a) $A(OH)_2 + 2HCl \to ACl_2 + 2H_2O$

$n_{HCl} = \dfrac{146.10\%}{36,5} = 0,4(mol)$

$n_{A(OH)_2} = \dfrac{1}{2}n_{HCl} = 0,2(mol)$
$\Rightarrow M_{A(OH)_2} = A + 3 4= \dfrac{18}{0,2} = 90$

$\Rightarrow A = 56(Fe)$

Vậy CT của hidroxit kim loại A là $Fe(OH)_2$
b) $m_{dd\ sau\ pư} = 18 + 146 =164(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{164}.100\% = 15,49\%$

1: =>(5x-3)(5x+3)=0

=>x=3/5 hoặc x=-3/5

2: =>x²+8x+16-x²+1=16

=>8x+17=16

=>x=-1/8

1, \(25x^2-9=0\)

\(\Leftrightarrow25x^2=9\) \(\Leftrightarrow\left(5x\right)^2=3^2=\left(-3\right)^2\)\(\Leftrightarrow\left[{}\begin{matrix}5x=3\\5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

2, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x+4\right)^2-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

3, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow2x+10+245=0\Leftrightarrow2x=-255\Leftrightarrow x=-\dfrac{255}{2}\)

Câu 4: 

a: =>(x+3)(x-1)=0

=>x=1 hoặc x=-3

b: =>(x-2)(x+2)=0

=>x=2 hoặc x=-2

c: =>(x+5)(2x-6+x-2)=0

=>(x+5)(3x-8)=0

=>x=8/3 hoặc x=-5

d: =>(x+4)(x+1)=0

=>x=-4 hoặc x=-1

Có trong câu hỏi tương tự ; k cần cảm ơn đâu 

Bài 2: 

a: \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

\(n_{CaCl_2}=\dfrac{22.2}{83}\simeq0.3\left(mol\right)\)

=>\(n_{Ag\left(NO_3\right)}=0.15\left(mol\right)\)

\(m_{AgNO_3}=0.15\cdot170=25.5\left(g\right)\)

b: \(m_{Ca\left(NO_3\right)_2}=0.3\cdot136=40.8\left(g\right)\)

\(m_{AgCl}=0.15\cdot143.5=21.525\left(g\right)\)

\(A=\left(2-1\right)\left(2+1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100-99\right)\left(100+99\right)\)

\(=1+2+...+99+100\)

=5050