95.8 L of fluorine gas is being held at a temperature of 24.5oC. If the temperature were raised to 46.9 oC, what would the new volume be?

103 litres

Explanation:

Given that,

Original volume of fluorine gas (V1) = 95.8L

Original temperature of fluorine gas (T1) = 24.5°C

[Convert 24.5°C to Kelvin by adding 273

24.5°C + 273 = 297.5K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 46.9°C

[Convert 46.9°C to Kelvin by adding 273

46.9°C + 273 = 319.9K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

95.8L/297.5K = V2/319.9K

To get the value of V2, cross multiply

95.8L x 319.9K = 297.5K x V2

30646.42L•K = 297.5K•V2

Divide both sides by 297.5K

30646.42L•K/297.5K = 297.5K•V2/297.5K

103L = V2

Thus, the new volume of fluorine gas will be 103 litres

1. they vary with a higher number because boiling is way hotter than melting

i think its 5049.

i hope this !