2 The meeting will be held before May Day

3 The engine of the car has to be repaired

4 A story is going to be told by the teacher

5 The cake is being cut with a sharp knife by Mary

6 Luckily for me, my name wasn't called

7 His story isn't believed

8 After class, the chalk board is always erased by one of the students

9 The victims have been provided with food and drink

11 Each of us were given 2 exercises books

12 He will be told that news

13 A lot of money was paid me to do the job

14 The window of the laboratory has been changed

C

C

B

A

C

A

a) \(Mg\left(OH\right)_2\) - Magie hidroxit

    \(MgCl_2\)       - Magie clorua

    \(MgS\)          - Magie sunfua

    \(Mg_3\left(PO_4\right)_2\) - Magie photphat

    \(Mg\left(HCO_3\right)_2\)  - Magie bicacbonat

a, \(Mg\left(OH\right)_2,MgCl_2\), \(MgS,Mg\left(PO_4\right)_2,Mg\left(HCO_3\right)_2\)

b, \(AlBr_3,Al\left(HS\right)_3,\) \(Al_2\left(SO_4\right)_3\), \(Al\left(HPO_4\right)_3\) , \(Al_2\left(CO_3\right)_3\)

c, \(Cu\left(OH\right)_2,CuCO_3,CuNO_3\)

d, \(HNO_2,HNO_3,H_2CO_3,H_3PO_4\) , \(H\left(H_2PO_4\right)\) , \(H\left(HCO_3\right),H\left(HS\right),H_2S,\)\(H\left(HSO_3\right)\)

b) \(AlBr_3\)  - Nhôm bromua

    \(Al\left(HS\right)_3\)  - nhôm hidrosunfua

     \(Al_2\left(SO_4\right)_3\)  - Nhôm sunfat

     \(Al_2\left(HPO_4\right)_3\)  - Nhôm hidrophotphat

     \(Al_2\left(CO_3\right)_3\)   - Nhôm cacbonat

c) \(Cu\left(OH\right)_2\)  - Đồng (II) hidroxit

    \(CuCO_3\)   -  Đồng (II) cacbonat

    \(Cu\left(NO_3\right)_2\)  - Đồng (II) nitrat

d)  \(HNO_2\)  - Axit nitro

     \(HNO_3\)  - Axit nitric

     \(H_2CO_3\)  - Axit cacbonic

     \(H_3PO_4\)  - Axit photphoric

1: Xét ΔABD vuông tại A và ΔDAC vuông tại D có

góc ABD=góc DAC

=>ΔABD đồng dạng với ΔDAC

2: ΔABD đồng dạng với ΔDAC

=>BD/AC=AB/DA=AD/DC

=>AD/16=BD/AC=18/DA

=>AD^2=16*18=288

=>AD=12căn 2(cm)

AC=căn AD^2+DC^2=4căn 82(cm)

Câu 10: 

\(a,Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ b,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ c,2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\\ d,2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+H_2O\\ Fe_2\left(SO_4\right)_3+BaCl_2\rightarrow FeCl_2+BaSO_4\left(PTHH.Sai\right)\\ 2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\\ 2Al+3MgO\rightarrow\left(t^o\right)Al_2O_3+3Mg\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\)

Câu 8,9 đã làm!

Câu 7:

\(\text{Đ}\text{ặt}:N_xO_y\left(x,y:nguy\text{ê}n,d\text{ươ}ng\right)\\ V\text{ì}:\dfrac{m_N}{m_O}=\dfrac{7}{20}\\ \Leftrightarrow\dfrac{14x}{16y}=\dfrac{7}{20}\\ \Rightarrow\dfrac{x}{y}=\dfrac{7.16}{20.14}=\dfrac{2}{5}\\ \Rightarrow x=2;y=5\\ \Rightarrow CTHH:N_2O_5\)

Câu 5,6 tương tự nha!

a) \(d\dfrac{CH_4}{kk}=\dfrac{M_{CH_4}}{M_{kk}}=\dfrac{16}{29}< 1\)

⇒ \(CH_4\) nhẹ hơn kk

b) \(d\dfrac{N_2O}{kk}=\dfrac{M_{N_2O}}{M_{kk}}=\dfrac{44}{29}=1,52>1\)

⇒ \(N_2O\) nặng hơn kk

\(15\left(\dfrac{m.}{s}\right)=54\left(\dfrac{km}{h}\right)\)

\(\Rightarrow v_{tb}=\dfrac{s}{\dfrac{\dfrac{1}{2}s}{v'}+\dfrac{\dfrac{1}{2}s}{v''}}=\dfrac{s}{\dfrac{s}{36}+\dfrac{s}{108}}=\dfrac{s}{\dfrac{4s}{108}}=\dfrac{108}{4}=27\left(\dfrac{km}{h}\right)\)

\(TBCv_1-v_2=\left(18+54\right):2=36\left(\dfrac{km}{h}\right)\)

\(\Rightarrow v_{tb}< TBCv_1-v_2\left(27< 36\right)\)

Vậy ta đc đpcm.