An unknown gas at 71.1 °c and 1.00 atm has a molar mass of 28.01 g/mol. what would the density of the gas be?

0.992 g/L

Step-by-Step Explanation:

We can use the Ideal Gas Law to calculate the density of the gas.

   pV = nRT

      n = m/M           Substitute for n

   pV = (m/M)RT     Multiply both sides by M

pVM = mRT            Divide both sides by V

  pM = (m/V) RT

     ρ = m/V            Substitute for m/V

 pM = ρRT             Divide each side by RT

    ρ = (pM)/(RT)

p = 1.00 atm

M = 28.01 g/mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 71.1 °C = 344.25 K            Solve for ρ

ρ = (1.00 × 28.01)/(0.082 06 × 344.25)

ρ = 28.01/28.25

ρ = 0.992 g/L

c is the answer because when something is left after something evaporates that shows the chemical change.