The Volume of a certain gas is 29.3 liters at STP. What would be the volume of the same gas at 1.39 atmospheres and -23.0oC? (R= 0.0821 L. atm/mol. k)

V₂ =19.30 L

Explanation:

Given data:

Initial volume of gas = 29.3 L

Initial temperature = standard = 273 K

Initial pressure = standard = 1 atm

Final volume = ?

Final temperature = -23°C (-23+273 =250 K)

Final pressure = 1.39 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 29.3 L × 250 K / 273 K × 1.39 atm

V₂ = 7325 atm.L.K / 379.47 K.atm

V₂ =19.30 L

answer: lol

explanation:

lol

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e