Ответ:
дано
m( ppa ZnCL2) = 75 g
W(ZnCL2) = 9%
---------------------
m(Zn(OH)2)-?
m(ZnCL2) = 75 * 9% / 100% = 6.75 g
ZnCL2+2KOH-->Zn(OH)2+2KCL
M(ZnCL2) = 136 g/mol
n(ZnCL2) = m/M = 6.75 / 136 = 0.05 mol
n(ZnCL2) = n(Zn(OH)2)
M(Zn(OH)2) = 99 g/mol
m(Zn(OH)2) = n*M = 0.05 * 99 = 4.95 g
ответ 4.95 г
Объяснение:
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Ответ:
дано
m( ppa ZnCL2) = 75 g
W(ZnCL2) = 9%
---------------------
m(Zn(OH)2)-?
m(ZnCL2) = 75 * 9% / 100% = 6.75 g
ZnCL2+2KOH-->Zn(OH)2+2KCL
M(ZnCL2) = 136 g/mol
n(ZnCL2) = m/M = 6.75 / 136 = 0.05 mol
n(ZnCL2) = n(Zn(OH)2)
M(Zn(OH)2) = 99 g/mol
m(Zn(OH)2) = n*M = 0.05 * 99 = 4.95 g
ответ 4.95 г
Объяснение: