I. Read the following text carefully and choose the correct answer A, B, C or D for each of the gap. When you are in Singapore, you can go about (1)__________ taxi, by bus, or by underground. I myself prefer the underground (2) __________ it is fast, easy and cheap. There are (3)__________ buses and taxis in Singapore and one cannot drive along the road (4) and without many stops, especially on Monday morning. The underground is therefore usually quicker (5)__________ taxis or buses. If...
Đọc tiếp
I. Read the following text carefully and choose the correct answer A, B, C or D for each of the gap.
When you are in Singapore, you can go about (1)__________ taxi, by bus, or by underground. I myself prefer the underground (2) __________ it is fast, easy and cheap. There are (3)__________ buses and taxis in Singapore and one cannot drive along the road (4) and without many stops, especially on Monday morning. The underground is therefore usually quicker (5)__________ taxis or buses. If you do not know Singapore very well, it is difficult (6)__________ the bus you want. You can take a taxi, but it is (7)__________ expensive than the underground or a bus. On the underground, you find good maps that (8)__________ you the names of the stations and show you (9)__________ to get to them, so (10)__________ it is easy to find your way.
1. A. by B. in C. at D. on
2. A. but B. because C. when D. so
3. A. few B. a lot C. many D. some
4. A. quick B. quickly C. quicker D. quickest
5. A. so B. like C. than D. as
6. A. find B. to find C. finding D. found
7. A. less B. more C. most D. much
8. A. tell B. told C. tells D. telling
9. A. who B. what C. when D. how
10. A. how B. that C. when D. where
Giúp mình với mai đi học phải nộp đề cương rồi
a) PTHH : \(2Al+6HCl-->2AlCl_3+3H_2\) (1)
\(Fe+2HCl-->FeCl_2+H_2\) (2)
\(H_2+CuO-t^o->Cu+H_2O\) (3)
b) Ta có : \(m_{CR\left(giảm\right)}=m_{O\left(lay.di\right)}\)
=> \(m_{O\left(lay.di\right)}=32-26,88=5,12\left(g\right)\)
=> \(n_{O\left(lay.di\right)}=\frac{5,12}{16}=0,32\left(mol\right)\)
Theo pthh (3) : \(n_{H_2\left(pứ\right)}=n_{O\left(lay.di\right)}=0,32\left(mol\right)\)
=> \(tổng.n_{H_2}=\frac{0,32}{80}\cdot100=0,4\left(mol\right)\)
Đặt \(\hept{\begin{cases}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{cases}}\) => \(27a+56b=11\left(I\right)\)
Theo pthh (1) và (2) : \(n_{H_2\left(1\right)}=\frac{3}{2}n_{Al}=\frac{3}{2}a\left(mol\right)\)
\(n_{H_2\left(2\right)}=n_{Fe}=b\left(mol\right)\)
=> \(\frac{3}{2}a+b=0,4\left(II\right)\)
Từ (I) và (II) => \(\hept{\begin{cases}a=0,2\\b=0,1\end{cases}}\)
=> \(\hept{\begin{cases}m_{Al}=27\cdot0,2=5,4\left(g\right)\\m_{Fe}=56\cdot0,1=5,6\left(g\right)\end{cases}}\)