alternatively

symbolizes

hiking

inexperienced

attendances

believable

28.alternatively

29.symbolizes

30.hiking

31.inexperienced

32.attendances

33.believable

alternatively

symbolizes

hiking

inexperienced

attendances

believable

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

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

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

1.A

2.A 
3.A

4.B 
5.B

6.C

7.C

8. A

9. C

10. B

11. D

12.A

13. B

14. C

15. A

16. C

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

C - A - D - B - A - C
she learnt Math as well as Hoa
eaten noodles for four years
after her mother very much
as good at chemistry as Susan

\(\left(-2023\right).2024-2023.\left(10-2024\right)\\ =-2023.2024-2023.10+2023.2024\\ =\left(2023.2024-2023.2024\right)-2023.10\\ =0-2023.10=-20230\\ ---\\ 198.246-246.\left(198-19\right)\\ =198.246-246.198+246.19\\ =0-246.19=-4674\\ ---\\ 297.\left(435-24\right)-435.\left(297-24\right)\\ =297.435-297.24-435.297+435.24\\ =\left(297.435-435.297\right)+\left(435.24-297.24\right)\\ =0+\left(435-297\right).24\\ =0+138.24=3312\\ ----\\ \left(-367\right).\left(19-198\right)+198.\left(19-367\right)\\ =-367.19+367.198+198.19-367.198\\ =\left(367.198-367.198\right)+\left(198.19-367.19\right)\\ =0+19.\left(198-367\right)=0+19.\left(-169\right)=-3211\)

1. B

2. C

3. C

4. B

5. A

6. D

7. C

8. C

9. A

10. B

1. B 2. C 3. C 4. B 5. A 6. D 7. C 8. C 9. A 10. B

Bài 8:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1--->0,15------>0,05---------->0,15

a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

b. \(V'=0,15.22,4=3,36\left(l\right)\)

c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)

Bài 9:

\(n_{Fe}=\dfrac{2,8}{56}=0,2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2-->0,4---->0,2------>0,2

a. \(V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

b.\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. \(CM_{FeCl_2}=\dfrac{0,2}{0,4}=0,5M\)

Bài 10:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1<--0,2<----0,1<----0,1

a. \(m=m_{Fe}=0,1.56=5,6\left(g\right)\)

b. \(V_{HCl}=\dfrac{0,2}{0,1}=2\left(l\right)=200\left(ml\right)\)

c.\(CM_{FeCl_2}=\dfrac{0,1}{2}=0,05M\)