Consider the equation. Which represents the correct mass-volume relationship at STP? 32.00 g of O2 react with an excess of H2 to produce (2 ´ 22.4) L of H2O. 2.00 g of H2 react with an excess of O2 to produce (2 ´ 22.4) L of H2O. 32.00 g of O2 react with an excess of H2 to produce 22.4 L of H2O. (2 ´ 2.00) g of H2 react with an excess of O2 to produce 22.4 L of H2O.

the answer is   c.) 1575 j.

explanation:

i think is conclusions.i think so

(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