Enumerating decidable languages
You can enumerate exactly the decidable languages. I've given this question as a homework problem so I'll just give a hint here: You can modify a TM $M$ to a machine $M'$ such that if $M$ is total (halts on all inputs) then $L(M')=L(M)$ and if $M$ is not total then $L(M')$ is finite.

By request I'm burning the homework question and putting in the full proof. I heard the problem from someone else (don't remember who) so this isn't original.

Define $M'(x)$ to accept if $M(x)$ accepts and $M(y)$ halts for all $y$ with $|y|\leq |x|$. Note $M'$ has the property mentioned above.

Let $M_1,M_2,\ldots$ be a standard enumeration of Turing machines. The enumeration $M'_1,M'_2,\ldots$ is an enumeration of Turing machines such that $D=\{L(M'_1),L(M'_2),\ldots\}$ is exactly the set of decidable languages.

DECIDABLE is contained in $D$: If $L$ is decidable then $L=M_i$ for some total $M_i$ so $L(M'_i)=L(M_i)=L$.

$D$ is contained in DECIDABLE: If $M_i$ is total then $L(M'_i)=L(M_i)$ is decidable. If $M_i$ is not total then $L(M'_i)$ is finite and thus decidable.