a) PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

Ta có : n_Al = \(\dfrac{5,4}{27}\) = 0,2 mol

\(n_{H_2SO_4}\) = \(\dfrac{30}{98}\) = 0,306 mol

Vì \(\dfrac{0,2}{2}< \dfrac{0,306}{3}\) => H₂SO₄ dư

b) Theo PT: \(n_{H_2}\) = 0,3 mol

=> \(V_{H_2}\) = 0,3 x 22,4 = 6,72l

c) \(m_{H_2SO_4}\) dư = \(\left(0,306-0,3\right)\times98\) = 0,588g

\(m_{Al_2\left(SO_4\right)_3}\) = \(0,1\times342\) = 34,2g

\(m_{H_2}\) = 0,3 x 2 = 0, 6g

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

=> \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2.\dfrac{abc}{abc}\) = 4 ( vì a+b + c = abc)

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) => đpcm

Ta có: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)

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