One way to represent this equilibrium is: 2 Al(s) 3 Br2(l)2 AlBr3(s) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

The question is incomplete, the complete question is;

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:

Al(s) + 3/2 Br2(l)AlBr3(s)

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)

2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)

3) AlBr3(s) Al(s) + 3/2 Br2(l)

See explanation

Explanation:  

We have that; Al(s) + 3/2 Br2(l)AlBr3(s)

So;

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [  AlBr₃] / [ Al] [  Br₂]³/²

K² =  [  AlBr₃]² / [  Al ] ² [ Br₂]³

Now;

1) 2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ =  [  Al ] ² [ Br₂]³ /  [  AlBr₃]²

K₁ =  ( 1 / K² ) = K⁻²

For the second reaction;

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [  Al ]² [  Br₂ ]³

K₂ = K²

For the third reaction;

3 )

AlBr₃(s) =   Al(s) + 3/2 Br₂(l)

K₃  = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

=  ( 1 / K ) = K⁻¹

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.

my dad did alot of meth and ace cheeks