Предмет: Физика, автор: xosrov

При электролизе сернокислого цинка в течение 5часов выделилось 3,06*10^-2кг Цинка при напрежении 10В.определите сопроттвление электролита Закон гей люссака

Ответы

Автор ответа: Аноним

ДАНО

t = 5часов = (5*3600) c

m = 3,06*10^-2 кг

U =10 B

k = 3.4*10^-7 кг/Кл - электрохимический эквивалент цинка

НАЙТИ

РЕШЕНИЕ

m = k*I*t = k*U/R*t

R = k*U/m*t = 3.4*10^-7*10/(3.06*10^-2) *(5*3600) = 2 Ом

ОТВЕТ 2 Ом

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