I am trying to understand some variations of singular chain complex and their homology.
Suppose that $X$ is a topological space. We have the abelian groups $C_n(X)$ of simplicial $n$-chains and the corresponding chain complex $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \xrightarrow{\enspace \partial \enspace} C_{n-1} \xrightarrow{\enspace \partial \enspace} \dotsb $$
Every abelian group can be identified with a $\mathbb Z$-module over the integers. So the above chain complex of abelian groups translates into a chain complex of $\mathbb Z$-modules. The homology groups of the chain complex above are abelian groups, too, and as $\mathbb Z$-modules they translate to the homology modules of the $\mathbb Z$-module complex.
Q1: Is this correct so far?
More akin to the textbooks, I can consider the tensor product of the original chain complex with an abelian group $G$, giving the chain complex with coefficients in $G$, $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \otimes G \xrightarrow{\enspace \partial \enspace} C_{n-1} \otimes G \xrightarrow{\enspace \partial \enspace} \dotsb $$
Q2: Are the homology groups of this complex isomorphic to the homology groups of the first complex?
Above we have two variations of the original chain complex. Now suppose that I have a ring $R$.
Let $C_n^R$ be the free $R$-module generated by the singular chains of $X$. We then have a chain complex of $R$-modules $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n^R \xrightarrow{\enspace \partial \enspace} C_{n-1}^R \xrightarrow{\enspace \partial \enspace} \dotsb $$
We can also forget the ring structure of $R$ and consider it an abelian group. We then have the chain complex with coefficients in $R$: $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \otimes R \xrightarrow{\enspace \partial \enspace} C_{n-1} \otimes R \xrightarrow{\enspace \partial \enspace} \dotsb $$
Q3: How do the homology groups / modules of these chain complex relate to each other?