Ответ:
Ca(OH)2 + SO2→ CaSO3↓ + H2O
CaSO3↓ + SO2 + H2O → Ca(HSO3)2
m(NaOH) = 80 г
m(соли) = ?
HCl + NaOH = NaCl + H2O
Mr(NaOH) = 23 + 16 + 1 = 40 г/моль
<span> M</span>r(NaCl) = 23 + 35,5 = 58,5 г/моль
80---x
40----58,5
х = 80 х 58,5/40 = 117 г
M р-ра = m(H2O) + m(соли) = 920 г
w(соли) = m(соли) / m(р-ра) * 100\%
W(соли) = 120 г / 920 г * 100 \%= 13 \%
Ответ: 13 \%
<span>4)Cl2O7
5)CO2
6)CrO3
)))))))))))</span>
Запишем уравнения Нернста для данной пары:
![\phi(Zn^2^+/Zn) = \phi^o(Zn^2^+/Zn) + \frac{0,0592}{2} lg [Zn^2^+] \\ \phi(Ag^+/Ag) = \phi^o(Ag^+/Ag) + \frac{0,0592}{1} lg [Ag+]](https://tex.z-dn.net/?f=%5Cphi%28Zn%5E2%5E%2B%2FZn%29+%3D+%5Cphi%5Eo%28Zn%5E2%5E%2B%2FZn%29+%2B++%5Cfrac%7B0%2C0592%7D%7B2%7D+lg+%5BZn%5E2%5E%2B%5D+%5C%5C+%5Cphi%28Ag%5E%2B%2FAg%29+%3D+%5Cphi%5Eo%28Ag%5E%2B%2FAg%29+%2B++%5Cfrac%7B0%2C0592%7D%7B1%7D+lg+%5BAg%2B%5D)
Значения
![\phi^o(Zn^2^+/Zn)](https://tex.z-dn.net/?f=%5Cphi%5Eo%28Zn%5E2%5E%2B%2FZn%29)
и
![\phi^o(Ag^+/Ag)](https://tex.z-dn.net/?f=%5Cphi%5Eo%28Ag%5E%2B%2FAg%29)
следующие:
![\phi^o(Zn^2^+/Zn)](https://tex.z-dn.net/?f=%5Cphi%5Eo%28Zn%5E2%5E%2B%2FZn%29)
= -0,763 В
![\phi^o(Ag^+/Ag)](https://tex.z-dn.net/?f=%5Cphi%5Eo%28Ag%5E%2B%2FAg%29)
= 0,799 В (значения из Вики)
Подставляя имеющиеся данные, находим потенциалы:
![\phi(Zn^2^+/Zn) = -0,763 +\frac{0,0592}{2} * lg1 = -0,763](https://tex.z-dn.net/?f=%5Cphi%28Zn%5E2%5E%2B%2FZn%29+%3D+-0%2C763++%2B%5Cfrac%7B0%2C0592%7D%7B2%7D+%2A+lg1+%3D+-0%2C763)
В.
![\phi(Ag^+/Ag) = 0,799 + \frac{0,0592}{1}*lg0,01 = 0,6806](https://tex.z-dn.net/?f=%5Cphi%28Ag%5E%2B%2FAg%29+%3D+0%2C799+%2B++%5Cfrac%7B0%2C0592%7D%7B1%7D%2Alg0%2C01+%3D+0%2C6806+)
В.
Находим ЭДС:
![\epsilon = \phi(Ag^+/Ag) - \phi(Zn^2^+/Zn) = 0,6806-(-0,763) = 1,4436](https://tex.z-dn.net/?f=%5Cepsilon+%3D+%5Cphi%28Ag%5E%2B%2FAg%29+-+%5Cphi%28Zn%5E2%5E%2B%2FZn%29+%3D+0%2C6806-%28-0%2C763%29+%3D+1%2C4436)
В.