the reaction a(b) = 2b(g) has an equilibrium constant of k = 0.045. what is the equilibrium constant for the reaction b(g) =1/2a
express your answer using two significant figures.
i

The   for the reaction will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for

                 = 0.045

The reverse will be

Then,      

                    =  

                    =

The equilibrium constant for will be

                    = 4.69

Therefore, for the reaction will be 4.69.

(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