дано
m(ppa AgNO3) = 20 g
W(AgNO3)= 12%
HCL
m(AgCL)-?
m(AgNO3) =20*12% / 100% = 2.4 g
AgNO3+HCL-->AgCL+HNO3
M(AgNO3)= 170 g/mol
n(AgNO3) = m/M = 2.4 / 170 = 0.014 mol
n(AgNO3) = n(AgCL) = 0.014 mol
M(AgCL) = 143.5 g/mol
m(AgCL) = n*M = 0.014 * 143.5 = 2 g
ответ 2 г
Надеюсь написала понятно)
дано
m(ppa AgNO3) = 20 g
W(AgNO3)= 12%
HCL
m(AgCL)-?
m(AgNO3) =20*12% / 100% = 2.4 g
AgNO3+HCL-->AgCL+HNO3
M(AgNO3)= 170 g/mol
n(AgNO3) = m/M = 2.4 / 170 = 0.014 mol
n(AgNO3) = n(AgCL) = 0.014 mol
M(AgCL) = 143.5 g/mol
m(AgCL) = n*M = 0.014 * 143.5 = 2 g
ответ 2 г
Надеюсь написала понятно)