\(\dfrac{3}{7}=\dfrac{3.5}{7.5}=\dfrac{15}{35}\)

\(\dfrac{2}{5}=\dfrac{2.7}{5.7}=\dfrac{14}{35}\)

\(\dfrac{15}{35}>\dfrac{14}{35}\) \(\rightarrow\dfrac{3}{7}>\dfrac{2}{5}\)

\(\dfrac{5}{9}< \dfrac{5}{8}\)

\(\dfrac{8}{7}=\dfrac{8.8}{7.8}=\dfrac{64}{56}\)

\(\dfrac{7}{8}=\dfrac{7.7}{8.7}=\dfrac{49}{56}\)

\(\dfrac{64}{56}>\dfrac{49}{56}\rightarrow\dfrac{8}{7}>\dfrac{7}{8}\)

\(25-\left\{-x+\left[x-x\right]\right\}=0\)

\(25-\left\{-x+0\right\}=0\)

\(25+x=0\)

x=-25

1C

3A

4C

5C

6A

9C

10C

He suggested going to Nha Trang next week

\(x_{max}=v_0.\sqrt{\dfrac{2h}{g}}=25.\sqrt{\dfrac{2.8,75}{10}}\approx33\)

\(n_P=\dfrac{46,5}{31}=1,5\left(mol\right)\)

PTHH : \(4P+5O_2\rightarrow2P_2O_5\)

            1,5      1,875              (mol)

\(m_{O_2}=1,875.32=60\left(g\right)\)

\(n_{Al}=\dfrac{67,4}{27}=2,496\left(mol\right)\)

PTHH : \(4Al+3O_2\rightarrow2Al_2O_3\)

           2,496    1,872                  (mol)

\(m_{O_2}=1,872.32=59,9\left(g\right)\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH : \(2H_2+O_2\rightarrow2H_2O\)

             1,5     0,75              (mol)

\(m_{O_2}=0,75.32=24\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH : \(2Fe+3Cl_2\rightarrow2FeCl_3\)

            0,2                      0,2      (mol)

\(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)

PTHH :                 \(S+O_2\rightarrow SO_2\)

Ban đầu :           0,1    0,12              (mol)

Phản ứng :        0,1    0,1      0,1     (mol)

Sau phản ứng :  0       0,02    0,1     (mol)

\(m_{O_2}=0,02.32=0,64\left(g\right)\)

\(m_{SO_2}=0,1.64=6,4\left(g\right)\)

\(V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

Gọi kim loại cần tìm là R

PTHH : \(R+2HCl\rightarrow RCl_2+H_2\)

\(\rightarrow n_R=n_{RCl_2}\)

\(\rightarrow\dfrac{1,3}{R}=\dfrac{2,72}{R+35,5.2}\)

\(\rightarrow R=65\)

Vậy R là Zn ( Kẽm )

\(P=\dfrac{3}{0,25}=12\left(kWh\right)\)