For the reaction A(g) 2B(g), what is
the equilibrium concentration of B if the
initial concentration of A is
1.0 M and Kc = 16.0.
B = [?]M
If needed, the quadratic formula is:
-b + b2 - 4ac
2a

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For the reaction A(g)

2B(g), what is

the equilibrium concentration of B if the

initial concentration of A is

1.0 M and Kc = 16.0.

B = [?]M

If needed, the quadratic formula is:

-b + b2 - 4ac

2a

it takes on an electronical energy

(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