 атм., зная стандартный электродный потенциал +0.2688 В реакции Hg2Cl2(т) + 2e– = 2Cl–(aq) + 2Hg(ж), при 25 °С.

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0732. Решение задачки по химии

  • Продавец: Derkach
  • Модель: 4020195
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