in the completion of a linear order of cardinality $\kappa$? Can there be more than $\kappa$?
To be more precise, suppose $L$ is a linear order of cardinality $\kappa$. Let $\overline L$ denote the completion of $L$. Can there be more than $\kappa$ different subspaces (suborders) of $\overline L$ which are dense in $\overline L$ and are pairwise nonisomorphic.
As noted in Eric Towers' answer, $\mathbb R$ is the completion of $\mathbb Q$. I will show that there are $2^{2^{\aleph_0}}$ nonisomorphic dense subsets of $\mathbb R$.
Let $\mathcal D$ be the set of all dense subsets of $\mathbb R$. It's clear that $|\mathcal D|=2^{2^{\aleph_0}}$; e.g., $\{X:\mathbb Q\subseteq X\subseteq\mathbb R\}\subset\mathcal D$. The relation "is order-isomorphic to" is an equivalence relation on $\mathcal D$; I claim that each equivalence class has cardinality at most $2^{\aleph_0}$. (To see this, observe that, if $X,Y\in\mathcal D$ are order-isomorphic, then any order-isomorphism between $X$ and $Y$ extends to a homeomorphism of $\mathbb R$ with itself; and of course there are only $2^{\aleph_0}$ such homeomorphisms.) It follows that there are $2^{2^{\aleph_0}}$ equivalence classes, i.e., there are $2^{2^{\aleph_0}}$ nonisomorphic dense subsets of $\mathbb R$.