Electrons after drop voltage I know that the voltage between two terminals of a battery causes the electrons to move through the wire, and I also know that the voltage in a circuit is distributed among the number of resistors in a circuit. If the electrons loses all their voltage after the last resistor, what will happen to the electrons? Will they stop moving or will they continue?
Does part of the potential energy each of the electrons has will convert to other kinds of energy such as kinetic energy? If some are converted to KE, will the KE be affected by the resistor?
 A: If the electron has crossed the last resistor, no extra energy is required, because there is no opposition (resistance) to overcome.
This can be imagined in a situation where a body is thrown in space. It can displace itself till infinity without any extra energy. Thus once electron has crossed the last hurdle, it can reach the positive terminal of battery without any energy. The portion of the circuit that the electron is passing has 0 resistance or is a superconductor. You must be knowing current doesn't require any potential difference in a superconductor to flow.
A: The simple answer is that although it is hard to track
the motion of a single electron, the fundamental
characteristic of a circuit is that electrons in any
point in the circuit acquire a drift current, based
on the accumulation of charges in the circuit due to material
properties and driving effects (e.g. the battery).
In a series circuit with minimal inductance, this drift
current rapidly becomes constant across the entire circuit.
By "drift current," I mean the average charge flux (or e.g. in
the case of a capacitor, displacement current) through the
cross-section of the wire or circuit element we are considering.
For example, many electrons might be passing left or right
through the cross section, but if more electrons are going right
than left, then we say the drift current is to the left.
(Noting that electrons are negatively charged, of course.)
It is this average charge flux—this drift current—that
equals the current $I$ through the circuit. So, to conclude
the "answer" part of this response: even after the last
resistor, electrons will continue to move toward the
battery terminal, on average.
(To be repetitive), this is because in steady-state for a DC-powered series circuit, electrons drift from the negative terminal toward the positive terminal no matter where they are in the series circuit (excluding the battery, which pulls negative charges from the positive terminal and puts it back on the negative terminal, expending its stored energy in the process).
(Note: I have some other thoughts on this, including a rough explanation thinking of resistance as being due to an effective drag force, but I will need some time to collect my thoughts here. I'll probably edit this answer later.)
A: The voltage between the battery terminals is a difference in potential, key word being difference. So if for example if an electron leaving one terminal was originally at 25 eV of energy and entered the other terminal at 17 eV of energy it would be a 9 volt battery, and if it left with 59 eV and entered with 50 eV it would still be a 9 volt battery. In each case the electron lost 9 eV of energy, and that energy lost is distributed among the resistors. They still have 17 or 50 or however many electron volts of energy necessary to keep them moving. In the resistors that energy might be converted into light energy, heat, motion, or other ways to dissipate or use energy. The remaining energy stays in the electrons as their kinetic energy, which is less than what it started because it lost 9 electron volts. 
