A study of the following system, 4 NH3(g) + 7 O2(g) <--> 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [N2O4] = [H2O] = 3.60 M as the only components initially. At equi­librium, [H2O] is 0.600 M. Calculate the equilibrium concentration of O2(g).

3.50 M

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

Step 2: Make an ICE chart

        4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

I              0               0              3.60            3.60

C         +4x            +7x               -2x             -6x

E            4x              7x           3.60-2x      3.60-6x

Step 3: Calculate the value of x

The concentration of water at equilibrium is 0.600 M. Then,

3.60-6x = 0.600 M

x = 0.500 M

Step 4: Calculate the concentration of O₂ at equilibrium

The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.