Câu 4:
$X:C_nH_{2n}$
$448ml=0,448l$
$n_{H_2}=\dfrac{0,448}{22,4}=0,02(mol)$
$C_nH_{2n}+H_2\xrightarrow{t^o,Ni}C_nH_{2n+2}$
$C_nH_{2n}+Br_2\to C_nH_{2n}Br_2$
Theo PT: $n_{C_nH_{2n}Br_2}=n_{H_2}=0,02(mol)$
$\to M_{C_nH_{2n}Br_2}=14n+160=\dfrac{4,32}{0,02}=216$
$\to n=4$
$\to X:C_4H_8$
Vậy X là $CH_3-CH=CH-CH_3$