What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K: S(s) + 3F₂(g) → SF₆ (g)

A volume of 2997 liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the given reaction takes place at 2.00 atm and 273.15 K.

Explanation:

The given reaction is as follows.

This show that 3 moles of fluorine is reacting to give 1 moles of sulfur hexafluoride.

According to the ideal gas formula,

PV = nRT

This means that volume of a gas is directly proportional to the number of moles. Hence, volume of fluorine required is calculated as follows.

Thus, we can conclude that a volume of 2997 liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the given reaction takes place at 2.00 atm and 273.15 K.

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

contour lines on a map indicate elevation, or depth. numeric notations on contour lines refer to the level of elevation, or height about sea level.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).