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  • 0732. Решение задачки по химии
1. Вычислить ЭДС ячейки Pt(т) | Pb(т)| PbI2(т) | HI(aq)| H2(г)| Pt(т) при активности HI(aq) равной 0.815 и фугитивности водорода 1 атм., зная стандартный электродный потенциал –0.364 В реакции PbI2(т) + 2e– = 2I–(aq) + Pb(т), при 25 °С.

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0732. Решение задачки по химии

  • Продавец: Derkach
  • Модель: 4020194
  • Наличие: Есть в наличии
  • 116р.

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