This may have been asked before, but I'm currently I'm an argument about whether or not passing 12v through a .1 ohm resistor would result in 120 amps due to Ohm's Law. (I=V/R) Providing that the power source has a negligible internal resistance
In theory, yes, exactly.
Provided the 12V source has negligible internal resistance and provided the 0.1 ohm resistor does not change resistance (or melt or explode) from dissipating 1.44kW.
If you assume internal resistance is negligible then yes that is exactly what will happen!
Of course you may start a fire or explode your resistor at that level of power...
If you can maintain 12v across a 0.1ohm resistor, then yes, it will conduct 120 Amps.
At these power levels and currents in the real world, we'd normally flip Ohm's Law around, and say that if a current of 120A was passing through a resistance of 0.1ohms, then it would drop 12v across it.
So for instance if you put your arc welder on the 120A setting, and used cables with a total resistance of 0.1ohms to connect to the torch and workpiece, 12v would get dropped on the cables, reducing the voltage available at the arc. The cables would get quite warm as well.
Rather than passing a voltage through a resistor (or circuit) it is more usual to describe it as applying a voltage. The term passing is more usually used in relation to current. i.e. apply a voltage and pass a current. The short answer is that according to Ohm's Law, V = I * R, so applying V = 12 V across a resistance of R = 0.1 Ohm, the resulting I must be 120 A. Likewise passing 120 A through a resistance of 0.1 Ohm must result in a voltage of 12 V across it.
As others have indicated, there are various practical considerations related to a real-world application of this particular scenario. The theory, however, is incontrovertible.
If the power of 0.1 ohm resistor > 1.44kW then you are right.
But based on my understanding, there is no such a big power resistor.