Đáp án:
\(\begin{array}{l}
14)\\
{Z_X} = 20\\
15)\\
Che:1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
14)\\
X + 2HCl \to XC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_X} = {n_{{H_2}}} = 0,1mol\\
{M_X} = \dfrac{4}{{0,1}} = 40dvC\\
\Rightarrow {Z_X} = 40 - 20 = 20\\
15)\\
2X + 6HCl \to 2XC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_X} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
{M_X} = \dfrac{{2,7}}{{0,1}} = 27dvC\\
\Rightarrow {p_X} = 27 - 14 = 13\\
Che:1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}
\end{array}\)