If f(n) = Ω(n) and g(n) = O(f(n)), what do we know about g? Let f(n) = Ω(n),
and g(n) = O(f(n)).Then g(n) = _______.  

I thought of it this way, since f(n) is Ω(n),then f(n) belongs to the set of functions defined by Ω(n), ie,{n,$n^2$,$n^3$ ....}.  So g(n) will change accordingly , ie , g(n) will be of complexity O(n),O($n^2$),O($n^3$) ...etc. Is it correct to directly substitute functional values into O(f(n)) to get a sense of the complexity?
Knowing that $f(n)=\Omega(n)$ and $g(n)=O(f(n))$ doesn't tell you anything about $g(n)$. For all we know, $f(n)$ could be $2^{(2^{(2^n)})}$ and then $g(n)$ could be almost any function. All we know is it is upper bounded by some function, which might grow arbitrarily fast. It's akin to saying "up to 50% off or more".