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\section{Common nuclear magnetic moments in organic conductors}  \subsection{H nuclear magnetic moment}  The predominant stable isotope of hydrogen is ${}^{\ 1}_{\ 1}\textrm{H}$, which has a nuclear spin $I = 1/2$ and a nuclear magnetic moment $\mu = 2.79278 $ \textrm{ [nm]}$ (nm =  nuclear magnetons, magneton),  resulting in a nuclear g-factor $g =\mu/I = 5.586$ \cite{Fuller_1976}. This can lead to a significant contribution to the magnetic field dependent nuclear specific heat at dilution refrigerator temperatures. \footnote{One way to correct for this term in H-rich organic conductors is to repeat the measurements using samples containing deuterated hydrogen ${}^{\ 2}_{\ 1}\textrm{H}$. This has a nuclear spin $I = 1$ and a nuclear magnetic moment $\mu = 0.85742 \textrm{[ nm]}$, \textrm{ [nm]}$,  resulting in a much smaller nuclear g-factor $g =I/\mu = 0.857$.} \subsection{Cu nuclear magnetic moment}  ${}^{\ 63}_{\ 29}\textrm{Cu}$ and ${}^{\ 65}_{\ 29}\textrm{Cu}$ are the two most common stable Cu isotopes. ${}^{\ 63}_{\ 29}\textrm{Cu}$ has a nuclear spin $I = 3/2$ and a nuclear magnetic moment $\mu = 2.2228 \textrm{[ nm]}$ \textrm{ [nm]}$  while ${}^{\ 65}_{\ 29}\textrm{Cu}$ has a nuclear spin $I = 3/2$ and a nuclear magnetic moment $\mu = 2.3812 \textrm{[ nm]}$ \textrm{ [nm]}$  \cite{Fuller_1976}. Here, $\mu_N$ is the nuclear magneton, $g = \mu / I$ and the resulting energy splitting in applied field is $\Delta = g \mu_n H / k$ (expressed in temperature units of Kelvin). k [\textrm{ Kelvin}$.  Numerically, the nuclear magneton $\mu_N = \frac{h e}{4 \pi M_p c} = 5.051 10^{-27} \textrm{ J/T}$ is a factor of 1836.1 smaller than the Bohr magneton $\mu_B$.