Bài 1 : 

$a) MgO + 2HCl \to MgCl_2 + H_2o$
$b) Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
$c) Fe +2 HCl \to FeCl_2 + H_2$
$d) Zn + H_2SO_4 \to ZnSO_4 + H_2$
 

Bài 2 : 

a)

$KOH + HCl \to KCl + H_2O$
$2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O$
$2KOH + CO_2 \to K_2CO_3 + H_2O$
$Fe(OH)_3 + 3HCl \to FeCl_3 + 3H_2O$
$2Fe(OH)_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 6H_2O$
 

Tham khảo nhé bạn:

a. Những chất điều chế bằng pứ hóa hợp: 

b. 

b. Những chất điều chế bằng pứ phân hủy: 

- Phản ứng hóa hợp là phản ứng gồm 2 hay nhiều chất tham gia và chỉ tạo thành 1 chất sản phẩm

- Phản ứng phân hủy là phản ứng gồm 1 chất tham gia  và chỉ tạo thành 2 hay  nhiều chất sản phẩm , phản ứng cần nhiệt độ

a, SO₂

b, CO

c, 

- Mn2O7

d, d, PbO₂

_{Bài 3:}

Giải thích các bước giải:

Bài 1 : 
$Ca + 2H_2O \to Ca(OH)_2 + H_2$

Khí ẩm $H_2$ được làm khô bằng canxi vì $H_2$ không phản ứng với $Ca(OH)_2$

Khí ẩm $CO_2,SO_2$ không thể làm khô bằng canxi vì chúng phản ứng tạo sản phẩm khác

Bài 2 : 

2 tấn = 2000 kg

$m_{CaCO_3} = 2000.80\% = 1600(kg)$
$CaCO_3 \xrightarrow{t^o} CaO + CO_2$
$n_{CaO} = n_{CaCO_3\ pư} = \dfrac{1600}{100}.90\% = 14,4(kmol)$
$m_{CaO} = 14,4.56 = 806,4(kg)$

Bài 3 : 

a) $SO_3 + H_2O \to H_2SO_4$

b)

$n_{H_2SO_4} = n_{SO_3} = \dfrac{8}{80} = 0,1(mol)$
$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,25} = 0,4M$

Vd5:

a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)

\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)

Ta có: \(23^0<30^0<38^0<48^0<75^0\)

=>Sin23<sin30<sin38<sin48<sin 75

=>cos67<sin30<sin38<cos42<sin75

b: \(\cot49^0=\tan\left(90^0-49^0\right)=\tan41^0\)

\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)

Ta có: 25<27<40<41<80

=>tan25<tan27<tan40<tan41<tan80

=>tan25<tan27<cot49<cot50<tan80

Vd6:

a: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)

=>cosa=0,6

tan a=sin a/cosa

=0,8/0,6

=4/3

cot a=cosa/sin a

=0,6/0,8=3/4

b: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-0,6^2=1-0,36=0,64\)

=>sin a=0,8

tan a=sin a/cosa

=0,8/0,6

=4/3

cot a=cosa/sin a

=0,6/0,8=3/4

c: \(\tan a\cdot\cot a=1\)

=>\(\cot a=\frac13\)

\(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+3^2=1+9=10\)

=>\(cos^2a=\frac{1}{10}\)

=>\(cosa=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)

=>\(\sin a=\frac{3}{\sqrt{10}}\)

d: \(\tan a\cdot\cot a=1\)

=>\(\tan a=\frac12\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=1+\frac14=\frac54\)

=>\(cos^2a=\frac45\)

=>\(cosa=\frac{2}{\sqrt5}\)

\(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac45=\frac15\)

=>\(\sin a=\frac{1}{\sqrt5}\)

Vd5:

a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)

\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)

Ta có: 23<30<38<48<75

=>sin23<sin30<sin38<sin48<sin75

=>cos67<sin30<sin38<cos42<sin75

b: \(\cot49^0=\tan\left(90^0-49^0\right)=tan41^0\)

\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)

Ta có: 25<27<40<41<80

=>tan 25<tan 27<tan 40<tan41<tan80

=>tan 25<tan27<cot49<cot50<tan80

Vd6:

a: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)

=>cosa=0,6

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

b: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)

=>sin a=0,8

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

c: \(\tan a\cdot\cot a=1\)

=>\(\cot a=\frac13\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+3^2=10\)

=>\(cos^2a=\frac{1}{10}\)

=>\(cosa=\frac{1}{\sqrt{10}}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)

=>\(\sin a=\frac{3}{\sqrt{10}}\)

d: \(\tan a\cdot\cot a=1\)

=>\(\tan a=\frac{1}{\cot a}=\frac12\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2a=\frac45\)

=>\(cosa=\frac{2}{\sqrt5}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(sin^2a=1-\frac45=\frac15\)

=>sin a=\(\frac{1}{\sqrt5}\)

Vd5:

a: \(cos42^0=\sin\left(90^0-42^0\right)=\sin48^0\)

\(cos67^0=\sin\left(90^0-67^0\right)=\sin23^0\)

Ta có: 23<30<38<48<75

=>sin23<sin30<sin38<sin48<sin75

=>cos67<sin30<sin38<cos42<sin75

b: \(\cot49^0=\tan\left(90^0-49^0\right)=tan41^0\)

\(\cot50^0=\tan\left(90^0-50^0\right)=\tan40^0\)

Ta có: 25<27<40<41<80

=>tan 25<tan 27<tan 40<tan41<tan80

=>tan 25<tan27<cot49<cot50<tan80

Vd6:

a: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)

=>cosa=0,6

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

b: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)

=>sin a=0,8

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

c: \(\tan a\cdot\cot a=1\)

=>\(\cot a=\frac13\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+3^2=10\)

=>\(cos^2a=\frac{1}{10}\)

=>\(cosa=\frac{1}{\sqrt{10}}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac{1}{10}=\frac{9}{10}\)

=>\(\sin a=\frac{3}{\sqrt{10}}\)

d: \(\tan a\cdot\cot a=1\)

=>\(\tan a=\frac{1}{\cot a}=\frac12\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2a=\frac45\)

=>\(cosa=\frac{2}{\sqrt5}\)

Ta có: \(\sin^2a+cos^2a=1\)

=>\(sin^2a=1-\frac45=\frac15\)

=>sin a=\(\frac{1}{\sqrt5}\)

vd5 :cos67^o,sin30^o,sin38^o,cos42^o ,sin75^o

        tan25^o,  tan27^o, cot50^o,cot49^o,tan80^o

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{3+\sqrt5}{\sqrt2}=\frac{3\sqrt2+\sqrt{10}}{2}\)

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

=>\(\sqrt{\left(x-1\right)^2}=7\)

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)