Question

what volume is occupied by 5.03g of O2 at 28C and a pressure of .998atm

Answer 1

Answer:

The answer is 117.13 m^3.

PV=nRT

0.998×V= (5.03/16) ×8.3×28

V=1168.97/9.98

V=117.13m^3

Please mark as brainliest. Follow me.

Answer 2

The volume of oxygen gas is 4 L.

Given:

A volume occupied by 5.03g of Oxygen gas at 28C and a pressure of .998atm.

To Find:

The volume is occupied by 5.03g of Oxygen gas at 28C and a pressure of .998atm.

Solution:

To find the volume occupied by 5.03g of Oxygen gas at 28C and a pressure of 0.998atm, we will follow the following steps:

As we know,

According to the ideal gas equation:

PV = nRT

Now,

According to the question:

Pressure (p) = 0.998 atm

Mass of oxygen(w) = 5.03 g

The molecular mass of oxygen(m) = 32g

Number of moles =

$\frac{w}{m} = \frac{5.03}{32} = 0.16 \: moles$

Temperature (T) = 28 + 273 = 301K

R = 0.0821 atm L

Now,

In putting values we get,

$v = \frac{nRt}{p} = \frac{0.16 \times 0.0821 \times 301}{0.998} = 4 \: l$

Henceforth, the volume of oxygen gas is 4 L.

#SPJ3