write the molecular, ionic and net ionic equation for the following

Ph(NO3)2 + NaOH
Pb(NO3)2 + Na2CO3

Ni(NO3)2 + NaOH
Ni(NO3)2 + Na2Co3

recognize what is going to precipate, that takes those ions in solution out.

For instance on first.

Pb(NO3)2+NaOH>> Pb(OH)2 precipates, so net ionic is ...
Pb++ + OH- >> Pb(OH)2

To write the molecular, ionic, and net ionic equations for the given reactions, we first need to identify the formulas of the reactants and products involved.

1. Ph(NO3)2 + NaOH:

The reactants are Ph(NO3)2 (phenyl nitrate) and NaOH (sodium hydroxide). To write the equation, we need to determine their formulas.

- Ph(NO3)2: The compound contains the Ph (phenyl) cation and the NO3 (nitrate) anion. The formula for phenyl nitrate is C6H5NO3.

- NaOH: This compound consists of the Na (sodium) cation and OH (hydroxide) anion. The formula for sodium hydroxide is NaOH.

The molecular equation for the reaction is:

C6H5NO3 + NaOH → C6H5NO3 + NaOH

Now, let's break the molecular equation into its ionic form:

C6H5NO3 + NaOH → C6H5NO3 + NaOH
Na+ + OH- → Na+ + OH-

The net ionic equation removes the spectator ions (those that appear on both sides of the equation):

OH- → OH-

2. Pb(NO3)2 + Na2CO3:

The reactants are Pb(NO3)2 (lead(II) nitrate) and Na2CO3 (sodium carbonate). Let's determine their formulas:

- Pb(NO3)2: The compound consists of the Pb2+ (lead) cation and the NO3- (nitrate) anion. The formula for lead(II) nitrate is Pb(NO3)2.

- Na2CO3: This compound contains Na+ (sodium) cation and CO3^2- (carbonate) anion. The formula for sodium carbonate is Na2CO3.

The molecular equation for the reaction is:

Pb(NO3)2 + Na2CO3 → Pb(NO3)2 + Na2CO3

Next, let's break it down into its ionic form:

Pb2+ + 2NO3- + 2Na+ + CO3^2- → Pb2+ + 2NO3- + 2Na+ + CO3^2-

The net ionic equation eliminates the spectator ions:

CO3^2- → CO3^2-

3. Ni(NO3)2 + NaOH:

The reactants are Ni(NO3)2 (nickel(II) nitrate) and NaOH (sodium hydroxide). We determine their formulas as follows:

- Ni(NO3)2: The compound contains the Ni2+ (nickel) cation and the NO3- (nitrate) anion. The formula for nickel(II) nitrate is Ni(NO3)2.

- NaOH: This compound has the Na+ (sodium) cation and OH- (hydroxide) anion. The formula for sodium hydroxide is NaOH.

The molecular equation for the reaction is:

Ni(NO3)2 + 2NaOH → Ni(NO3)2 + 2NaOH

Breaking it down into its ionic form:

Ni2+ + 2NO3- + 2Na+ + 2OH- → Ni2+ + 2NO3- + 2Na+ + 2OH-

The net ionic equation removes the spectator ions:

OH- → OH-

I hope this helps you understand how to write the molecular, ionic, and net ionic equations for these reactions!