a) PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)

b+c+d) Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,2\cdot40=8\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=204,4\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{8}{204,4}\cdot100\%\approx3,91\%\)

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

200 (ml hay l) dung dịch nước ạ?

😥😥😥😥😥

em chưa đăng bài lên nhé

\(R_{13}=\dfrac{4\cdot2}{4+2}=\dfrac{4}{3}\Omega\)

\(R_N=\dfrac{\dfrac{4}{3}\cdot6}{\dfrac{4}{3}+6}=\dfrac{12}{11}\Omega\)

\(U_{AB}=12+8=20V\)

\(I=\dfrac{U}{R}=\dfrac{20}{\dfrac{12}{11}}=\dfrac{55}{3}A\)

\(I_{13}=I_m=\dfrac{55}{3}A\Rightarrow U_{13}=\dfrac{55}{3}\cdot\dfrac{4}{3}=\dfrac{220}{9}V\)

\(\Rightarrow U_3=\dfrac{220}{9}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{220}{9}}{2}=\dfrac{110}{9}A\)

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giúp em với ạ ! 

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\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)

\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)

Câu 3:

a: \(25x-30x^2=5x\left(5-6x\right)\)

b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)

c: \(24x^8y^7-15x^6y^5+18x^6y^3\)

\(=3x^6y^3\left(8x^2y^4-5y^2+6\right)\)