dài thế sao làm đc

Trường bn gắt quá

I

1 works - will pass

2 had - would buy

3 come - will phone

4 will feel - don't give

5 closed

6 goes - will be

7 will know - don't reveal

8 have 

9 won

10 will stop

II

1 if I wanted to dance

2 when I had come

3 if John had arrived

4 where Maria parked her car

5 If I had watched the latest film

6 if she could help me

7 they had cleaned the windows

8 he hadn't had time to do his homework

9 her mother would be 50 years old

10 the boss had to sign the letter

Phần trên mình làm ở câu hỏi trước của bạn rồi nhớ.
III.
1. doesn't never drink -> never drinks
2. dressing -> dressed
3. Yes -> No
4. are -> will
5. nine -> ninth
6. Who -> How
7. tooth -> teeth
8. many -> much
9. do -> does
10. at -> in
V.
1. worse
2. as good as
3. wider
4. bigger
5. harder
6. tougher
7. fewer
8. narrower
9. the fastest
10. happier
D.
I.
1. A
2. B
3. D
4. D
5. B
II.
1. It is 3 years since we last went to Spain.
2. We have arrived in Budapest for 2 days.
3. This is the second time I have visited the Stone Forest.
4. Dung is keen on travelling.
5. If you don't turn off all the lights, you will pay more money.

III

1 never => bỏ

2 dressing => dressed

3 Yes => no

4 are you doing => do you do

5 nine => ninth

6 Who => how

7 tooth => teeth

8 many => much

9 do => does

10 at => in

V

1 worse

2 better than

3 wider

4 bigger

5 harder

6 tougher

7 fewer

8 narrower

9 faster

10 happier

30 nha

\(A=\dfrac{3}{2\cdot2}=\dfrac{3}{4}\\ A=\dfrac{3}{2\cdot5}=\dfrac{3}{10}\\ A=\dfrac{3}{2\cdot3}=\dfrac{1}{2}\)

bn đang thi thì tự lm nhé, kể cả KT hay là Thi thử

thi tự lm nhé

2 A

3 A

4 C

5 A

6 D

7 A

8 B

9 C

10 A

Câu 1.

Tờ vé số có dạng \(\overline{a_1a_2a_3a_4a_5a_6}\in A=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

\(;a_i\ne a_j\)

Chọn \(a_1\ne0\) nên \(a_1\) có 9 cách chọn.

5 số còn lại là chỉnh hợp chập 5 của 8 số còn lại \(\in A\backslash\left\{a_1\right\}\)

\(\Rightarrow\)Có \(A_8^5\) cách.

Vậy có tất cả \(A_8^5\cdot9=60480\) vé số.

c

Bài 1:

Ta có : 1 mol muối RCO₃ (có khối lượng = R + 60n) chuyển thành 1 mol RCl_n (có khối lượng = R + 71n)

=> khối lượng tăng = 71n – 60n = 11n gam

=> Khi chuyển 1 mol gốc CO₃ thành 2 mol gốc Cl và tạo ra 1 mol CO₂ thì khối lượng tăng 11 gam

a) Ta có công thức tính nhanh sau : \(m_{muốiclorua}=n_{muốicacbonat}+11.n_{CO_2}\)

=> \(n_{CO_2}=\dfrac{11,1-10}{11}=0,1\left(mol\right)\)

=> \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

b) \(MCO_3+2HCl\rightarrow MCl_2+CO_2+H_2O\)

\(n_{HCl}=2n_{CO_2}=0,2\left(mol\right)\)

=> \(m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)

c) \(m_{ddsaupu}=10+200-0,1.44=205,6\left(g\right)\)

\(C\%_{muối}=\dfrac{11,1}{205,6}.100=5,4\%\)

d) \(n_{MCO_3}=n_{MCl_2}\)

=> \(\dfrac{10}{M+60}=\dfrac{11,1}{M+71}\)

=> \(M=40\left(Ca\right)\)