Consider the following reaction: ch3br(aq)+oh−(aq)→ch3oh(aq)+br−(aq) the rate law for this reaction is first order in ch3br and first order in oh−. when [ch3br] is 5.0×10−3m and [oh−] is 0.050 m, the reaction rate at 298 k is 0.0432 m/s.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter  

(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