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?

7.10
Để chuyển thành thể lỏng thì nhiệt độ bằng 0

=>Cần tăng thêm 4,5 độ C

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

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

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

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

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

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

d d d a d a a d a b c a d c b a

a: Chiều dài là x+4

Chu vi là (x+x+4)x2=4x+8(m)

b: Theo đề, ta có; 4x+8=20

=>4x=12

hay x=3(m)

=>Chiều dài là 7(m)