\(m_{NaOH}=\dfrac{200\cdot10\%}{100\%}=20g\) \(\Rightarrow n_{NaOH}=0,5mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25   \(\leftarrow\)  0,25  \(\leftarrow\)  0,5

\(m_{Na_2O}=0,25\cdot62=15,5g\)

\(m_{H_2O}=0,25\cdot18=4,5g\)

\(\left(7x+3\right)^2-\left(7x-1\right)\left(7x-3\right)=-12\)

\(\Rightarrow49x^2+42x+9-\left(49x^2-21x-7x+3\right)=-12\)

\(\Rightarrow70x+18=0\) \(\Rightarrow x=-\dfrac{18}{70}=-\dfrac{9}{35}\)

\(\left(5x^2-4x\right)\left(x-3\right)=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\)

\(m_{MgSO_4}=\dfrac{75\cdot40}{100}=30g\) \(\Rightarrow n_{MgSO_4}=0,25mol\)

\(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+K_2SO_4\)

 0,25          0,5

\(m_{KOH}=0,5\cdot56=28g\) 

\(m_{ddKOH}=\dfrac{28}{11,2\%}\cdot100\%=250g\)

a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)

b) \(x^2-2xy-x+2y=\left(x^2-2xy\right)-\left(x-2y\right)=x\left(x-2y\right)-\left(x-2y\right)=\left(x-1\right)\left(x-2y\right)\)

Chọn D.

Điện lượng cần đặt:

   \(Q=C\cdot U=2\cdot10^{-6}\cdot4=8\cdot10^{-6}C\)

\(Al_2\left(SO_x\right)_3=342đvC\) \(\Rightarrow2\cdot27+3\cdot\left(32+16\cdot x\right)=342\)

\(\Rightarrow x=4\)

Vậy  \(Al_2\left(SO_4\right)_3\)

\(v_{tb}=\dfrac{S}{\dfrac{0,5S}{30}+\dfrac{0,5S}{v_2}}=40\) 

\(\Rightarrow40\cdot0,5\cdot\left(\dfrac{1}{30}+\dfrac{1}{v_2}\right)=1\)

\(\Rightarrow v_2=60\)km/h

\(v=54\) km/h=15m/s; \(t=1'=60s\)

a) Gia tốc xe:  \(v=v_0+at=0+60a=15\Rightarrow a=0,25\)m/s²

b) Quãng đường xe đi được sau 1 phút:

    \(S=v_0t+\dfrac{1}{2}at^2=60\cdot0+\dfrac{1}{2}\cdot0,25\cdot60^2=450m\)

     Vận tốc vật sau 1 phút:   \(v=v_0+at=0+0,25\cdot60=15\)m/s