Giúp mình vớiii

n_CO2 = 0.1 (mol) 

Ba(OH)2 + CO2 => BaCO3 + H2O 

0.1________0.1_____0.1

CM Ba(OH)2 = 0.1/0.2 = 0.5 M 

mBaCO3 = 0.1*197 = 19.7 (g) 

Chúc học tốt <3 

Để loại bỏ được CO₂ , SO₂ ta dùng một chất vô cùng rẻ và dể tìm kiếm là : Ca(OH)₂ để hấp thụ lại và tạo ra những chất không gây ô nhiễm môi trường. 

CO₂ + Ca(OH)₂=> CaCO₃ + H₂O

SO₂ + Ca(OH)₂ => CaSO₃ + H₂O

Ta có:  là góc nội tiếp chắn nửa đường tròn (O)

Xét tứ giác HKBI ta có:

 Tứ giác BKHI là tứ giác nội tiếp (dhnb).

b) Xét TGiac AHI và Tgiac AKB có:

    ^AKB = ^AHI ( do cùng =90 độ)

    ^A chung

=> tam giác AHI đồng dạng với AKB (g - g)

=> AH/AB = AI/AK (cặp cạnh tg ứg tỉ lệ)

=> AH.AK = AI.AB

Mà AI; AB cố định

=> AH.AK không phụ thuộc vào vị trí điểm K (đpcm)

Xét (O): Dây CD vuông góc AB tại I (gt)

=> I là TĐ của CD  và OA ( .....)

Xét Tứ giác MDNC có:

^MCN = 90 độ ( do ^MCN là góc nội tiếp chắn nửa đường tròn tâm O)

^CMD =90 độ ( vì CM vg góc MD)

^CND  =90 độ ( vì DN vg góc CB)

=>MNDC là hcn (dhnb)

=> CD và MN cắt nhau tại TĐ của mỗi đường (tc hcn)

Mà I là trung điểm của CD (cmt)

=> CD và MN cắt nhau tại I 

Mà  I thuộc AB (gt)

=> MN, AB, CD đồng quy (đpcm)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

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