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a) PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
b+c+d) Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,2\cdot40=8\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=204,4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{8}{204,4}\cdot100\%\approx3,91\%\)
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\(R_{13}=\dfrac{4\cdot2}{4+2}=\dfrac{4}{3}\Omega\)
\(R_N=\dfrac{\dfrac{4}{3}\cdot6}{\dfrac{4}{3}+6}=\dfrac{12}{11}\Omega\)
\(U_{AB}=12+8=20V\)
\(I=\dfrac{U}{R}=\dfrac{20}{\dfrac{12}{11}}=\dfrac{55}{3}A\)
\(I_{13}=I_m=\dfrac{55}{3}A\Rightarrow U_{13}=\dfrac{55}{3}\cdot\dfrac{4}{3}=\dfrac{220}{9}V\)
\(\Rightarrow U_3=\dfrac{220}{9}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{220}{9}}{2}=\dfrac{110}{9}A\)
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