Đáp án:
\(\begin{array}{l}
1\,A\\
2\,A\\
3\,C\\
4\,A\\
5\,B\\
6\,B\\
7\,A\\
8\,D\\
9\,D\\
10\,B
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
CTHH:R{H_4}\\
\% {m_R} = 75\% \Rightarrow \dfrac{{{M_R}}}{{{M_R} + 4}} = 75\% \\
\Rightarrow {M_R} = 12\,g/mol \Rightarrow R:Cacbon(C)\\
3)\\
4K + {O_2} \to 2{K_2}O\\
{n_K} = \dfrac{{7,88}}{{39}} = 0,2\,mol\\
{n_{{K_2}O}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_{{K_2}O}} = 0,1 \times 94 = 9,4g\\
5)\\
1 \le \dfrac{{13 - 2{p_X}}}{{{p_X}}} \le 1,5 \Leftrightarrow 3,71 \le {p_X} \le 4,33\\
{p_X} = 4 \Rightarrow X:1{s^2}2{s^2}\\
6)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_R} = 2{n_{{H_2}}} = 0,2\,mol\\
{M_R} = \dfrac{{4,6}}{{0,2}} = 23\,g/mol \Rightarrow R:Natri(Na)\\
8)\\
CTHH:M{O_3}\\
\% {m_M} = 40\% \Leftrightarrow \dfrac{{{M_M}}}{{{M_M} + 48}} = 40\% \\
\Rightarrow {M_M} = 32\,g/mol \Rightarrow M:S\\
CTHH:S{O_3}
\end{array}\)
