Câu 2:

 \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

  2a           3a              a                 3a

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 b         b                 b           b

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25mol\)

Ta có: \(\left\{{}\begin{matrix}54a+65b=9.2\\3a+b=0.25\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.05\\b=0.1\end{matrix}\right.\)

a.\(\%m_{Al}=\dfrac{0.05\times54\times100}{9.2}=29.3\%\)

\(\%m_{Zn}=100-29.3=70.7\%\)

Vdd sau phản ứng = 9.2 + 600 - 0.0056 = 609.2ml

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0.05}{0.6092}=0.08M\)

\(CM_{ZnSO_4}=\dfrac{0.1}{0.6092}=0.16M\)

Câu 3: 

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

 0.2       0.2            0.2          0.2

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2mol\)

a. \(\%m_{Mg}=\dfrac{0.2\times24\times100}{12}=40g\)

\(\%m_{FeO}=100-40=60\%\)

b. \(n_{FeO}=\dfrac{12-0.2\times24}{72}=0.1mol\)

m muối khan \(=m_{MgSO_4}+m_{FeSO_4}=0.2\times120+0.1\times152=39.2g\)

Bài 3: 

a: Thay x=4 vào A, ta được:

\(A=\dfrac{2\cdot4}{4-9}=\dfrac{8}{-5}=-\dfrac{8}{5}\)

b: Ta có: \(B=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{5}{\sqrt{x}+3}+\dfrac{2x+12}{9-x}\)

\(=\dfrac{2x+6\sqrt{x}-5\sqrt{x}+15-2x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{1}{\sqrt{x}-3}\)

Task 2

1 annoying

2 meaningful

3 hopeless

4 volunteer

5 needy 

6 disadvantaged

7 donate

8 successful

Task 3

1 Did you watch

2 did - break

3 came

4 were you doing

5 were

6 called

7 was playing - arrived

8 were swimming - were sunbathing

Luôn cả câu 4 nữa ạ

Bài 2: 

a: \(\Leftrightarrow x^2+2x-x^2=0\)

=>2x=0

hay x=0

b: \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

hay \(x\in\left\{-3;5\right\}\)

Wlk=(Z*mp-(A-Z)*mn-mCo)*c^2

=(27*1,0073u +31*1,0087u-56,9540u)*c^2=1,5128u*c^2=1,5128*931,5

=1409,1732MeV

--->Wr=Wlk/A=1409,1732/58=24,296MeV

bài 1: 

\(\left\{{}\begin{matrix}x+y=57\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=228\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6y=234\\x+y=57\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=39\\x=18\end{matrix}\right.\)

Bài 1: 

c: Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là 

x1=1; \(x2=\dfrac{c}{a}=\dfrac{3\sqrt{2}+1}{1-\sqrt{2}}\)