It's easy enough to show that the theory of algebraically closed fields of characteristic p is decidable (since its complete). But does it follow from this that the theory of algebraically closed fields of any characteristic is? I suspect that the answer is "yes", but I'm not quite sure why.
The reason for my suspicion is that we can simply that the greatest lower bound of each of these theories and arrive at the right set. But I'm not sure if decidable sets are closed under countable intersections.
Thinking about this a little in the shower, I realized that the fact that $Th(ACF_p)$ for any $p=0$ or $p$ is complete does entail that $Th(ACF)$ is decidable. $Th(ACF)$ is recursively enumerable in the obvious way, and since each $Th(ACF_p)$ is decidable so is it's complement.