Sure, I can help with that.
The dehydration of copper(II) sulfate pentahydrate ([tex]$CuSO_4 \cdot 5H_2O$[/tex]) involves the removal of water molecules from the hydrated compound. This process results in anhydrous copper(II) sulfate ([tex]$CuSO_4$[/tex]) and water ([tex]$H_2O$[/tex]). The balanced chemical equation for this reaction is:
[tex]\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \rightarrow \text{CuSO}_4 + 5\text{H}_2\text{O} \][/tex]
Here’s a breakdown of the equation:
- [tex]$CuSO_4 \cdot 5H_2O$[/tex]: Copper(II) sulfate pentahydrate, which is the hydrated form of copper(II) sulfate.
- [tex]$CuSO_4$[/tex]: Anhydrous copper(II) sulfate, which means it no longer has water molecules attached.
- [tex]$5H_2O$[/tex]: Five molecules of water are released during the dehydration process.
So, the molecule of copper(II) sulfate pentahydrate releases five water molecules when it is dehydrated, resulting in anhydrous copper(II) sulfate.
