Answer:
[tex]2K^+\text{ + 2Br}^-\text{ }\rightarrow\text{ 2KBr}_{(s)}[/tex]Explanation:
Here, we want to write the net ionic equation for the given equation;
[tex]SrBr_2\text{ + K}_2SO_4\text{ }\rightarrow\text{ SrSO}_4\text{ + 2KBr}[/tex]Now, what is left is to write the participating ions:
[tex]Sr^{2+}\text{ + 2Br}^-\text{ + 2K}^+\text{ + SO}_4^{2-}\text{ }\rightarrow\text{ Sr}^{2+}\text{ + SO}_4^{2-}\text{ + 2KBr}[/tex]We must identify that the Potassium Bromide is the precipitate and thus would not be dissociated into ions
Thus, we have it that:
The Strontium ion cancels out, including the sulphate
We have left:
[tex]2K^+\text{ + 2Br}^-\text{ }\rightarrow\text{ 2KBr}[/tex]