Đáp án:
\(\begin{array}{l}
1)\,C\\
2)\,B\\
3)\,C\\
4)\,B\\
5)\,C\\
6)\,B\\
7)\,B\\
8)\,B\\
9)\,C\\
10)D
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
7)\\
\left\{ \begin{array}{l}
2{p_Y} + {n_Y} = 82\\
2{p_Y} - {n_Y} = 22
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4{p_Y} = 104\\
{n_Y} = 2{p_Y} - 22
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{p_Y} = 26\\
{n_Y} = 30
\end{array} \right.\\
{Z_Y} = {p_Y} = 26\\
{M_Y} = {p_Y} + {n_Y} = 26 + 30 = 56\\
8)\\
\overline {{M_C}} = 12 \times 98,89\% + 13 \times 1,11\% = 12,011\\
9)\\
{}_{29}^{63}Cu\\
{}_{29}^{65}Cu\\
\left\{ \begin{array}{l}
a + b = 1\\
63a + 65b = 63,54
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 73\% \\
b = 27\%
\end{array} \right.\\
10)\\
\overline {{M_{{\rm{Ar}}}}} = 36 \times 0,337\% + 38 \times 0,063\% + 40 \times 99,6\% = 39,98526g\\
{V_{{\rm{Ar}}}} = \dfrac{{20}}{{39,98526}} \times 22,4 = 11,204\,d{m^3}
\end{array}\)
