Đáp án:
\(\begin{array}{l}
1)\,A\\
2)\,C\\
3)\,C\\
4)\,C\\
5)\,B\\
6)\,D\\
7)\,A\\
8)\,A\\
9)\,D\\
10)\,A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
{A_{{H_2}O}} = 6 \times {10^{23}} \times 1 = 6 \times {10^{23}}\\
2)\\
{n_{Fe}} = \dfrac{{280}}{{56}} = 5\,mol\\
{A_{Fe}} = 6 \times {10^{23}} \times 5 = 30 \times {10^{23}}\\
3)\\
{n_{{H_2}O}} = \dfrac{{36}}{{18}} = 2\,mol\\
4)\\
{n_{Na}} = {n_H} = {n_C} = {n_{NaHC{O_3}}} = 0,5\,mol\\
{n_O} = 3{n_{NaHC{O_3}}} = 1,5\,mol\\
{A_{NaHC{O_3}}} = (0,5 + 0,5 + 0,5 + 1,5) \times 6 \times {10^{23}} = 18 \times {10^{23}}\\
5)\\
{n_{{H_2}}} = \dfrac{4}{2} = 2\,mol\\
{V_{{H_2}}} = 2 \times 22,4 = 44,8l\\
{n_{{N_2}}} = \dfrac{{2,8}}{{28}} = 0,1\,mol\\
{V_{{N_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{{O_2}}} = \dfrac{{6,4}}{{32}} = 0,2\,mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
{n_{C{O_2}}} = \dfrac{{22}}{{44}} = 0,5\,mol\\
{V_{C{O_2}}} = 0,5 \times 22,4 = 11,2l\\
6)\\
{M_{C{H_4}}} = 12 + 1 \times 4 = 16g/mol\\
{M_{CO}} = 12 + 16 = 28g/mol\\
{M_{He}} = 4g/mol\\
{M_{{H_2}}} = 1 \times 2 = 2g/mol\\
7)\\
{V_{C{H_4}}} = 0,15 \times 22,4 = 3,36l\\
8)\\
{M_X} = 14 \times 2 = 28g/mol\\
\Rightarrow X:{N_2}\\
9)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 2 \times \dfrac{{2,24}}{{22,4}} = 0,2\,mol\\
{m_{KMn{O_4}}} = 0,2 \times 158 = 31,6g\\
10)\\
2{H_2}O \to 2{H_2} + {O_2}\\
{n_{{H_2}O}} = \dfrac{{54}}{{18}} = 3\,mol\\
{n_{{H_2}}} = {n_{{H_2}O}} = 3\,mol\\
{V_{{H_2}}} = 3 \times 22,4 = 67,2l
\end{array}\)
