9) a 22.8 ml volume of argon gas is collected at 48.0o c. at constant pressure, what volume would this same sample of gas occupy at standard temperature?

19.39 mL.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

(V₁T₂) = (V₂T₁)

V₁ = 22.8 mL, T₁ = 48°C + 273 = 321 K,

V₂ = ??? mL, T₂ = 273 (standard temperature = 273 K),

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (22.8 mL)(273 K)/(321 K) = 19.39 mL.

b) mutualism

hope this : )