$@Heley$

$9.$

$1.B$

$n_{CO_{2}}=$$\frac{12.10^{23}}{6.10^{23}}=2(mol)$ 

$2.C$

$m_{CO_{2}}=n_{CO_{2}}.M_{CO_{2}}=2.44=88(g)$

$3.A$

$V_{CO_{2}}=n_{CO_{2}}.22,4=2.22,4=44,8(l)$

$10.$

$1.C$

$n_{CO_{2}}=$$\frac{0,72.10^{23}}{6.10^{23}}=0,12(mol)$ 

$m_{CO_{2}}=n_{CO_{2}}.M_{CO_{2}}=0,12.44=5,28(g)$

$2.A$

$V_{CO_{2}}=n_{CO_{2}}.22,4=0,12.22,4=2,688(l)$

$11.C$

phân tử$_{SO_{2}}=n_{SO_{2}}.6.10^{23}=1.6.10^{23}=6.10^{23}$(phân tử)

$12.C$

$M_{X}=\frac{m_{X}}{n_{X}}=\frac{4,8}{0,2}=24(g/mol)$

$⇒X :Mg$

$13.D$

$n_{Fe_{2}(SO_{4})_{3}}=\frac{m_{Fe_{2}(SO_{4})_{3}}}{M_{Fe_{2}(SO_{4})_{3}}}=\frac{40}{400}=0,1(mol)$

phân tử$_{Fe_{2}(SO_{4})_{3}}=n_{Fe_{2}(SO_{4})_{3}}.6.10^{23}=0,1.6.10^{23}=0,6.10^{23}$(phân tử)

$14.B$

$n_{hhX}=n_{H_{2}}.n_{O_{2}}=0,2+0,3=0,5(mol)$

phân tử$_{hhX}=n_{hhX}.6.10^{23}=0,5.6.10^{23}=3.10^{23}$(phân tử)

#$BTS$

Đáp án:

\(\begin{array}{l}
9,\\
1.B\\
2.C\\
3.A\\
10,\\
1.C\\
2.A\\
11,C\\
12,C\\
13,D\\
14,B
\end{array}\)

Giải thích các bước giải:

9,

\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{12 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 2mol\\
 \to {m_{C{O_2}}} = 88g
\end{array}\)

\({V_{C{O_2}}} = 44,8l\)

10,

\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{0,72 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,12mol\\
 \to {m_{C{O_2}}} = 5,28g\\
 \to {V_{C{O_2}}} = 2,688l
\end{array}\)

12,

\({M_X} = \dfrac{{4,8}}{{0,2}} = 24\)

13,

\(\begin{array}{l}
{n_{F{{\rm{e}}_2}{{(S{O_4})}_3}}} = 0,1mol\\
 \to 0,1 \times 6 \times {10^{23}} = 0,6 \times {10^{23}}nguyên tử
\end{array}\)

14,

\((0,2 + 0,3) \times 6 \times {10^{23}} = 3 \times {10^{23}}\)