At $\alpha = 0.05$, the significance cutoff of a Chi-squared distribution with 1 df is approximately 3.84, and that if we take the square root of 3.84, it's approximately 1.96, which is the 97.5 percentile of a N(0,1) normal distribution. I understand why this is the case for 1 df, as the Chi-squared distribution is nothing more than a squared N(0,1) distribution.
I also understand that for 2 dfs the Chi-squared distribution is just follows the distribution of the expected sum of two squared random samples from N(0,1), and has a significance threshold of approximately 6.0. Following the logic of the case with 1 df, in the case of 2 dfs we can also be expressed the threshold as the 97.5 percentile of a N(0, ~1.25) Normal distribution, which is about the square-root of 6.0.
Is there a general equation we can use to express a Chi-squared significance threshold (at say alpha = 0.05) for an arbitrary degrees of freedom (x), in terms of a Normal distribution with mean = 0 and standard deviation = y? I was only able to solve the above example for 2 dfs (i.e. x=2, y=1.25) though testing, but if there is some way to generalize this translation between Chi-squared and Normal, I think it would help me to gain more insights into the Chi-squared.