* Một số câu trực tiếp gián tiếp mình chưa học nên không làm được, cậu tự làm nhé :

1. Mary (be) _________has been__________ seriously ill since last night.

2. They sometimes (go) _________go___________ school by bus.

3. How long ____have______ you (study) _____studied_____ English?

- I (study) ___________have studied_________ it for 7 years.

4. We last (see) _______saw______ them last Sunday.

5. I (not meet) _________haven't met___________ him since he moved to Hanoi.

6. Now they (play) __________are playing__________ happily in the school yard.

7. We (not finish) _________haven't finished___________ our exercises yet.

9. Mr. David (just buy) __________just has bought__________ a new house.

10. They (build) __________will buid__________ a new hospital in my neighborhood next month.

11. It (not rain) ________hasn't rained____________ since the beginning of this month.

14. ________Have__ you ever (be) _____been_____ to Ho Chi Minh City?

15. The telephone (ring) ____rang______ when we (watch) ______were watching_____ television last night.

16. Mrs Hoa (teach) ____taught_____ English in this school since she (leave) ___has left___ the university.

17. Look! That car (run) ______is running___________ very fast.

18. The Browns (have) ________is having____________ dinner at the moment.

19. He (live) _________has lived___________ in Hanoi for 20 years.

20. Alexander Graham Bell (invent) ______invented_______ the telephone many years ago.

21. She (be) _____is_____ very ill now. She (not go) ______hasn't gone______ to school since Monday.

22. + _____Have_____ you (be) ____been______ to Phong Nha before?

+ Yes. I (spend) ____spent______ my holiday there last summer.

23. I must go now. I (phone) _____phoned___ you when I (arrive) ____was arriving__________ there.

24. When I last (see) ____saw______ them, they ( work) _____were working______ in a factory in Hanoi.

25. Ba's teacher (give) ____gave______ him a lot of homework last week.

26. Tom hopes he (meet) ________will meet__ many interesting people when he goes to Hanoi next month.

27. Mai and her friends (walk) __________are walking__________ to school now.

28. They (go) ______went____ to Sao Mai Movie Theater last night.

29. She (not see) _________hasn't seen___________ that film before.

30. My mother was waiting for me when I (come) ________came______ home.

31. Ann and Susan (cook) ___________were cooking__________ dinner when Martin arrived home.

32. The light (go) ______was going________ out while we were having dinner.

33. + What _____were_____ you (do) ______doing________ at this time yesterday?

+ I (play) ____________was playing_________ games on the computer.

34. Tom! Turn off the TV. You (always watch) _________always watch___________ TV.

35. Lan often (do) ____does______ her homework carefully.

36. This is the first time Hoa (meet) __________has met__________ Tim's family.

38. Yesterday (be) ______was____ a beautiful day. The Sun (shine) _______was shining_____________ brightly when we (leave) _______left_______ home.

40. Listen! Mai (play) ________is playing____________ the piano in her room.

41. + What kind of music __do__________ you (lik e) ___lik e_______ ?

+ I pop music.

42. In summer, the Sun usually (shine) _________shines_________ brightly.

43. I (not finish) ___________haven't finished____________ reading the novel you lent me yet.

44. Julia (write) _____________is writing___________ an email to her boy friend now.

45. Jack (not meet) ______________didn't meet_________ Tom the last time he came to America.

+) Oxit chứa 50% khối lượng oxi :

Gọi công thức hóa học của oxit là \(M_xO_y\) (x,y nguyên dương tối giản)

Ta có : \(\frac{16y}{M_M\cdot x+16y}\cdot100\%=50\%\)

=> \(32y=M_M\cdot x+16y\)

=> \(M_M=8\cdot\frac{2y}{x}\)

Ta có bảng sau : 

=> oxit cần tìm là SO₂

=> \(M_{SO_2}=32+16\cdot2=64\) (g/mol)

+) Oxit chứa 60% khối lượng oxi :

Gọi công thức của oxit chứa 60% khối lượng oxi là \(M_aO_b\) (a,b nguyên dương tối giản)

Ta có : \(\frac{16b}{M_M\cdot a+16b}\cdot100\%=60\%\)

=> \(16b=0,6M_M\cdot a+9,6b\)

=> \(M_M=\frac{16}{3}\cdot\frac{2y}{x}\) 

Ta có bảng sau : 

=> oxit cần tìm là SO₃

=> \(M_{SO_3}=32+16\cdot3=80\)(g/mol)

 Postscript : khi tìm đc ngto là lưu huỳnh, có thể lắp xuống dưới để tìm cthh của oxit còn lại, có vẻ như thế sẽ dễ hơn 

bán kính hình tròn đó là

7,536 : 2 : 3,14 = 1,2 [m]

diện tích hình tròn đó là

1,2 nhân 1,2 nhân 3,14 =4,5216 [m ²]

đáp số : 4,5216m²

chúc bạn học tốt

ồ vậy hả .

id  miniworld của mình này:133652328/KindFabian

Các đồng phân

+ C2H4O2: CH3COOH , HCOOCH3 , CH2(OH) CHO. 

+ C3H8O: CH3CH2CH2OH , CH3CH(OH) CH3 , CH3-O-CH2CH3. 

+ C5H10: CH2= CHCH2CH2CH3 , CH2= CH-CH(CH3)CH3 ,CH2= C(CH3) -CH2CH3 ,CH3-CH=CH-CH2CH3 , CH3CH=C(CH3)2.

. Các đồng phân

+ C2H4O2: CH3COOH , HCOOCH3 , CH2(OH) CHO. 

+ C3H8O: CH3CH2CH2OH , CH3CH(OH) CH3 , CH3-O-CH2CH3. 

+ C5H10: CH2= CHCH2CH2CH3 , CH2= CH-CH(CH3)CH3 ,CH2= C(CH3) -CH2CH3 ,CH3-CH=CH-CH2CH3 , CH3CH=C(CH3)2.

Bài 1 :

a) \(Al_2O_3\)

b) \(Na_2SO_4\)

c) \(Ba\left(OH\right)_2\)

Bài 2 :

Phân tử khối NaOH :\(23+16+1=40\)đvC

Phân tử khối FeCl₃ : \(56+35,5\times3=162,5\)đvC

đó là :8HS

Số học sinh 10 tuổi của lớp 5A là:

                      32:100x75=24(học sinh)

Số học sinh 11 tuổi của lớp 5A là:

                      32-24=8 (học sinh )

                                    Đáp Số : 8 học sinh

chịu,hỏi linh tinh vãi lozz

1948 nhé bạn !

tìm google dịch trên mạng mà dịch

fortune, nha

\(1,2H_2+O_2\underrightarrow{t}2H_2O\)

\(2Mg+O_2\underrightarrow{t}2MgO\)

\(2Cu+O_2\underrightarrow{t}2CuO\)

\(S+O_2\underrightarrow{t}SO_2\)

\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)

\(C+O_2\underrightarrow{t}CO_2\)

\(4P+5O_2\underrightarrow{t}2P_2O_5\)

\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)

c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O₂ phản ứng hết, Bài toán tính theo O₂

\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)

\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)

\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)

\(4,a,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)

\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)

\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)

\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)

\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)

\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)

\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=0,46875\left(mol\right)\)

\(n_{SO_2}=0,3\left(mol\right)\)

Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.

\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)

\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)

\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)

\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)

\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)

\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)

\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)

\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)

\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)

\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.

\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)

\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)

đủ cả 9 câu bạn nhé,