c d d d d

1 D

2 B

3 A

4 A

5 C

6 B

7 B

8 B

9 B

10 B

11 C

12 C

13 C

14 C

15 D

D B A A C B B B B B C C C C D

1 environmental

2 Polluting

3 conservation

4 consumption

5 appliances

6 innovation

7 Ultimately

8 suggestion

9 heated

10 environmentalists

11 worried

12protection

13 seriously

14 dripping

15 reducing

environmental

polluted

conservation

consumption

appliances

innovation

ultimately

suggestion

heated

environmentalists

worried

seriously

dripping

reducing

13. C

14.C

15.A

20A 21C 22B

Bạn chụp mặt trước nữa thì mình mới giúp được hết nha

13.D

14.B

15.A

 Câu 15:

Gọi CT chung là \(C_nH_{2n+2}\left(n>0\right)\)

\(M_{C_nH_{2n+2}}=18,5.2=37\left(g/mol\right)\)

\(\Rightarrow14n+2=37\Leftrightarrow n=2,5\)

`=>` 2 ankan là \(C_2H_6;C_3H_8\)

Áp dụng sơ đồ đường chéo:

\(\dfrac{n_{C_2H_6}}{n_{C_3H_8}}=\dfrac{44-37}{37-30}=\dfrac{1}{1}\Rightarrow\%V_{C_2H_6}=\%V_{C_3H_8}=\dfrac{1}{2}.100\%=50\%\)

Chọn C

Câu 16:

\(n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right);n_{H_2O}=\dfrac{0,21}{18}=\dfrac{7}{600}\left(mol\right)\)

Vì \(n_{H_2O}>n_{CO_2}\left(\dfrac{7}{600}>0,01\right)\Rightarrow\) A là ankan có CTPT là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

\(n_A=n_{H_2O}-n_{CO_2}=\dfrac{1}{600}\left(mol\right)\)

\(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,01\left(mol\right)\\n_H=2n_{H_2O}=\dfrac{7}{300}\left(mol\right)\end{matrix}\right.\Rightarrow m_A=0,01.12+\dfrac{7}{300}=\dfrac{43}{300}\left(g\right)\)

`=>` \(M_{C_nH_{2n+2}}=\dfrac{\dfrac{43}{300}}{\dfrac{1}{600}}=86\left(g/mol\right)\)

`=>` \(14n+2=86\Leftrightarrow n=6\)

Vậy A là \(C_6H_{14}\)

Mà A tạo được 2 dẫn xuất monoclo

`=>` CTCT của A là \(CH_3-CH\left(CH_3\right)-CH\left(CH_3\right)-CH_3\) (2,3 - đimetylbutan0

Chọn C

Câu 17:

\(n_{C_4H_{10}}=\dfrac{5,8}{58}=0,1\left(mol\right)\\ \xrightarrow[]{BTNT:H}n_{H_2O}=5n_{C_4H_{10}}=0,5\left(mol\right)\\ \Rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)

1D 2B 3C 4A 5A 6C 7A 8A 9B 10A 11C 12D 13D 14D 15A 16D 17A 18C 19C 20D 21D 22C 23C 24A 25B 26C 27D 28B 29C

\(n_{H_2SO_4}=\dfrac{50\cdot98\%}{98}=0.5\left(kmol\right)\)

\(n_{FeS_2}=\dfrac{0.5}{2}=0.25\left(kmol\right)\)

\(m_{FeS_2}=0.25\cdot120=30\left(kg\right)\)

\(m_{quặng}=\dfrac{30}{80\%}=37.5\left(kg\right)\)

\(\)