5A

1B

3C

4D

\(#TyHM\)

1 do you do

2 had done - went

3 went - had read

4 will attend

5 hadn't worn

6 to be

7 weren't sleeping - were playing

8 to be

9 had lived - moved

10 locking

11 had work - retired 

12 told - had learned

13 won't call

14 had met

do you do

had done-went

went- had read

will attend

hadn't worn

to be

12.2 cắt quá bạn đăng lại nhé

13

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

a.

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,4->1,2----->0,4-------->0,6

\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

b. \(C\%_{HCl}=\dfrac{1,2.36,5.\left(100+20\right)\%}{200}=26,28\%\)

c. \(n_{HCl.dư}=(\dfrac{200.26,28\%}{100\%}:36,5)-1,2=0,24\left(mol\right)\)

\(C\%_{HCl.dư}=\dfrac{0,24.36,5.100\%}{10,8+200-0,6.2}=4,18\%\)

\(C\%_{AlCl_3}=\dfrac{0,4.133,5.100\%}{10,8+200-0,6.2}=25,48\%\)

14

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

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

a. \(V=V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b, \(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\)

15

\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

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

a. \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b. \(CM_{KOH}=\dfrac{0,1}{0,2}=0,5M\)

16

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

\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

0,1--->0,2----->0,1-------->0,1

a. \(m_{Ca}=40.0,1=4\left(g\right)\)

b, \(CM_{Ca\left(OH\right)_2}=\dfrac{0,1}{0,3}=\dfrac{1}{3}M\)

a: Xét tứ giác AEHF có 

\(\widehat{AEH}=\widehat{AFH}=\widehat{FAE}=90^0\)

Do đó: AEHF là hình chữ nhật

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone

b) Áp dụng hệ thức lượng trong tam giác vuông vào ΔAHB vuông tại H có HM là đường cao ứng với cạnh huyền AB, ta được:

\(AM\cdot AB=AH^2\)(1)

Áp dụng hệ thức lượng trong tam giác vuông vào ΔAHC vuông tại H có HN là đường cao ứng với cạnh huyền AC, ta được:

\(AN\cdot AC=AH^2\)(2)

Từ (1) và (2) suy ra \(AM\cdot AB=AN\cdot AC\)

\(\left(2119-x\right)\cdot2,23=174\)

\(\Rightarrow219-x=\frac{174}{2,23}\)

\(x=219-\frac{174}{2,23}\)

X=1 ung ho nha