When we talk about entangling properties do we really mean we are entangling states, since properties are a classical description? I am attempting to be more concise in my own explanations and have been advised that properties is the incorrect term to refer to when preparing quantum states. Is it more accurate to refer to a property as a state and are properties just used as a term for ease of communication?
 A: I don't know if this what you meant to ask, but to some degree, I think your statement is correct. What talking about entanglement, one needs to specify the bipartite structure with respect to which entanglement is being discussed.
This is more obvious when discussing "entanglement of particles". Particles have multiple degrees of freedom, so a given two-particle quantum state can be entangled with respect to some degrees of freedom and not others.
A standard case study in this context is a single-photon state in a superposition of different position states, something you might often see written as $(a_1^\dagger+a_2^\dagger)|\mathrm{vec}\rangle$. Is this an "entangled state"? It would appear not, as there is not even any obvious bipartite structure in the space considered. But at the same time, you could equally write the state as $|10\rangle+|01\rangle$, where in this notation $|nm\rangle$ represents a two-mode, $(n+m)$-photon state with $n$ photons in the first mode and $m$ in the second one. In this form, the state certainly looks entangled, and in fact this entangled can be transferred (at least in principle) to, say, a spin system, which you would certainly call entangled. A canonical reference is here (van Enk 2005).
Similarly, one can have a quantum state which is entangled only with respect to some choices of bipartition. Consider for example a three-qubit state such as $|000\rangle+|111\rangle$. This is an entangled state with respect to any bipartition, but also, if you only have access to the last two qubits, it will look like the separable state $|00\rangle\!\langle00|+|11\rangle\!\langle11|$. Or as another example, consider $|0\rangle(|00\rangle+|11\rangle)$. This clearly has some degree of entanglement, but it is also a separable (and product) state between first and last two qubits.
