$n_{K_{2}O}=\frac{14,1}{94}=0,15\left(mol\right)$

PTHH: K₂O + H₂O --> 2KOH
            0,15---------->0,3

=> m_KOH = 0,3.56 = 16,8 (g)

m_{dd sau pư} = 14,1 + 41,9 = 56 (g)

=> $C{\%}_{dd.KOH}=\frac{16,8}{56}.100\%=30\%$

$(x-2)\times (2x^2+x-1)-2x\times (x^2-\frac{3}{2}x)=22$

$2x^3+x^2-x-4x^2-2x+2-2x^3+3x^2=22$

$(2x^3-2x^3)+(x^2-4x^2+3x^2)+(-x-2x)+2=22$

$-3x+2=22$

(x - 2) x (2x2 + x - 1) - 2x . (x2 - 3/2x ) =22

=> 2x³ + x² - x - 4x² - 2x + 2 - 2x³ + 3x² = 22

=> - 3x = 20

hay x = - 20/3

vậy x = [20/3 ; - 20/3]

$\iff 2x^3+x^2-x-4x^2-2x+2-2x^3+3x^2=22$

=>-3x=20

hay x=-20/3

$\left(x^2-1\right).\left(x+2\right)-x^2.\left(x+2\right)=21$

$\iff \left(x+2\right).\left(x^2-1-x^2\right)=21$

$\iff \left(x+2\right).\left(-1\right)=21$

$\iff -x-2=21$

$\iff -x=21+2$

$\iff -x=23$

$\iff x=-23$

(x2−1).(x+2)−x2.(x+2)=21⇔x3+2x2−x−2−x3−2x2=21⇔x−2=21⇔x=23

a) $2NaOH+C{O}_2\to N{a}_{2}C{O}_3+H_{2}O$

b) 

$n_{C{O}_2}=\frac{1,12}{22,4}=0,05\left(mol\right)$

Theo PTHH: $n_{NaOH}=2.n_{C{O}_2}=0,1\left(mol\right)$

=> $C_{M\left(dd.NaOH\right)}=\frac{0,1}{0,1}=1M$