Câu 4:

4.1/ Ta có: \(n_{NaCl}=2,5.0,4=1\left(mol\right)\)

\(\Rightarrow m_{NaCl}=1.58,5=58,5\left(g\right)\)

4.2/ Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,2___________0,2____0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

Bạn tham khảo nhé!

a)\(R_{23}=R_2+R_3=2+4=6\Omega\)

   \(R_{123}=\dfrac{R_1\cdot R_{23}}{R_1+R_{23}}=\dfrac{2\cdot6}{2+6}=1,5\Omega\)

   \(R_{tđ}=R_4+R_{123}=4,4+1,5=5,9\Omega\)

   \(I_m=\dfrac{\xi}{r+R_N}=\dfrac{12}{0,1+5,9}=2A\)

   \(U_{AB}=2\cdot5,9=11,8V\)

b)\(I_4=I_{123}=I_m=2A\)

   \(U_1=U_{23}=U_{123}=2\cdot1,5=3V\)

   \(I_1=\dfrac{U_1}{R_1}=\dfrac{3}{2}=1,5A\)

   \(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{3}{6}=0,5A\)

  \(U_{AC}=U_4+U_2=I_4\cdot R_4+I_2\cdot R_2=2\cdot4,4+0,5\cdot2=9,8V\)

\(\dfrac{2020}{2019}>\dfrac{2019}{2020}\Rightarrow0< a< 1\)

\(log_ba< 1\Rightarrow b>1\)

\(P=log_b^2a+log_b^22-\dfrac{m^2log_2b}{log_2a}+2\left(log_ba-2log_b2\right)-\dfrac{4^{ab^2}-2m.2^{ab^2}}{log_ba}\)

\(=log_b^2a+log_b^22+2log_ba-4log_b2-\dfrac{4^{ab^2}-2m.2^{ab^2}+m^2}{log_ba}\)

\(=\left(log_ba+1\right)^2+\left(log_b2-2\right)^2+\dfrac{\left(2^{ab^2}-m\right)^2}{-log_ba}-5\ge-5\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}log_ba=-1\\log_b2=2\\2^{ab^2}=m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{\sqrt{2}}\\b=\sqrt{2}\\m=2^{ab^2}=2^{\sqrt{2}}\end{matrix}\right.\)

Sau khi tính lại thì không có đáp án nào đúng :(