дано
m(Cu) = 6.4 g
η(CuO)) = 94%
m практ (CuO) - ?
2Cu+O2-->2CuO
M(Cu) = 64 g/mol
n(Cu) = m/M = 6.4 / 64 = 0.1 mol
2n(Cu) = 2n(CuO) = 0.1 mol
M(CuO) = 80 g/mol
m(CuO) = n*M = 0.1 * 80 = 8 g
m прак (CuO) = 8*94% / 100% = 7.52 g
ответ 7.52 гр
Объяснение:
дано
m(Cu) = 6.4 g
η(CuO)) = 94%
m практ (CuO) - ?
2Cu+O2-->2CuO
M(Cu) = 64 g/mol
n(Cu) = m/M = 6.4 / 64 = 0.1 mol
2n(Cu) = 2n(CuO) = 0.1 mol
M(CuO) = 80 g/mol
m(CuO) = n*M = 0.1 * 80 = 8 g
m прак (CuO) = 8*94% / 100% = 7.52 g
ответ 7.52 гр
Объяснение: