Ответ:
дано
m (Ag2O) = 10 g
W(AgNO3) = 20%
----------------------
m(ppa AgNO3) - ?
2AgNO3+2NaOH-->2NaNO3+Ag2O+H2O
M(Ag2O) = 232 g/mol
n(Ag2O) =m(Ag2O) / M(Ag2O) = 10 / 232 = 0.043 mol
2n(AgNO3) = n(Ag2O)
n(AgNO3) = 0.043 * 2 = 0.084 mol
M(AgNO3) = 170 g/mol
m(AgNO3) = n*m = 0.084 * 170 = 14.28 g
m(рра AgNO3) = 14.28 * 100% / 20% = 71.4 g
ответ 71.4 гр
Объяснение:
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Answers & Comments
Ответ:
дано
m (Ag2O) = 10 g
W(AgNO3) = 20%
----------------------
m(ppa AgNO3) - ?
2AgNO3+2NaOH-->2NaNO3+Ag2O+H2O
M(Ag2O) = 232 g/mol
n(Ag2O) =m(Ag2O) / M(Ag2O) = 10 / 232 = 0.043 mol
2n(AgNO3) = n(Ag2O)
n(AgNO3) = 0.043 * 2 = 0.084 mol
M(AgNO3) = 170 g/mol
m(AgNO3) = n*m = 0.084 * 170 = 14.28 g
m(рра AgNO3) = 14.28 * 100% / 20% = 71.4 g
ответ 71.4 гр
Объяснение: