Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow. NaBr(s)

The balanced dissociation equation for the compound, NaBr is:

NaBr(s) —> Na⁺(aq) + Br¯(aq)

NOTE:

Dissociation will only occurs when the compound is dissolved in water

The dissociation equation for NaBr can be written as follow:

From the equation above, we can see that the ions that makes up NaBr are clearly seen. Hence, the above equation is the dissociation equation for NaBr.

Learn more: link

NaBr(s) → Na+ (aq) + Br– (aq)

Explanation:

Dissociation refers to the separation of an ionic compound or molecule into its constituent ions when dissolved in water. The ions with their respective charges make up the ionic compound, hence, an ionic compound is said to dissociate when the respective cations (+ charge) and anions (- charge) are released in an aqeous solution.

In this case, NaBr(s) or Sodium bromide is an ionic solid {signified by the (s) sign}. Upon dissolving in water, NaBr(s) breaks down into the ionic charges that makes it up i.e dissociates. The balanced ionic equation is as follows:

NaBr(s) → Na+ (aq) + Br– (aq)

Note: 1 mole of the solid ionic compound dissociates into one sodium ion (Na+) and one bromide ion (Br–).

indium, and thallium have three electrons i think