Mgo(s) + h2o(ℓ) ←→ mg(oh)2(aq) if you remove mg(oh)2 from the system, how will the system respond?

The equilibrium position will shift to the right i.e favours the forward reaction and thus producing more Mg(OH)2.

Explanation:

The equation for the reaction is given below:

MgO(s) + H2O(l) ←→ Mg(OH)2(aq)

According to Le Chatelier's principle, the equilibrium position will shift so as to neutralize the effect when an external constrain such as change in concentration, pressure or temperature is imposed on a chemical system in equilibrium.

Now removing Mg(OH)2 will cause the equilibrium position to shift to the right i.e favours the forward reaction and thus producing more Mg(OH)2.

Further more, removing Mg(OH)2 means the reaction now have more reactants that product so the reactants react more to produce Mg(OH)2 thereby shifting the equilibrium position to the right.

one state would change   like louisiana to texas. thats a change

explanation:

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890