Đáp án:
\(\begin{array}{l}
a)\\
{d_K} = 1,93\,g/c{m^3}\\
b)\\
m = 10,92g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{V_K} = \dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi \times {(0,2 \times {10^{ - 7}})^3} = 3,351 \times {10^{ - 23}}(\,c{m^3})\\
{m_K} = 39\,dvC = 39 \times 1,6605 \times {10^{ - 24}} = 6,47595 \times {10^{ - 23}}(g)\\
{d_K} = \dfrac{{6,47595 \times {{10}^{ - 23}}}}{{3,351 \times {{10}^{ - 23}}}} = 1,93\,g/c{m^3}\\
b)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
KOH + HCl \to KCl + {H_2}O\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = 0,2 + 0,04 \times 2 = 0,28\,mol\\
{n_K} = {n_{KOH}} = 0,28\,mol\\
{m_K} = 0,28 \times 39 = 10,92g
\end{array}\)