Bài 1:

Ta có: \(n_{HCl}=1,5\cdot0,08=0,12\left(mol\right)\) \(\Rightarrow V_{HCl\left(2M\right)}=\dfrac{0,12}{2}=0,06\left(l\right)=60\left(ml\right)\)

Bài 2:

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{HCl}=2,5\cdot2=5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5}{3}\left(mol\right)\\n_{H_2}=2,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=\dfrac{5}{3}\cdot27=45\left(g\right)\\V_{H_2}=2,5\cdot22,4=56\left(l\right)\end{matrix}\right.\) 

 C NHA BN CÂU 45 KO LÀM ĐC

a) Ta có: AB//CD.

=>ABH=BDC (2 góc so le trong).

=> ∆AHB~∆BCD(g.g).

b) ∆ABD có :  DB²=AB²+AD²( Định lý Pitago)

=> DB= 15(cm).

Ta có ∆ABH~∆BCD(cmt).

=>AH/BC=AD/BD.

Hay AH=9.12/15=7,2(cm).

c)Ta có ∆AHB~∆BCD cmt.

=> HBA=CBD. (1)

Ta lại có : CBD= ADH (AB//CD).(2)

Từ 1 và 2 => HAB=ADH.

=>∆DHA~∆AHB(g.g).

S∆DHA/S∆AHB=(AD/AB)²=9/16

d) từ câu (a) và (b) => ∆BCD~∆DHA.

Cm ∆DHA~∆MDA(g.g)

Từ đó  suy ra ∆BDC~∆MDA.

Sau đó cm ∆BCD~∆ADC(g.g).

=> ∆MDA~∆ADC(g.g).

=>Ad/DC=DM/DC.

=>Đpcm.

Câu 2.

\(C=\dfrac{N}{20}=\dfrac{4080}{20}=204\)(chu kì)

\(L=\dfrac{N}{2}\cdot3,4=\dfrac{4080}{2}\cdot3,4=6936A^o\)

\(G=X=30\%\cdot4080=1224nu\)

\(A=T=\dfrac{4080-2\cdot1224}{2}=816nu\)

\(a.C=\dfrac{N}{20}=204\)

\(b.L=\dfrac{Nx3,4}{2}=6936A\)

\(c.A=T=20\%N=816;G=X=30\%N=1224\)

\(d.rN=\dfrac{N}{2}=2040\)

b.L=Nx3,4/2=6936A

1. English is more interesting than music.

2. Today they are not as happy as they were yesterday.

3. Ha Noi is not as small as Hai Duong.

4. Mai's sister is not as pretty as her.

6. You have got more money than me.

7. Art is not as difficult as French.

8. Nam's father is more careful than him.

9. No one in our town is as rich as Mr Ron.

10. He is the most intelligent in my class.

11. Everest is the highest mountain in the world.

12. Minh is the fattest person in my group.

13. I can't swim as far as Jan.

14B 15C 16A 17C 18B 19C 20B

bạn có thể chụp thẳng đc hong?????

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3mol\Rightarrow m_C=3,6g\)

\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\Rightarrow n_H=0,4\cdot2=0,8\Rightarrow m_H=0,8g\)

Nhận thấy: \(m_C+m_H=4,4=m_A\)

\(\Rightarrow A\) chỉ chứa hai nguyên tố C và H.

Gọi CTHH là \(C_xH_y\).

\(\Rightarrow x:y=n_C:n_H=0,3:0,8=3:8\)

\(\Rightarrow C_3H_8\)

Gọi CTĐGN là \(\left(C_3H_8\right)_n\)

Mà \(M=44\)g/mol\(\Rightarrow44n=44\Rightarrow n=1\)

Vậy CTPT là \(C_3H_8\)

A không làm mất màu dung dịch brom.

8D 5D 6B