- solve what volume is occupied by 5.024g of O2 at 28.0 C and a pressure of .998 atm? -

V = 3.89 L

Explanation:

Given data:

Mass of oxygen  = 5.024 g

Temperature = 28°C (28+273 = 301 k)

Pressure = 0.988 atm

Volume occupy by oxygen = ?

Solution:

PV = nRT

We will calculate the number of moles;

Number of moles = mass/ molar mass

Number of moles = 5.024 g/ 32 g/mol

Number of moles = 0.157 mol

Now we will calculate the volume,

PV = nRT

V = nRT/P

V = 0.157 mol × 0,.0821 atm.L /mol.K × 301 K / 0.998 atm

V = 3.88 L / 0.998

V = 3.89 L

the volume of fluid displaced would equal the volume of the gas. use a gas syringe - this is a laboratory instrument that can be used to withdraw a certain volume of gas from a closed system or to measure the volume of gas from a chemical reaction.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3