heh

1.

\(\dfrac{M_{NxOy}}{H2}=23\Rightarrow M_{NxOy}=46\Rightarrow CT:NO_2\)

⇒ Chọn B

2.

\(2KMnO4->K2MnO4+MnO2+O2\)

\(nKMnO4=\dfrac{47,4}{158}=0,3\left(mol\right)\)

\(\Rightarrow nO2=0,15\left(mol\right)\)

\(\Rightarrow V=0,15.22,4=3,36\left(l\right)\)

⇒ Chọn C

PTHH đủ nè :

\(2C_xH_y+\left(2x+\dfrac{1}{2}y\right)O_2->2xCO_2+yH_2O\)

\(2CxHy\left\{{}\begin{matrix}C:x\\H:y\end{matrix}\right.+O2->2xCO2+yH2O\)

\(nCO_2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow2x=0,1\Leftrightarrow x=0,05\left(mol\right)\)

\(nH2O=\dfrac{1,8}{18}=0,1\left(mol\right)\)

\(\Rightarrow y=0,1\left(mol\right)\)

\(\Rightarrow m=12.0,05+1.0,1=0,7\left(g\right)\)

4Al + 3O2 ---> 2Al2O3

nAl2O3 = 10,2 / 102 = 0,1 ( mol )

=> nAl = 2.0,1 = 0,2 ( mol )

=> mAl = 0,2 . 27 = 5,4 g

=> D

Tỉ lệ gen dị hợp :

1/2^2 = 1/4 = 25%