1 a

2 an

3 a

4 the

5 the

6 the

7 the

8 an

B

1 a

2 -

3 a 

4 the

5 the

6 the

7 the

8 the

9 the

10 a

C

1 the

2 the

3 the cars => cars

4 a

5 the

6 the

7 a

8 an

 \(=>Qthu1=0,2.340000=68000J\)

\(=>Qthu2=2100.0,2.20=8400J\)

\(=>Qtoa=2.4200.25=210000J\)

\(=>Qthu1+Qthu2< Qtoa\)=>đá nóng chảy hoàn toàn

\(=>0,2.2100.20+0,2.340000+0,2.4200.tcb=2.4200\left(25-tcb\right)\)

\(=>tcb=14,5^oC\)

\(m_{H_2O}=1,62\left(g\right)\Rightarrow n_{H_2O}=0,09\left(mol\right)\Rightarrow n_H=0,18\left(mol\right);m_H=0,18.1=0,18\left(g\right)\\ n_{CO_2}=\dfrac{2,64}{44}=0,06\left(mol\right)\Rightarrow n_C=n_{CO_2}=0,06\left(mol\right);m_C=0,06.12=0,72\left(g\right)\\ Vây:m_C+m_H=0,72+0,18=0,9< 1,38\\ \Rightarrow X.có.chứa.O\\ m_O=1,38-0,9=0,48\left(g\right);n_O=\dfrac{0,48}{16}=0,03\left(mol\right)\\ Đặt.X:C_aH_bO_c\left(a,b,c:nguyên,dương\right)\\ Ta.có:a:b:c=0,06:0,18:0,03=2:6:1\\ \Rightarrow CTĐG:C_2H_6O\\ M_X=23.2=46\left(\dfrac{g}{mol}\right)=M_{C_2H_6O}\\ \Rightarrow X:C_2H_6O\)

37

ta thấy khi cân bằng nhiệt mực nước giảm 0,5cm chứng tỏ đá tan

\(=>\Delta h=0,45-0,25=0,2m\)

\(=>Dđ.V2=Dn.V1=>900.S.h=Dn.S\left(h-0,005\right)\)

\(=>h=0,05m< 0,25m\)=>đá chưa tan hết\(=>tcb=0^oC\)

\(=>Qtoa=Dn.S.\Delta h.t1.4200=1000.S.0,2.t1.4200=840000St1\left(J\right)\)

\(=>Qthu1=0,25.S.Dđ.2100.20=9450000S\left(J\right)\)

\(=>Qthu2=S.0,05.900.340000=15300000S\left(J\right)\)

\(=>840000St1=24750000S=>t1=29,5^oC\)

5. Tea is not as good as coffee.

6. Lan is not as tall as Lan.

7. The bus isn't as fast as the train.

8. Motorbikes aren't as convenient as cars.

9. Her suitcase is not as big as his one.

10. Your backpack is not as new as mine.

35. Although Ben wanted to study in France, his French was bad.

36. The fastest way to travel from Ho Chi Minh City to Ha Noi is flying.

37. Jane has a small old white house.

38. Your brother is not as young as mine.

39. They don't have to work at weekends.

40. He was late again, so I was really angry.

a, khi cân bằng nhiệt ta có \(0,5.3,4.10^5+0,5.\left(4200+2100+400\right).t=1.\left(50-t\right).4200\Rightarrow t=5,3^oC\)

b, để nhiệt cân bằng hệ bằng 0 thì lượng nước đá p tan vừa đủ

\(m_đ.3,4.10^5=1.50.4200\Rightarrow m_đ\approx0,617\left(kg\right)\)

chữ bạn khó nhìn quá :))) 

chữ bn khó nhìn qué^^

a: \(BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\)

b: Xét ΔABI có

AH là đường cao

AH là đường trung tuyến

Do đó: ΔABI cân tại A

hay AB=AI

\(12:\left\{390:\left[500-\left(5^3+7^2.5\right)\right]\right\}\)

\(=12:\left\{390:\left[500-\left(125+49.5\right)\right]\right\}\)

\(=12:\left\{390:\left[500-\left(125+245\right)\right]\right\}\)

\(=12:\left[390:\left(500-370\right)\right]\)

\(=12:\left(390:130\right)\)

\(=12:3\)

\(=4\)

   12 : { 390 : [ 500 - ( 5³  + 7²  . 5 ) ] }

= 12 : { 390 : [ 500 - (125 + 49 . 5 ) ] }

= 12 : { 390 : [ 500 - (125 + 245 ) ] }

= 12 : { 390 : [ 500 - 370] }

= 12 : { 390 : 130}

= 12 : 3

= 4

Xin mời XD

Trường hợp 1: với thì tương lai, hiện tại đơn, hiện tại tiếp diễn:

C1:S+be+Vp2+ that +S-V

C2:S+be+VP2+to+ V

Trường hợp 2: với các thì Hiện tại honaf thành, quá khứ

C1:S+be+VP2+ that +S-V

C2:S+be+VP2+to have +VP2