Bài 2:

\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,05(mol)\\ \Rightarrow m_{Mg}=24.0,05=1,2(g)\\ \Rightarrow m_{MgO}=9,2-1,2=8(g) b,\%_{Mg}=\dfrac{1,2}{9,2}.100\%=13,04\%\\ \Rightarrow \%_{MgO}=100\%-13,04\%=86,96\%\\ c,n_{MgO}=\dfrac{8}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,5(mol)\\ \Rightarrow \Sigma m_{HCl}=0,5.36,5=18,25(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{18,25}{14,6\%}=125(g)\)

Bài 5 :

 \(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2            1              1

      0,05      0,1                          0,05

        \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

         1              2             1            1

        0,2           0,4

a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1,2\left(g\right)\)

\(m_{MgO}=9,2-1,2=8\left(g\right)\)

b) ⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

    ⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

c) Có : \(m_{MgO}=8\left(g\right)\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

 Chúc bạn học tốt

Bài 3:

\(a,PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)\\ \Rightarrow n_{CO_2}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\\ c,m_{dd_{HCl}}=1,15.300=345(g)\\ n_{NaCl}=0,4(mol);n_{H_2O}=n_{CO_2}=0,2(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\\ m_{H_2O}=18.0,2=3,6(g)\\ m_{CO_2}=44.0,2=8,8(g)\\ m_{dd_{NaCl}}=21,2+345-3,6-8,8=353,8(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{23,4}{353,8}.100\%=6,61\%\)

12

Làm mẫu câu a nhé:

Ta có: \(2x=3y\)

   \(\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x^2}{9}=\frac{y^2}{4}\)

Áp dụng t/c dãy tỉ số = nhau ta có:

\(\frac{x}{3}=\frac{y}{2}=\frac{x^2}{9}=\frac{y^2}{4}=\frac{x^2-y^2}{9-4}=5\)

\(\Rightarrow x=3.5=15\)

\(y=5.2=10\)

Ý 1:

\(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)

Áp dụng t/c DTSBN ta có : \(\frac{x}{3}=\frac{y}{2}=\frac{x^2-y^2}{3^2-2^2}=\frac{25}{5}=5\)

=> x,y=...

\(\frac{x}{3}=\frac{y}{4}\)

Áp dụng t/c DTSBN ta có : \(\frac{x}{3}=\frac{y}{4}=\frac{3x-2y}{3.3-2.4}=\frac{5}{1}=5\)

=>x,y=...

\(3x=2y=5z\Leftrightarrow\frac{x}{2}=\frac{y}{5}=\frac{z}{3}\)

Áp dụng t/c DTSBN ta có : \(\frac{x}{2}=\frac{y}{5}=\frac{z}{3}=\frac{y-2x}{5-2.2}=\frac{5}{1}=5\)

=>x,y,z=....

1a.

$x^2-5x+6=x^2-2x-(3x-6)=x(x-2)-3(x-2)=(x-2)(x-3)$

1b.

$3x^2+9x-30=3(x^2+3x-10)=3(x^2-2x+5x-10)$

$=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$

1c.
$x^2-3x+2=(x^2-x)-(2x-2)=x(x-1)-2(x-1)=(x-1)(x-2)$

1d.

$x^2-9x+18=x^2-3x-(6x-18)=x(x-3)-6(x-3)=(x-3)(x-6)$

1e.

$x^2-6x+8=x^2-2x-(4x-8)=x(x-2)-4(x-2)=(x-2)(x-4)$

1f.
$x^2-5x-14=x^2-7x+2x-14=x(x-7)+2(x-7)=(x+2)(x-7)$

1g.

$x^2+6x+5=(x^2+x)+(5x+5)=x(x+1)+5(x+1)=(x+1)(x+5)$

1h.

$x^2-7x+12=x^2-3x-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$

1i.

$x^2-7x+10=(x^2-2x)-(5x-10)=x(x-2)-5(x-2)=(x-2)(x-5)$

2a.

$3x^2-5x-2=(3x^2-6x)+(x-2)=3x(x-2)+(x-2)$

$=(x-2)(3x+1)$

2b. 

$2x^2+x-6=(2x^2+4x)-(3x+6)=2x(x+2)-3(x+2)$

$=(2x-3)(x+2)$

2c.

$7x^2+50x+7=(7x^2+49x)+(x+7)$

$=7x(x+7)+(x+7)=(7x+1)(x+7)$

2d. 

$12x^2+7x-12=(12x^2-9x)+(16x-12)$

$=3x(4x-3)+4(4x-3)=(4x-3)(3x+4)$

2e.

$15x^2+7x-2=15x^2-3x+10x-2$

$=3x(5x-1)+2(5x-1)=(5x-1)(3x+2)$

4.

\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)

\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)

\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)

\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

3.

\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)

\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)

\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)