A) \[{{H}_{2}}\]
B) He
C) \[C{{H}_{4}}\]
D) \[S{{O}_{2}}\]
Correct Answer: C
Solution :
\[{{V}_{rms}}\propto \sqrt{\frac{T}{{{M}_{\infty }}}}\] \[{{V}_{rms}}=\sqrt{2}\,{{({{V}_{rms}})}_{{{O}_{2}}}}\] \[\sqrt{\frac{T}{{{M}_{\infty }}}}=\sqrt{2}\,\sqrt{\frac{T}{32}}\] \[{{M}_{w}}=16\,gm\] so gas is \[C{{H}_{4}}\]You need to login to perform this action.
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