28.D

29.B

30.B

31.C

32.D

33.A

34.B

35.A

35C....bôi dầu vào cửa khó chịu lắm em ơi

18A 19D 20D

33A

34B

35C

36D

37D

38A

39B

40.Natri

PTHH: Al2O3+6HCl➝2AlCl3+3H2O(1)

a)nAl2O3=\(\dfrac{10,2}{102}\)=0,1(mol)

   mHCl=\(\dfrac{5\%.219}{100\%}\)=10,95(g)

   ⇒nHCl=\(\dfrac{10,95}{36,5}\)=0,3(mol)

Xét tỉ lệ Al2O3:\(\dfrac{0,1}{1}\)=0,1

Xét tỉ lệ HCl:\(\dfrac{0,3}{6}\)=0,05

⇒HCl pứng hết,Al2O3 còn dư

Theo PTHH(1) ta có nAl2O3 pứng=\(\dfrac{nHCl}{6}\)=\(\dfrac{0,3}{6}\)=0,05(mol)

⇒nAl2O3 dư=nAl2O3ban đầu-nAl2O3 pứng=0,1-0,05=0,05(mol)

⇒mAl2O3 dư=0,05.102=5,1(g)

b) C%HCl=\(\dfrac{0,3.36,5}{219+10,2}\).100%=4,8%

     nAlCl3=0,1(mol)

⇒C%AlCl3=\(\dfrac{0,1.136,5}{10,2+219}\).100%=6%

Bài 2:

a: \(\left(2x-5\right)^2-4x\left(x-3\right)=0\)

=>\(4x^2-20x+25-4x^2+12x=0\)

=>-8x+25=0

=>-8x=-25

=>\(x=\frac{25}{8}\)

b: \(2\left(x+5\right)-x^2-5x=0\)

=>2(x+5)-x(x+5)=0

=>(x+5)(2-x)=0

=>\(\left[\begin{array}{l}x+5=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=2\end{array}\right.\)

c: \(6x^2-7x+2=0\)

=>\(6x^2-3x-4x+2=0\)

=>3x(2x-1)-2(2x-1)=0

=>(2x-1)(3x-2)=0

=>\(\left[\begin{array}{l}2x-1=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=\frac23\end{array}\right.\)

Bài 1:

a: x(x-5)+(x+3)(x-3)

\(=x^2-5x+x^2-9\)

\(=2x^2-5x-9\)

b: \(\frac{x}{x-1}+\frac{2x-4}{x^2-1}-\frac{5}{x+1}\)

\(=\frac{x\left(x+1\right)+2x-4-5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x+2x-4-5x+5}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x-1}{x+1}\)

c: \(\left(20x^2+7x-6\right):\left(5x-2\right)\)

\(=\left(20x^2-8x+15x-6\right):\left(5x-2\right)\)

\(=\frac{4x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=4x+3\)

PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(\left\{{}\begin{matrix}m_{H_2SO_4}=588\cdot5\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{3}\) \(\Rightarrow\) Al₂O₃ còn dư

\(\Rightarrow n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)=n_{Al_2O_3\left(dư\right)}\)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1\cdot342}{20,4+588-0,1\cdot102}\cdot100\%\approx5,72\%\)

My favourite folk tale names Cinderella. There is Cinderella who is the main character, her two step-sisters and stepmother who are the villains. It's about a beautiful girl by the name of Cinderella lived with her wicked stepmother and step-sisters. She had a really unlucky life unless she met the Prince and they had fallen in love each other. Finally, they had a successful marriage and lived happily ever after. The theme of the story "Cinderella" is that kindness will be rewarded, whereas selfishness will not. The reason why i like this folk tale most it because it gave me a meaningful lesson and it's also romantic, because of the love story of Cinderella and the Prince.
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