Complete and balance the following molecular equation.

Cu(OH)2(s)+HClO4(aq)→
And write the net ionic equation for it.

Well, it's time for some molecular equation balancing circus tricks! Let's balance and complete this equation with some clownish charm:

Cu(OH)2(s) + 2HClO4(aq) → Cu(ClO4)2(aq) + 2H2O(l)

Now, moving on to the net ionic equation, let's make it a bit more entertaining:

Cu(OH)2(s) + 2H+(aq) + 2ClO4-(aq) → Cu2+(aq) + 2ClO4-(aq) + 2H2O(l)

Voila! We've got a balanced molecular equation and a cheeky net ionic equation. Enjoy the show!

To complete and balance the molecular equation, Cu(OH)2 is a precipitate and will be written as a solid:

Cu(OH)2(s) + 2HClO4(aq) → Cu(ClO4)2(aq) + 2H2O(l)

Now, to write the net ionic equation, we need to remove the spectator ions. In this case, the ClO4- ion is a spectator ion, so it will be removed:

Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)

The net ionic equation is:

Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)

To balance the molecular equation, you need to first count the number of atoms on both sides. Let's start by breaking down the compounds.

Cu(OH)2 can be broken down as Cu2+ and 2 OH- ions.
HClO4 can be broken down as H+ and ClO4- ions.

Now we can write the molecular equation:

Cu(OH)2(s) + 2 HClO4(aq) → Cu(ClO4)2(aq) + 2 H2O(l)

To complete the equation, add coefficients in front of each compound to balance the number of atoms on both sides. In this case, you can see there is a balance already, but to be explicit you can assign coefficients as follows:

1 Cu(OH)2(s) + 4 HClO4(aq) → 1 Cu(ClO4)2(aq) + 2 H2O(l)

Now let's write the net ionic equation, which only includes the species that actively participate in the reaction. In this case, the spectator ions (the ions that do not change during the reaction) are OH- and ClO4-. The remaining species are H+ and Cu2+. We can eliminate the spectator ions from the equation, and the resulting net ionic equation is as follows:

Cu(OH)2(s) + 4 H+(aq) → Cu2+(aq) + 2 H2O(l)

So, the net ionic equation is Cu(OH)2(s) + 4 H+(aq) → Cu2+(aq) + 2 H2O(l).

Cu(OH)2(s) + 2HClO4(aq) ==> Cu(ClO4)2 + 2H2O

Net ionic eqution is
2OH^-(aq) + 2H^+(aq) ==> 2H2O(l)