nNaOH = 16/40 = 0,4 (mol)

PTHH: 2NaOH + SO2 -> Na2SO3 + H2O

Mol: 0,4 ---> 0,2 ---> 0,2

VSO2 = 0,2 . 22,4 = 4,48 (l)

mNa2SO3 = 0,2 . 126 = 25,2 (g)

Câu 1: D

Câu 2: C

Câu 3: C

Câu 4: D

Câu 5: A

 1: D

 2: C

 3: C

 4: D

 5: A

Bạn tham khảo:

Làm hộ mình câu 3 hình với ạ ^^ - Hoc24

3.

a. $2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}$

$=2\sqrt{6}-4\sqrt{2}+(1+8+4\sqrt{2})-2\sqrt{6}$

$=2\sqrt{6}-2\sqrt{6}-4\sqrt{2}+4\sqrt{2}+9$

$=9$

b.

$\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}$

$=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}$

$=\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}$

$=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}$

c.

\(\sqrt{\frac{4}{(2-\sqrt{5})^2}}-\sqrt{\frac{4}{(2+\sqrt{5})^2}}=\sqrt{(\frac{2}{2-\sqrt{5}})^2}-\sqrt{(\frac{2}{(2+\sqrt{5}})^2}\)

\(=|\frac{2}{2-\sqrt{5}}|-|\frac{2}{2+\sqrt{5}}|=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2(\sqrt{5}+2)-2(\sqrt{5}-2)}{(\sqrt{5}-2)(\sqrt{5}+2)}=\frac{8}{5-2^2}=8\)

Bài 3:

a) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(2\sqrt{2}+1\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}-2\sqrt{6}+9+4\sqrt{2}\)

=9

b) Ta có: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\sqrt{6}\)

Bài 2: 

a) Ta có: \(A=\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)

\(=-\sqrt{3}\)

a) 4x² + 4xy + y² = (2x + y)²

c) 8a³ - 36a²b + 54ab² - 27b² = (2a - 3b)³

d) 8x³ + 12x²y + 6xy² + y³ = ( 2x - y)³

5. 

$|E|=|x_1-x_2|=\sqrt{(x_1-x_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2}$

$\Rightarrow E=\pm \sqrt{(x_1+x_2)^2-4x_1x_2}$

6.

$|F|=|x_1^2-x_2^2|=|x_1-x_2|.|x_1+x_2|=\sqrt{(x_1+x_2)^2-4x_1x_2}.|x_1+x_2|$

$\Rightarrow F=\pm \sqrt{(x_1+x_2)^2-4x_1x_2}(x_1+x_2)$

7.

$G=x_1^3-x_2^3=(x_1-x_2)(x_1^2+x_1x_2+x_2^2)$

$=(x_1-x_2)[(x_1+x_2)^2-x_1x_2]=E[(x_1+x_2)^2-x_1x_2]$

8.

$H=x_1^4+x_2^4=(x_1^2+x_2^2)(x_1^2-x_2^2)=AF$

9.

$I=\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}$10.

$K=\frac{1}{x_1-1}+\frac{1}{x_2-1}=\frac{x_2-1+x_1-1}{(x_1-1)(x_2-1)}=\frac{(x_1+x_2)-2}{x_1x_2-(x_1+x_2)+1}$

11.

$L=\frac{x_2}{x_1-1}+\frac{x_1}{x_2-1}=\frac{x_2^2-x_2+x_1^2-x_1}{(x_1-1)(x_2-1)}$

$=\frac{(x_1^2+x_2^2)-(x_1+x_2)}{(x_1-1)(x_2-1)}$

$=\frac{A-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}$

12.

M=\frac{1}{x_1}-\frac{1}{x_2}=\frac{x_2-x_1}{x_1x_2}=\frac{-E}{x_1x_2}$

13.

$N=|x_1-x_2|=\sqrt{(x_1-x_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2}$

14.

$P=|x_1|+|x_2|=\sqrt{(|x_1|+|x_2|)^2}=\sqrt{x_1^2+x_2^2+2|x_1x_2|}$

$=\sqrt{(x_1+x_2)^2-2x_1x_2+2|x_1x_2|}$

15.

$Q=\sqrt{x_1}+\sqrt{x_2}=\sqrt{(\sqrt{x_1}+\sqrt{x_2})^2}=\sqrt{x_1+x_2+2\sqrt{x_1x_2}}$

16.

$T=x_1\sqrt{x_1}+x_2\sqrt{x_2}$

$=\sqrt{x_1x_2}(\sqrt{x_1}+\sqrt{x_2})=\sqrt{x_1x_2}.Q$

câu j hả bạn

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