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

Gas state

Step-by-step 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