\(\dfrac{AB}{CD}=\dfrac{30}{50}=\dfrac{3}{5}\)

\(\text{Đổi:7dm=70cm}\)

\(\dfrac{AB}{CD}=\dfrac{40}{70}=\dfrac{4}{7}\)

\(\text{Ta có:}\dfrac{AB}{CD}=\dfrac{3}{5}\left(gt\right)\)

\(\text{ Mà AB=15cm(gt)}\)

\(\Rightarrow\dfrac{15}{CD}=\dfrac{3}{5}\Rightarrow CD=\dfrac{15.5}{3}=25\left(cm\right)\)

\(\text{Ta có:}\dfrac{AB}{CD}=\dfrac{4}{3}\left(gt\right)\)

\(\text{Mà CD=12cm}\left(gt\right)\)

\(\Rightarrow\dfrac{AB}{12}=\dfrac{4}{3}\)

\(\Leftrightarrow AB=\dfrac{12.4}{3}=16\left(cm\right)\)

Nhìn nhầm:v

\(6Fe+2O_2\underrightarrow{t^o}2Fe_3O_4\)

\(6mol\)  \(2mol\)   \(2mol\)

\(0,1mol\) \(\dfrac{1}{30}mol\)  \(\dfrac{1}{30}mol\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\text{Ta thấy }O_2\text{ dư,}Fe\text{ phản ứng hết}\)

\(m_{Fe_3O_4}=n.M=\dfrac{1}{30}.232\approx7,73\left(g\right)\)

\(\text{Đổi:}3h48p=\dfrac{19}{5}\left(h\right)\)

\(\text{Vận tốc lúc về là:50-5=45}\left(km/h\right)\)

\(\text{Gọi quãng đường AB là x(x>0)(km)}\)

\(\text{Thời gian lúc đi là:}\dfrac{x}{50}\left(h\right)\)

\(\text{Thời gian lúc về là:}\dfrac{x}{45}\left(h\right)\)

\(\text{Theo đề ta có:}\dfrac{x}{50}+\dfrac{x}{45}=\dfrac{19}{5}\)

\(\Leftrightarrow\dfrac{9x}{450}+\dfrac{10x}{450}=\dfrac{1710}{450}\)

\(\Leftrightarrow19x=1710\)

\(\Leftrightarrow x=1710:19=90\left(km\right)\)

\(\text{Vậy quãng đường AB dài là }90km\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1mol\)   \(2mol\)                    \(1mol\)

\(0,1mol\)  \(0,2mol\)              \(0,1mol\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=n.M=0,2.36,5=7,3\left(g\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(1mol\)               \(1mol\)     \(1mol\)

\(0,1mol\)         \(0,1mol\)   \(0,1mol\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(m_{Fe_2O_3}=n.M=0,1.127=12,7\left(g\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(6H_2+2Fe_2O_3\underrightarrow{t^o}4Fe+6H_2O\)

\(6mol\)   \(2mol\)        \(4mol\)

\(0,1mol\)  \(\dfrac{1}{30}mol\)   \(\dfrac{1}{15}mol\)

\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\text{Ta thấy }H_2\text{ dư,}Fe_2O_3\text{ phản ứng hết}\)

\(m_{Fe}=n.M=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)

\(2Fe_2O_3+6H_2\underrightarrow{t^o}4Fe+6H_2O\) Thế

\(Fe+2HCl\rightarrow FeCl_2+H_2\)   Thế

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)  Hóa hợp