Help writing complete ionic and net ionic equations.?

Hg2(NO3)2 (aq) + CuSO4 ---> Cu(NO3)2 (aq) + Hg2SO4 (s)

K2CO3 (aq) + MgI2 --> MgCO3 (s) + 2KI

3 NaCrO4 (aq) +2 AlBr3 ---> Al2(CrO4)3 (s) + 6NaBr

i got down writing the formula equation but i would like help with the complete and net ionic equations i just don't get it and would like help
thanks.

Here is what you do. First write and balance the molecular equation. I think you have done that.

K2CO3(aq) + MgI2(aq) --> MgCO3(s) + 2KI(aq)

Step 2. Convert those aq solns to ions. I will omit the aq part on each ion t save typing but you should put it in. If it isn't aq (for examle the MgCO3 is (s), you leave it as is. Slightly ionized materials such as H2O(l) would not be changed.) This gives you the COMPLETE ionic equation.
2K^+ + CO3^2- + Mg^2+ + 2I^- ==> MgCO3(s) + 2K^+ + 2I^-

Step 3. Cancel ions common to each side. I see 2K^+ on both sides and 2I^- on both sides. Cancel those. Those are the spectator ions. What is left is the NET ionic equation.
Mg^2+(aq) + CO3^2-(aq) ==> MgCO3(s)

Three steps will do them all.
Write and balance the molecular equation.
Separate the (aq) materials (those soluble in water) into ions. This give you the complete ionic equation.
Last step is to cancel the ions common to both sides. What is left is the net ionic equation.

Sure, I'd be happy to help you with the complete and net ionic equations for these reactions!

1. Hg2(NO3)2 (aq) + CuSO4 ---> Cu(NO3)2 (aq) + Hg2SO4 (s)

Complete Ionic Equation:
2 Hg^2+ (aq) + 2 NO3^- (aq) + Cu^2+ (aq) + SO4^2- (aq) ---> Cu^2+ (aq) + 2 NO3^- (aq) + Hg2SO4 (s)

Net Ionic Equation:
Hg^2+ (aq) + SO4^2- (aq) ---> Hg2SO4 (s)

2. K2CO3 (aq) + MgI2 ---> MgCO3 (s) + 2 KI

Complete Ionic Equation:
2 K^+ (aq) + CO3^2- (aq) + Mg^2+ (aq) + 2 I^- (aq) ---> MgCO3 (s) + 2 K^+ (aq) + 2 I^- (aq)

Net Ionic Equation:
CO3^2- (aq) + Mg^2+ (aq) ---> MgCO3 (s)

3. 3 NaCrO4 (aq) + 2 AlBr3 ---> Al2(CrO4)3 (s) + 6 NaBr

Complete Ionic Equation:
3 Na^+ (aq) + CrO4^2- (aq) + 2 Al^3+ (aq) + 6 Br^- (aq) ---> Al2(CrO4)3 (s) + 6 Na^+ (aq) + 6 Br^- (aq)

Net Ionic Equation:
CrO4^2- (aq) + 2 Al^3+ (aq) ---> Al2(CrO4)3 (s)

Hope that helps! Let me know if you have any more questions.

To write the complete ionic equation, you need to break down all the soluble compounds into their respective ions. The insoluble compounds remain in their molecular form.

1) Hg2(NO3)2 (aq) + CuSO4 (aq) ---> Cu(NO3)2 (aq) + Hg2SO4 (s)

The soluble compounds dissociate into ions as follows:
Hg2(NO3)2 (aq) --> 2Hg2+ (aq) + 2NO3- (aq)
CuSO4 (aq) --> Cu2+ (aq) + SO4^2- (aq)

The ionic equation becomes:
2Hg2+ (aq) + 2NO3- (aq) + Cu2+ (aq) + SO4^2- (aq) ---> Cu(NO3)2 (aq) + Hg2SO4 (s)

2) K2CO3 (aq) + MgI2 (aq) ---> MgCO3 (s) + 2KI

The soluble compounds dissociate into ions as follows:
K2CO3 (aq) --> 2K+ (aq) + CO3^2- (aq)
MgI2 (aq) --> Mg2+ (aq) + 2I- (aq)

The ionic equation becomes:
2K+ (aq) + CO3^2- (aq) + Mg2+ (aq) + 2I- (aq) ---> MgCO3 (s) + 2K+ (aq) + 2I- (aq)

3) 3 NaCrO4 (aq) + 2 AlBr3 (aq) ---> Al2(CrO4)3 (s) + 6 NaBr

The soluble compounds dissociate into ions as follows:
3 NaCrO4 (aq) --> 3 Na+ (aq) + 3 CrO4^2- (aq)
2 AlBr3 (aq) --> 2 Al^3+ (aq) + 6 Br- (aq)

The ionic equation becomes:
3 Na+ (aq) + 3 CrO4^2- (aq) + 2 Al^3+ (aq) + 6 Br- (aq) ---> Al2(CrO4)3 (s) + 6 Na+ (aq) + 6 Br- (aq)

To write the net ionic equation, you only include the ions that undergo a chemical change. In this case, it's easiest to see by cancelling out ions that appear on both sides of the equation.

1) Net ionic equation:
2Hg2+ (aq) + Cu2+ (aq) + 2NO3- (aq) + SO4^2- (aq) ---> Cu(NO3)2 (aq) + Hg2SO4 (s)

2) Net ionic equation:
CO3^2- (aq) + Mg2+ (aq) ---> MgCO3 (s)

3) Net ionic equation:
3 CrO4^2- (aq) + 2 Al^3+ (aq) ---> Al2(CrO4)3 (s)

Please note that in some cases, if all the ions remain in solution or if there are no insoluble compounds formed, there might not be a net ionic equation.

Sure, I can help you with writing the complete and net ionic equations for the given reactions.

To start, let's write the formula equation for each of the reactions:

1) Hg2(NO3)2 (aq) + CuSO4 --> Cu(NO3)2 (aq) + Hg2SO4 (s)

2) K2CO3 (aq) + MgI2 --> MgCO3 (s) + 2KI

3) 3 NaCrO4 (aq) + 2 AlBr3 --> Al2(CrO4)3 (s) + 6 NaBr

Now, let's break down each equation and identify the spectator ions to write the complete ionic equation.

1) In the first equation, both Hg2(NO3)2 and CuSO4 are in aqueous form, which means they dissociate into their respective ions when dissolved in water. The complete ionic equation would be:

Hg^2+ (aq) + 2NO3^- (aq) + Cu^2+ (aq) + SO4^2- (aq) --> Cu^2+ (aq) + 2NO3^- (aq) + Hg2SO4 (s)

In this equation, the spectator ions are NO3^- and Cu^2+. They are present on both sides of the equation and don't participate in the actual reaction. Therefore, the net ionic equation can be written as:

Hg^2+ (aq) + SO4^2- (aq) --> Hg2SO4 (s)

2) In the second equation, K2CO3 and MgI2 are also in aqueous form. The complete ionic equation would be:

2K^+ (aq) + CO3^2- (aq) + Mg^2+ (aq) + 2I^- (aq) --> MgCO3 (s) + 2K^+ (aq) + 2I^- (aq)

The spectator ions are 2K^+ and 2I^-. So, the net ionic equation is:

CO3^2- (aq) + Mg^2+ (aq) --> MgCO3 (s)

3) In the third equation, NaCrO4 and AlBr3 are aqueous, and they dissociate as follows:

3Na^+ (aq) + CrO4^2- (aq) + 2Al^3+ (aq) + 6Br^- (aq) --> Al2(CrO4)3 (s) + 6Na^+ (aq) + 6Br^- (aq)

The spectator ions are 6Na^+ and 6Br^-. Hence, the net ionic equation is:

CrO4^2- (aq) + 2Al^3+ (aq) --> Al2(CrO4)3 (s)

So, these are the complete and net ionic equations for the given reactions.