When calcium carbide (CaC2) comes into contact with water, it produces ethyne (C2H2) according to this equation: CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s). Suppose this reaction takes place in a sealed container in which the gas occupies a volume of 1.50 liters. The container is kept at a temperature of 0°C. If 5.18 grams of calcium carbide reacts with excess water, what is the partial pressure of the ethyne gas produced, in atmospheres? 1.21

(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