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1. send - will come
2. will not understand - whisper
3. will not survive - do not take
4. press - will save
5. will cross - fly
6. will answer - has
7. wears - will not stay
8. touch - will not scream
9. will forget 0 do not phone
10. will remember - give
//
1. study - will improve
2. will go - gets
3. does not call - will leave
4. rings - will you answer
5. will you do - do not find
6. will find - keeps
7. feel - take
8. do not tell - will keep
9. Will you let - promise
10. will eat - feels
//
1. study - will pass
2. shines - will walk
3. has - will see
4. come - will be
5. earns - will fly
6. travel - will visit
7. wear - will slip
8. forget - will give
9. go - will listen
10. wait - will ask
//
1. B => come
2. D => can ask
3. D => can't continue
4. B => phones
5. D => do not solve
Exercise 1. Put the verbs in brackets in the correct form of conditional sentence type 1.
1. If we _____send________ (send) an invitation, our friends __will come___________ (come) to our party.
2. He ________won't understand_____ (not/ understand) you if you ____whisper_________ (whisper).
3. They _____won't survive_______ (not/ survive) in the desert if they ___don't take__________ (not/ take) extra water with them.
4. If you _______press______ (press) CTRL + S, you __________will save___ (save) the file.
5. You ______will cross_______ (cross) the Channel if you __________fly___ (fly) from Paris to London.
Bước 1: Chọn Insert -> Page Number -> Format Page Numbers. Bước 2: Một bảng công cụ sẽ hiện ra, bạn chú ý ở mục Page Numbering có 2 ô lựa chọn, ô bên dưới Start at: … chính là số trang mà bạn muốn bắt đầu đánh số, chỉ cần điền vào đó số trang bạn muốn bắt đầu đánh, Microsoft Word sẽ tự động đánh số từ trang đó cho bạn.
\(a)n_{HCl}=\dfrac{47,45}{36,5}=1,3mol\\ n_{CuO}:n_{Fe_2O_3}=1:1\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}\\ 80n_{CuO}+160n_{Fe_2O_3}=24\\ \Rightarrow80n_{CuO}+160n_{CuO}=24\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}=0,1mol\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ 0,1.......0,2.........0,1..........0,1\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ 0,1..........0,6.........0,2..........0,3\\ n_{HCl.pứ}=0,2+0,6=0,8< n_{HCl}\left(1,3\right)\)
Vậy hh X tan hết
\(b)m_{CuCl_2}=0,1.135=13,5g\\
m_{FeCl_3}=0,2.162,5=32,5g\\
c)n_{HCl.dư}=1,3-0,8=0,5mol\\
ZnO+2HCl\rightarrow ZnCl_2+H_2\\
n_{Zn}=\dfrac{0,5}{2}=0,25mol\\
m_{ZnO}=0,25.81=20,25g\)
\(a)n_C=\dfrac{30}{12}=2,5mol\\ n_P=\dfrac{37,2}{31}=1,2mol\\ n_{O_2}=\dfrac{80}{22,4}=\dfrac{25}{7}mol\\ C+O_2\xrightarrow[]{t^0}CO_2\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\\ n_{O_2.cần,.dùng}=2,5+1,2\cdot\dfrac{5}{4}=4mol< n_{O_2}\left(\dfrac{25}{7}\right)\)
Vậy hh Y không cháy hết
\(b)2H_2O\xrightarrow[điện]{phân}2H_2+O_2\\ n_{H_2O}=4.2=8mol\\ m_{H_2O}=8.16=128g\)
\(A-B=35^2+33^2+31^2+....+3^2+1^2-\left(34^2+32^2+30^2+....+4^2+2^2\right)\\ =\left(35^2-34^2\right)+\left(33^2-32^2\right)+\left(31^2-30^2\right)+...+\left(3^2-2^2\right)+1^2\\ =\left(35-34\right)\left(35+34\right)+\left(33-32\right)\left(33+32\right)+\left(31-30\right)\left(31+30\right)+....+\left(3-2\right)\left(3+2\right)+1\\ =1.\left(35+34\right)+1.\left(33+32\right)+1.\left(31+30\right)+....+1.\left(3+2\right)+1\\ =1+2+3+....+30+31+32+33+34+35\\ =\dfrac{\left(1+35\right).35}{2}=630\)
\(=630\)