Non-realizability of $\mathbb{Q}$ as a cohomology group (This is a re-post of 1)
In the paper "On the realizability of singular cohomology groups" by Kan and Whitehead, it is shown that there is no space $X$ and integer $n\geq 1$ such that $H^{n−1}(X)=0$ and $H^n(X)=\mathbb{Q}$ (cohomology with integral coefficients).
At the very end of the article there is a remark where it is stated that, at the time of writing (around 1960, I suppose), it was not known whether $\mathbb{Q}$ could be a (integral) singular cohomology group at all.
My question is: is this still not known?
 A: EDIT: If I also did more than 1 second worth of checking homological algebra books, I wouldn't immediately find a desired UCT for homology that involves Ext (instead of Tor). So there's a potential gap here, as Oscar points out in the comment:
This is known (and given be a 1 second google search). The assumption on $H^{n-1}(X)$ can be removed, and it has a cute proof:
Use the UCT for cohomology to see that $H^n$ has an Ext term from $H_{n-1}$, and use UCT for homology to see that $H_{n-1}$ has an Ext term from $H^n$, and then note that $Ext(\prod_\mathbb{Z}\mathbb{Q},\mathbb{Z})$ is uncountable (while $\mathbb{Q}$ is countable).
One Remark on the Realizability of Singular Cohomology Groups 
A: This may depend on your axioms, see
S. Shelah "The consistency of Ext(G,Z)=Q", Israel J. Math. 39 (1981), no. 1-2, 74–82. 
There it is shown that it is consistent with the generalised continuum hypothesis that there exists a group $G$ having $Ext(G, \mathbb{Z})=\mathbb{Q}$. Then a Moore space $M(G,n-1)$ has the required property.
