Prove/disprove that the class of decidable (resp. partially decidable) languages is closed under symmetric difference
Assume partially decidable languages are closed under symmetric difference, and let $A$ be a language that is partially decidable but not decidable.  Then, where $U$ is the language that includes everything,

$A \otimes U = (A - U) \cup (U - A) = \overline{A}$.

Then $\overline{A}$ must also be partially decidable, and $A$ is decidable; this is a contradiction.  (This builds on an answer above, which suggested investigating complementation.)