дано
m(ppa NaCL) = 73 g
W(NaCL) = 15%
+AgNO3
m(AgCL)-?
m(NaCL) = 73 * 15% / 100% = 10.95 g
NaCL+AgNO3-->NaNO3+AgCL
M(NaCL) = 58.5 g/mol
n(NaCL) = m/M =10,95 / 58.5 = 0.187 mol
n(NaCL) = n(AgCL) = 0.187 mol
M(AgCL) = 143.5 g/mol
m(AgCL) = n*M = 0.187 * 143.5 = 26.83 g
ответ 26.83 г
Объяснение:
дано
m(ppa NaCL) = 73 g
W(NaCL) = 15%
+AgNO3
m(AgCL)-?
m(NaCL) = 73 * 15% / 100% = 10.95 g
NaCL+AgNO3-->NaNO3+AgCL
M(NaCL) = 58.5 g/mol
n(NaCL) = m/M =10,95 / 58.5 = 0.187 mol
n(NaCL) = n(AgCL) = 0.187 mol
M(AgCL) = 143.5 g/mol
m(AgCL) = n*M = 0.187 * 143.5 = 26.83 g
ответ 26.83 г
Объяснение: