a. Đổi 100 m = 0,1 lít

\(n_{CO_2}=\dfrac{1.12}{22,4}=0,05\left(mol\right)\)

PTHH : CO₂ + 2NaOH -> Na₂CO₃ + H₂O

          0,05    0,1

\(C_M=\dfrac{0.1}{0,1}=1\left(M\right)\)

\(\left(x^2-1\right).\left(x+2\right)-x^2.\left(x+2\right)=21\\ \Leftrightarrow x^3+2x^2-x-2-x^3-2x^2=21\\ \Leftrightarrow x-2=21\\ \Leftrightarrow x=23\)

\(\left(x+2\right).\left(x^2-4x+1\right)-\left(x^3-2x^2+2\right)=-18\\ \Leftrightarrow x^3-4x^2+x+2x^2-8x+2-x^3+2x^2-2=-18\\ \Leftrightarrow-7x=-18\\ \Leftrightarrow x=\dfrac{18}{7}\)

D

\(\left(x^2-3x\right).\left(2x-1\right)-x.\left(2x^2-7x\right)=12\\ \Leftrightarrow2x^3-x^2-6x^2+3x-2x^3+7x^2=12\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)

\(\left(2x+3\right).\left(x^2+x\right)-2x\left(x^2+2,5x\right)=9\\ \Leftrightarrow2x^3+2x^2+3x^2+3x-2x^3-5x^2=9\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)

Câu 3 :

a. \(n_{H_2}=\dfrac{3.36}{22,4}=0,15\left(mol\right)\)

PTHH : Fe + H₂SO₄ --t^o---> FeSO₄ + H₂

          0,15                    0,15     0,15

b. \(m_{Fe}=0,15.56=8,4\left(g\right)\)

c. \(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

d. PTHH : CuO + H₂ -> Cu + H₂O

            0,15   0,15

\(m_{CuO}=0,15.80=12\left(g\right)\)

nãy thấy có D j đó nx mà =))

đề có thiếu j ko á nhỉ