Is algebraic manifold always orientable?
For example, unorientable Mobius strip $M$ can be represented as $$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u$$ $$y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u$$ $$z(u,v)= \frac{v}{2}\sin \frac{u}{2}$$ Hence there exists algebraic equation $Z(x, y, z) = 0$ : $(a, b, c) \in M \Rightarrow Z(a, b, c) = 0$. But fact that $M$ is equal to set of all solutions of $Z$ is not obvious, even if it is true.
Algebraic means given by polynomial equations. Over $\Bbb C$, yes, it'll be orientable. Not true over $\Bbb R$: The hyperplane section of $\Bbb RP^3$ (given by a linear equation) is $\Bbb RP^2$, which is not orientable.