Exponential tail bounds automatically imply moment bounds and vice versa. That is to say, $(a)$ is equivalent to $(A)$ for $a\in \{j,k,l\}$ below where $X$ is a nonnegative random variable and $\|X\|_p = (\mathbb{E} X^p)^{1/p}$. $C,c>0$ are universal constants that may change from line to line.

$(j)$ For all $p\ge 1$, $\|X\|_p \le c\sigma\sqrt{p}$

$(J)$ For all $\lambda>0$, $\mathbb{P}(X > \lambda) \le C \cdot e^{-c\lambda^2/\sigma^2}$

$(k)$ For all $p\ge 1$, $\|X\|_p \le cKp$

$(K)$ For all $\lambda>0$, $\mathbb{P}(X > \lambda) \le C\cdot e^{-c\lambda/K}$

$(l)$ For all $p\ge 1$, $\|X\|_p \le c\sigma\sqrt{p} + cKp$

$(L)$ For all $\lambda>0$, $\mathbb{P}(X > \lambda) \le C\cdot\max\left\{e^{-c\lambda^2/\sigma^2}, e^{-c\lambda/K}\right\}$

One can come up with other examples. Now, $(a)$ implies $(A)$ by Markov's inequality on the $p$th moment. $(A)$ implies $(a)$ using integration by parts (let $\varphi$ be the pdf in what follows and $\Phi$ be the cdf):
$$
\mathbb{E} X^p = \int_0^\infty \lambda^p \varphi(\lambda)d\lambda = -\int_0^\infty \lambda^p (-\varphi(\lambda))d\lambda = [\lambda^p\cdot(1 - \Phi(\lambda))]_{\lambda=0}^\infty + \int_0^\infty p\lambda^{p-1}(1 - \Phi(\lambda))d\lambda
$$
Now note $1-\Phi(\lambda) = \mathbb{P}(X > \lambda)$, so as long as $\mathbb{P}(X>\lambda)$ decays fast enough, the $[\lambda^p\cdot(1 - \Phi(\lambda))]_{\lambda=0}^\infty$ term goes to $0$ and we've now expressed the $p$th moment in terms of something depending on the tail bound (i.e. $(A)$ implies $(a)$ as claimed).

Now back to your question. Whatever Chernoff bound you're thinking of, it gives a tail bound. Thus, by the above, it automatically implies a moment bound. So if your random variables are $p$-wise independent for $p = \sqrt{n}$ or whatever other value, you can see what tail bound you get by doing Markov's inequality on the $p$th moment.