5. a) Fe + H2SO4 --------> FeSO4 + H2

Mg + H2SO4 --------> MgSO4 + H2

Gọi x, y lần lượt là số mol của Fe, Mg

Ta có \(\left\{{}\begin{matrix}56x+24y=12\\x+y=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,15\\y=0,15\end{matrix}\right.\)

=> \(\%m_{Fe}=\dfrac{0,15.56}{12}.100=70\%\)

=> %m_Mg = 100 -70 = 30%

b) Ta có n _H2SO4 = n _Fe + n_Mg = 0,15 + 0,15 =0,3 (mol)

=>a = \(CM_{H_2SO_4}=\dfrac{0,3}{0,2}=1,5M\)

c) \(CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\)

\(CM_{MgSO_4}=\dfrac{0,15}{0,2}=0,75M\)

6. a) 2Al + 6HCl → 2AlCl3 + 3H2

Zn  + 2HCl →ZnCl2 + H2

Gọi x, y lần lượt là số mol của Al, Zn

\(\left\{{}\begin{matrix}27x+65y=11,9\\\dfrac{3}{2}x+y=\dfrac{8,96}{22,4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> \(\%m_{Al}=\dfrac{27.0,2}{11,9}.100=45,38\%\)

=> %m _Zn = 100- 45,38=54,62%

b) \(a=C\%_{HCl}=\dfrac{\left(0,2.2+0,1.2\right).36,5}{200}.100=10,95\%\)

c) \(m_{ddsaupu}=11,9+200-\left(0,2.2+0,1.2\right)=211,3\left(g\right)\)

=> \(C\%_{AlCl_3}=\dfrac{0,2.133,5}{211,3}.100=12,64\%\)

\(C\%_{ZnCl_2}=\dfrac{0,1.136}{211,3}.100=6,44\%\)

câu 6 thể tích 1kg khí ở 0 độ 1,29 m³ 

áp dụng đẳng quá trình \(\dfrac{P_0V_0}{T_0}=\dfrac{PV}{T}\Leftrightarrow\dfrac{1,01.10^5.1,29}{0+273}=\dfrac{3.10^5.V}{70+273}\Rightarrow V\approx0,5456\left(m^3\right)\)

khối lượng riêng 0,5456kg/m³

$n_{CaO} = \dfrac{5,6}{56} =0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH : 

$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$

\(n_{CaO}\) = 0,1 mol

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

0,1 mol → 0,2 mol

\(\Rightarrow m_{HCl}\) = 0,2.36,5 = 7,3 gam

\(\Rightarrow\) Khối lượng dd HCl đã dùng là:  \(m_{HCl}\)= \(\dfrac{7,3\times100\%}{14,6\%}\)=50(g)

Câu 3: 

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)

Do đó: x=54; y=36

=2001

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)

1 interested in travelling by plane

2 15 minutes riding his bike to school

3 is made by my mother everymorning

4 take care of her little brother

5 able to speak 2 languages when he was young

6 able to play the piano

7 is played all over VN

8 to play voleyball very well when he was young

9 to go fishing when he was young

10 way to the shopping mall