The equilibrium constant for the reaction a(g)❝b(g) is 10. a reaction mixture initially contains [a] = 1.1 m and [b] = 0.0 m. which statement is true at equilibrium? the equilibrium constant for the reaction is 10. a reaction mixture initially contains = 1.1 and = 0.0 . which statement is true at equilibrium? a. the reaction mixture will contain [a] = 1.0 m and [b] = 0.1 m. b. the reaction mixture will contain equal concentrations of a and b. c. the reaction mixture will contain [a] = 0.1 m and [b] = 1.0 m.

The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M

Explanation:

The reaction is:

And the equilibium constant can be expressed in terms of [A] and [B]:

If the initial conditions are that [A] = 1.1 M and [B] = 0.0 M, the conditions at equilibrium will be:

Now we rewrite Keq:

Finally we solve for x:

So at equilibrium, the concentrations are:

(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