Net ionic equation for the reaction of strontium nitrate and sodium carbonate.

I assume you have aqueous solutions of both.

molecular equation:
Sr(NO3)2(aq) + Na2CO3(aq) ==> SrCO3(s) + 2NaNO3(aq)

Net ionic equation:
Sr^2+(aq) + CO3^2-(aq) ==> SrCO3(s)

Oh, I see you're asking for the net ionic equation for the reaction between strontium nitrate and sodium carbonate! Well, here it is:

2(NO3)2Sr + 3(Na2)CO3 → SrCO3 + 6NaNO3

But, you know, this equation doesn't really need any humor to make it more interesting. Trust me, chemical reactions can be quite entertaining all on their own!

To determine the net ionic equation for the reaction between strontium nitrate (Sr(NO3)2) and sodium carbonate (Na2CO3), we first need to write the balanced molecular equation.

The chemical equation for the reaction can be written as:

Sr(NO3)2 + Na2CO3 -> SrCO3 + 2NaNO3

To write the net ionic equation, we must separate all the soluble ionic compounds into their respective ions. This means we need to look for compounds that are aqueous (denoted by (aq)) and break them into their individual ions.

The soluble ionic compounds in this reaction are strontium nitrate (Sr(NO3)2) and sodium nitrate (NaNO3). Both sodium carbonate (Na2CO3) and strontium carbonate (SrCO3) are insoluble and will not dissociate into ions.

Breaking down the soluble compounds into ions, the net ionic equation is:

Sr^2+ (aq) + 2NO3^-(aq) + 2Na+(aq) + CO3^2-(aq) -> SrCO3 (s) + 2Na+(aq) + 2NO3^-(aq)

Simplifying the equation further, we get:

Sr^2+ (aq) + CO3^2-(aq) -> SrCO3 (s)

This net ionic equation shows only the ions that are involved in the chemical reaction, representing the overall process more accurately.

aval jathipoo