= Solution
At $T=310.15$ K, $RT/F=26.72666$ mV. Always retain the valence sign. For the two monovalent positive <ions>,
$$
\boxed{E_K=26.72666\ln(4/155)=-97.74\ \mathrm{mV},\qquad
E_{\rm Na}=26.72666\ln(145/12)=+66.60\ \mathrm{mV}.}
$$
For <calcium>, $z=2$ and the specified <voltage> is positive:
$$
\boxed{c_{{\rm Ca},\rm in}=1.5\exp[-2(130)/26.72666]\ \mathrm{mM}
=8.938\times10^{-5}\ \mathrm{mM}\simeq89.4\ \mathrm{nM}.}
$$
For chloride, $z=-1$ reverses the sign in the <Nernst equation>:
$$
-90=-26.72666\ln(c_{{\rm Cl},\rm out}/4),\qquad
\boxed{c_{{\rm Cl},\rm out}=116.0\ \mathrm{mM}.}
$$
These values use <thermodynamic activities> approximated by the given <concentrations>. <Calcium> is the \b[free] <concentration> relevant to its <electrochemical potential>, rather than total bound plus free intracellular <calcium>. Ratios use matching <concentration> units, so a ratio of millimolar values is already dimensionless. Failure to include the factor of two for <calcium> or the negative charge of chloride would change the answer substantially.
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