Em tham khảo nha :

\(\begin{array}{l}
6)\\
CTHH:{M_2}{O_3}\\
\text{Phân tử khối là 102dvC}\\
2{M_M} + 3{M_O} = 102\\
 \Rightarrow {M_M} = \dfrac{{102 - 3 \times 16}}{2} = 27dvC\\
 \Rightarrow M:\text{Nhôm}(Al)\\
7)\\
{M_{Na}} = 23dvC\\
 \Rightarrow {m_{Na}} = 23 \times 1,6605 \times {10^{ - 24}} = 3,81915 \times {10^{ - 23}}g\\
{M_O} = 16dvC\\
 \Rightarrow {m_O} = 16 \times 1,6605 \times {10^{ - 24}} = 2,6568 \times {10^{ - 23}}g\\
{M_N} = 14dvC\\
 \Rightarrow {m_N} = 14 \times 1,6605 \times {10^{ - 24}} = 2,3247 \times {10^{ - 23}}g\\
8)\\
\text{Đơn chất :}Fe,{O_2},{N_2},Pb,Au,{H_2},{O_3},C{l_2},B{r_2},{I_2}\\
\text{Hợp chất :}C{O_2},A{l_2}{(S{O_4})_3},CuO,CuC{l_2},{H_2}O,Ca{(N{O_3})_2},N{a_2}S{O_3}\\
{H_3}P{O_4},HCl,NaOH,Mg{(OH)_2}
\end{array}\)

Câu 6/

$M_{M_2O_3}=2M+3.16=102(đvC)$

$\leftrightarrow 2M=54$

$\leftrightarrow M=27(đvC)$

$\to M$ là nguyên tố Nhôm ($Al$)

Câu 7/

$M_{Na}=23.1,6605×10^{-24}=3,81915×10^{-23}$

$M_{O}=16.1,6605×10^{-24}=1,0378125×10^{-25}$

$M_{N}=14.1,6605×10^{-24}=2,3247×10^{-23}$

Câu 8/

Đơn chất: $Fe,O_2,N_2,Pb,Au,H_2,O_3,Cl_2,Br_2,I_2$

Hợp chất: $Al_2(SO_4)_3,CO_2,CuO,CuCl_2,H_2O,Ca(NO_3)_2,Na_2SO_3,H_3PO_4,HCl,NaOH,Mg(OH)_2$