The enthalpy change for the following reaction is -483.6KJ: 2H2(g)+O(g) --> 2H2O(g) therefore the enthalpy change for the following reaction is ---KJ 4H2 (g)+2O2(g)-->4H2O(g)

i think   the answer is c hope i'm right

(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

its pwirhfpiwe

explanation:

because dgjpwojefopjwp