What volume of oxygen gas at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53g of propane

14.7 L of oxygen gas at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53g of propane

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number of particles.

According to stoichiometry :

1 mole of propane requires 5 moles of oxygen  

Thus 0.125 moles of propane require= moles of oxygen  

According to the ideal gas equation:'

P = Pressure of the gas = 1.04 atm

V= Volume of the gas = ?

T= Temperature of the gas = 25°C = 298 K  

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= 0.625

elements

explanation:

yes the other ones are right just show me a pic of ur face cuz u look hot

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