Disconnected graphs. A question asked before 
 (There exists no zero-order or first-order theory for connected graphs) proved that no theory exists such that axiomtizes connected graphs. 
Using the same type of argument (Compactness theorem for first order logic), is it possible to prove the disconnected graphs are axiomtizable?
 A: There is no theory in the language of graphs whose models are exactly the disconnected (by which I assume you mean "not connected") graphs. It is possible to prove this fact simultaneously with the fact that there is no theory of connected graphs: the problem is that we can build graphs that are elementarily equivalent (i.e., have the same theory), but where one is connected and the other disconnected.
Let $G_0$ be the graph whose vertices are the integers, with edges between adjacent integers. Let $L$ be the language of graphs, and $T$ the complete theory of $G_0$ as an $L$ structure. Let $L'$ be $L$ together with two new constant symbols, and let $T'$ be $T$ together with an axiom schema asserting that there is no path of length $n$ connecting the two constant symbols, for any $n$. $T'$ is finitely consistent, using $G_0$ as the graph and interpreting the constant symbols as integers that are sufficiently far apart. By compactness, $T'$ has a model. That model will be a graph $G_1$ that is elementarily equivalent to $G_0$ in the graph language, but the new constant symbols from $L'$ must be interpreted as vertices in different connected components.
So we have $G_0$ elementarily equivalent to $G_1$, with $G_0$ connected and $G_1$ disconnected, so you can't tell whether a graph is connected by looking at its theory. Consequently, you can't axiomatize either "connected graphs" or "disconnected graphs" (in the language of graphs).
