We know
[tex]\boxed{\sf V\propto T}[/tex]
[tex]\boxed{\sf V\propto \dfrac{1}{m}}[/tex]
Lets find one by one
[tex]\\ \tt\hookrightarrow Kr=83.7g/mol[/tex]
[tex]\\ \tt\hookrightarrow CH_4=16g/mol[/tex]
[tex]\\ \tt\hookrightarrow N_2=28g/mol[/tex]
[tex]\\ \tt\hookrightarrow CH_2Cl_2=86g/mol[/tex]
[tex]\\ \tt\hookrightarrow CO_2=44g/mol[/tex]
Option B is correct.
Answer:
B) Methane, CH4
Explanation:
» For a molecule to have a great molecular speed, it mas have less molecular mass.
• From the samples given,
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