How many liters of water vapor can be produced if 8.9 liters of methane gas (ch4) are combusted, if all measurements are taken at the same temperature and pressure? show all of the work used to solve this problem. ch4 (g) + 2o2 (g) yields co2 (g) + 2h2o (g)

The normal condition of the mole of a gas is 22.44L.

The volume of CH4 = 8.9L

To get the number of moles we get,

8.9 / 22.44 = 0.397 moles.

The ratio of CH4 : H2O = 1: 2

The mole of water is 0.379 × 2 = 0.794 moles.

While we convert it into liters it will be

0.794 × 22.414 = 17.80L

The answer is 17.80 liters of water vapour which will be produced.

a electronegativity decreases beacuse the size of nucleus is bigger.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied