Câu 10: D

Câu 11:  

$d_{CH_4/kk}=\dfrac{18}{29}$

$d_{NO_2/kk}=\dfrac{46}{29}$

$d_{NH_3/kk}=\dfrac{17}{29}$

$d_{CO_2/kk}=\dfrac{44}{29}$

$→$ A

Câu 12:

$m_{Cu}=\dfrac{80.80}{100}=64(g)$

$→n_{Cu}=\dfrac{64}{64}=1(mol)$

$m_{O}=\dfrac{20.80}{100}=16(g)$

$→n_{O}=\dfrac{16}{16}=1(mol)$

Suy ra CTHH: $CuO$

$→$ C

Câu 13: Gọi CTHH $Fe_xO_y$

Theo QTHT:

$x.II=y.II$

$→\dfrac{x}{y}=\dfrac{1}{1}$

$→$ CTHH: $FeO$

$→$ C

Câu 14: $n_{CuSO_4}=\dfrac{m_{CuSO_4}}{M_{CuSO_4}}=\dfrac{160}{160}=1(mol)$

$→$ A

Câu 15: $\%Cu=\dfrac{64.100}{160}=40\%$

$→$ C

Câu 16: 

$n_{CO_2}=\dfrac{V_{CO_2}}{22,4}=\dfrac{3,36}{22,4}=0,15(mol)$

$→m_{CO_2}=0,15.44=6,6(g)$

$n_{CO}=\dfrac{V_{CO}}{22,4}=\dfrac{8,96}{22,4}=0,4(mol)$

$→m_{CO}=0,4.28=11,2(g)$

$m_X=m_{CO_2}+m_{CO}=6,6+11,2=17,8(g)$

$→$ C

Câu 17: Gọi CTHH $Fe_xO_y$

Theo QTHT:

$x.III=y.II$

$→\dfrac{x}{y}=\dfrac{2}{3}$

$→$ CTHH: $Fe_2O_3$

$→$ B

Câu 18: $d_{A/H_2}=14$

$→M_{A}=28=M_{N_2}(g/mol)$

$→$ C

câu 19: Theo ĐLBTKL:

$n_{O_2}=\dfrac{4,8}{32}=0,15(mol)$

$n_{N_2}=\dfrac{7}{28}=0,25(mol)$

$→n_A=0,4(mol)$

$→V=0,4.22,4=8,96(l)$

$→$ D

câu 20: $\%K=\dfrac{39.100}{101}=38,6\%$

Câu 21: C

Đáp án:

Giải thích các bước giải: