Turing recognizable -decidable languages-
There are two directions here. One is trivial: if $C$ is indeed of the above form, then it is clearly recognizable: given $x$ just run $D$ on all possible $y$'s in a dovetailing manner (see, e.g., here, or search in this site).

The other direction is less obvious, but also not too difficult: Assume that $C$ is recognizable. Then, there exists a machine that halts and accepts any $x\in C$. Thus, you can write the sequence of configurations of $M$ on input $x$, and this sequence is finite! This sequence will be the $y$ that exists if $x$ is in the language. You should be able to complete the details from here.