This is a claim in Loday's Cyclic Homology as follows: Let $k$ be a commutative ring, $A$ be a $k$-algebra, and $M$ be a bimodule over $A$. Then, when $A$ is unital (has a unit $1\in A$ s.t. $1m= m1=1$), the bimodule $M$ is equivalent to a right $A\otimes A^{op}$-module via $m(a\otimes a')= ama'$.
There seems to be a simple observation to make this work but I can't see it. Any hints are appreciated.
Edit regarding the comments below: By definition, an $A-A-$bimodule has the following action $(am)a'=a(ma')$. I don't see why one would rephrase this action as a right action of $A\otimes A^{op}$ via $m(a\otimes a')= ama'$, and not for example a right $A\otimes A$-module action given by the same formula. That is, why $A^{op}$?