a) PT: MgO + 2HCl \(\rightarrow\) MgCl₂ + H₂O

--------0,2------0,4-------------------------

b) n_MgO \(=\dfrac{8}{40}=0,2\left(mol\right)\)

Đổi 200ml = 0,2l

\(C_M=\dfrac{n}{V}=\dfrac{0.04}{0,2}=0,2\left(M\right)\)

-10x² + 11x + 6

= -10x² + 15x - 4x + 6

= -5x(2x - 3) - 2(2x - 3)

= (2x - 3)(-5x - 2)

-8x² + 5x + 3

= -8x² + 8x - 3x + 3

= -8x(x - 1) - 3(x -1)

= (x - 1)(-8x - 3)

1+2+3+...+100=100x(100+1):2=5050

3. Ta có:

\(\dfrac{U_1}{U_2}=\dfrac{I_1}{I_2}\Leftrightarrow\dfrac{4,5}{U_1-0,5}=\dfrac{0,36}{I_2}\Leftrightarrow\dfrac{4,5}{4,5-0,5}=\dfrac{0,36}{I_2}\Leftrightarrow I_2=0,32\left(A\right)\)

Đáp án: B

Ta có:

\(\dfrac{U_1}{U_2}=\dfrac{I_1}{I_2}\Leftrightarrow I_2=\dfrac{I_1.U_2}{U_1}=\dfrac{0,24.2,5}{4}=0,15\left(A\right)\)

Đề: \(\left(5-\sqrt{2}\right)^2=27-10\sqrt{2}\)

Biến đổi vế trái ta có:

\(\left(5-\sqrt{2}\right)^2=5^2-2.5.\sqrt{2}+2=25-10\sqrt{2}+2=27-10\sqrt{2}=VP\left(đpcm\right)\)

Yêu cầu của đề bạn ơi.

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

a) PT: 2Na + 2H₂O \(\rightarrow\) 2NaOH + H₂

--------0,1----------------0,1--------0,05 (mol)

b) m_NaOH = 0,1.40 = 4(g)

c) đktc: \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

d) 200 ml = 0,2 l

\(C_M=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

e) 4x² + 15x + 9

= 4x² + 12x + 3x + 9

= 4x(x + 3) + 3(x + 3)

= (x+3)(4x+3)