\(\Rightarrow U=IR=24.0,5=12V\)

vay phai dat vao 2 dau bong den 1 HDT=12V

Bài 8:

a: Thay x=16 vào P, ta được:

\(P=\dfrac{4+2}{4-1}=\dfrac{6}{3}=2\)

b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}=\dfrac{3+\sqrt{2}}{\sqrt{2}}=3\sqrt{2}+2\)

Ta có: \(\sin\left(5x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{3}\right)\)

=>\(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{2}-2x+\frac{\pi}{3}\right)=\sin\left(-2x+\frac56\pi\right)\)

=>\(\left[\begin{array}{l}5x+\frac{\pi}{3}=-2x+\frac56\pi+k2\pi\\ 5x+\frac{\pi}{3}=\pi+2x-\frac56\pi+k2\pi=2x+\frac16\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}7x=\frac56\pi-\frac{\pi}{3}+k2\pi=\frac12\pi+k2\pi\\ 3x=\frac16\pi-\frac{\pi}{3}+k2\pi=-\frac16\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{1}{14}\pi+\frac{k2\pi}{7}\\ x=-\frac{1}{18}\pi+\frac{k2\pi}{3}\end{array}\right.\)

Ta có: \(x=\frac{1}{14}\pi+\frac{k2\pi}{7}\)

mà \(x\in\left\lbrack0;\pi\right\rbrack\)

nên \(x\in\left\lbrace\frac{1}{14}\pi;\frac{5}{14}\pi;\frac{9}{14}\pi;\frac{13}{14}\pi\right\rbrace\)

Ta có: \(x=-\frac{1}{18}\pi+\frac{k2\pi}{3}\)

mà \(x\in\left\lbrack0;\pi\right\rbrack\)

nên \(x\in\left\lbrace\frac{11}{18}\pi\right\rbrace\)

Tổng các nghiệm thỏa mãn là:

\(\frac{1}{14}\pi+\frac{5}{14}\pi+\frac{9}{14}\pi+\frac{13}{14}\pi+\frac{11}{18}\pi=2\pi+\frac{11}{18}\pi=\frac{47}{18}\pi\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(a.PTHH:\)

\(Mg+2HCl--->MgCl_2+H_2\left(1\right)\)

\(CuO+2HCl--->CuCl_2+H_2O\left(2\right)\)

b. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,25.24=6\left(g\right)\)

\(\Rightarrow m_{CuO}=24,25-6=18,25\left(g\right)\)

c. Ta có: \(n_{CuO}=\dfrac{18,25}{80}=\dfrac{73}{320}\left(mol\right)\)

\(\Rightarrow n_{hh}=\dfrac{73}{320}+0,25=0,478125\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,478125=0,95625\left(mol\right)\)

Đổi 300ml = 0,3 lít

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,95625}{0,3}=3,1875M\)