True/False If 10.3 g of Ne and 10.3 g of N2 are put into a 7.0 L container, the partial pressure of N2 will be less than the partial pressure of Ne in the container.

Question

asked Mar 2, 2020 214k views

1 Answer

Josh Delsman · Answer 1 · 2020-03-09T03:22:48+0000

Answer:

True.

Step-by-step explanation:

Moles =mass/molarmass

Moles Ne =mass/molarmass

$=\frac {10.3g}{(20.18g per mole)}=0.510mol$

Moles
N_2 =mass/molarmass

$=\frac {10.3g}{(28g per mole)}=0.368mol$

Let Temperature be 273K (assumed)

PV=nRT

$P=\frac {nRT}{V}$

$P_(Ne)=\frac {0.510mol * 0.08206L atm K^(-1) mol^(-1) * 273K}{7.0L}$

P_(Ne)=1.63atm

$P_(N_2)=\frac {0.368mol * 0.08206L atm K^(-1) mol^(-1) * 273K}{7.0L}$

P_(N_2)=1.17atm

We see here partial pressure of
(1.17atm) is less than that of partial pressure of Ne (1.63 atm).