Answer:
True.
Explanation:
Moles =mass/molarmass
Moles Ne =mass/molarmass
[tex]=\frac {10.3g}{(20.18g per mole)}=0.510mol[/tex]
Moles [tex]N_2[/tex] =mass/molarmass
[tex]=\frac {10.3g}{(28g per mole)}=0.368mol[/tex]
Let Temperature be 273K (assumed)
[tex]PV=nRT[/tex]
[tex]P=\frac {nRT}{V}[/tex]
[tex]P_{Ne}=\frac {0.510mol \times 0.08206L atm K^{-1} mol^{-1} \times 273K}{7.0L}[/tex]
[tex]P_{Ne}=1.63atm[/tex]
[tex]P_{N_2}=\frac {0.368mol \times 0.08206L atm K^{-1} mol^{-1} \times 273K}{7.0L}[/tex]
[tex]P_{N_2}=1.17atm[/tex]
We see here partial pressure of [tex]N_2[/tex] (1.17atm) is less than that of partial pressure of Ne (1.63 atm).