Tham khảo nha 

1B 2D 3D 4B 5A 6C 7A 8B 9D 10C 11D 12C 13C 14C 15D 16A 17B 

34 A

35 B

36 C

37 A

38 A

39 B

40 C

41 B

42 A

43 B

44 D

45 B

46 B

47 B

48 A

49 C

50 D

34 A

35 B

36 C

37 A

38 A

39 B

40 C

41 B

42 A

43 B

44 D

45 B

46 B

47 B

48 A

49 C

50 D

Câu 18:

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{HCl}=0,15.2=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\), ta được HCl dư.

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(C_{M_{ZnCl_2}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\)

Em chỉ cần đáp án hay cần giải luôn nhỉ?

11 spacious

12 thoughtful

13 competitive

14 original

15 equipment

16 installation

17 reduction

18 pleasant

19 security

20 transmission

\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}\)

\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-\left(\dfrac{4}{8}+\dfrac{10}{8}\right)+\dfrac{19}{18}\)

\(=\dfrac{5}{5}-\dfrac{14}{8}+\dfrac{19}{18}\)

\(=1-\dfrac{7}{4}+\dfrac{19}{18}\)

\(=-\dfrac{3}{4}+\dfrac{19}{18}=\dfrac{11}{36}\)

\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}=\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{5}{4}=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{19}{18}-\dfrac{1}{2}\right)-\dfrac{5}{4}=1+\dfrac{5}{9}-\dfrac{5}{4}=\dfrac{36}{36}+\dfrac{20}{36}-\dfrac{45}{36}=\dfrac{11}{36}\)

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