Question

identify propane's physical state at room tempurture (20 degrees F) if propane's melting point is -190 degrees and boiling point is -42 dgreess

Answer 1

Answer:

Gas state

Explanation:

Propane

• Melting point: -190°C
• Boiling point: -42°C

Room temperature: 20°F

We need to calculate the room temperature in °C

$T_{F}=1.8*T_{C}+32$

$20=1.8*T_{C}+32$

$T_{C}=-6.7 C$

Given that the room temperature is above the propane's boiling point it is in gas state