Can Replacement schema be put in a single statement? is the following a theorem of ZFC?
$$\forall \varphi \forall A  \ [\forall x \in A \exists \theta: V_\theta  \models \exists! y \varphi(x,y) \to \\  \exists \alpha: \forall x \in A \exists ! y \in V_\alpha: V_\alpha \models \varphi(x,y)]$$
Notice that the above is a single sentence!
 A: No, this is in fact disprovable. The point is that we can use $\varphi$ to relate elements of $A$ to specific leveles of the cumulative hierarchy, preventing "simultaneous unique existence."
Take for example the formula $\varphi(x,y)$ which is the conjunction of the following clauses:

*

*There are either $17$ or $42$ ordinals in the whole universe.


*$x\in\{0,1\}$ and $y\in\{0,1\}$.


*If $x=0$ then: $y=1$ only if there are $17$ ordinals in the universe.


*If $x=1$ then: $y=1$ only if there are $42$ ordinals in the universe.
Consider what happens if $A=\{0,1\}$. For $x=0$ we get that $V_\alpha$ thinks that there is a unique $y$ with $\varphi(x,y)$ iff $\alpha=42$ (since this prevents $y=1$ via the third bulletpoint), and for $x=1$ we get that $V_\alpha$ thinks that there is a unique $y$ with $\varphi(x,y)$ iff $\alpha=17$ (since this prevents $y=1$ via the fourth bulletpoint).
Now use the fact that $17\not=42$.

You can avoid this if you require the set of $\alpha$s providing "unique existence" to be appropriately large. Specifically, say you demand that for each $x\in A$ the set of $\alpha$ such that $V_\alpha\models\exists!y\varphi(x,y)$ is a club. Then since $\mathsf{ZFC}$ proves that the intersection of set-many clubs is again club, $\mathsf{ZFC}$ would prove that there is a club of ordinals $\beta$ such that ($A\in V_\beta$ and) $V_\beta\models\forall a\in A\exists!y\varphi(a,y)$.
However, there's a subtlety here. Note that I used the hypothesis that the set of relevant $V$-indices is a club. A more natural hypothesis might be "contains a club." However, this isn't expressible in $\mathsf{ZFC}$ since we can't quantify over classes! We could, fixing $n$, say "contains a $\Sigma_n$-definable club" but that wouldn't be nearly as rich.

EDIT: in fact, $\mathsf{ZF}$ is not finitely axiomatizable over $\mathsf{Z}$ at all! This is a quick variation of the usual reflection argument that $\mathsf{ZF}$ is not finitely axiomatizable. Specifically, we use the fact that $\mathsf{ZF}$ proves "$V_\beta\models\mathsf{Z}$ whenever $\beta$ is a strong limit cardinal." We then argue in $\mathsf{ZF}$, for a fixed $\mathsf{ZF}$-theorem $\theta$, that there is a strong limit cardinal $\beta$ with $V_\beta\models\theta$ and hence $V_\beta\models\mathsf{Z}+\theta$. Consequently, there is no single sentence $\theta$ such that $\mathsf{Z}+\theta\dashv\vdash\mathsf{ZF}.$
And the situation is identical with $\mathsf{ZFC}$ vs. $\mathsf{ZC}$.
