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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j