Resistance-capacitance (RC) circuit 
My textbook says: "This continues until the voltage across the capacitor matches the emf of the battery"
If I were to have a resistor in series with a capacitor, would the voltage across the capacitor still be equal to the emf of the battery?
My thinking process: The voltage across the capacitor will not be the same due to KVL as there would be a voltage drop across the resistor as well.
 A: There will be a voltage drop across the resistor only when current flows. When the capacitor gets charged to equal the battery voltage the current stops and there is no longer a voltage drop across the resistor.
Hope this helps.
A: 
My question: If I were to have a resistor in series with a capacitor,
would the voltage across the capacitor still be equal to the emf of
the battery?

In an ideal circuit theory context, the answer is no except in the trivial case that the initial condition is just that.
That is, for the ideal RC circuit drawn, the voltage across the capacitor asymptotically approaches $V$, the voltage across the voltage source, if the initial condition is otherwise.
As Bob D very correctly points out, the voltage across the resistor is zero unless there is a non-zero current through. However, in the ideal case, the current through goes to zero only at $t=\infty$, i.e., the capacitor never stops charging.
Practically, however, and for the zero capacitor voltage initial condition, the capacitor is considered 'fully charged' after $t=5\cdot\tau= 5\cdot RC$ seconds.
A: If you set up a circuit like this, with reasonably high values for the resistor and the capacitor, a measurable current will flow for at least a few seconds. Initially the current is quite high, but as the charge and voltage on the capacitor increase, the voltage across the resistor decreases and so the current decreases. After a time the current will have dropped to a value which is too small to measure, and the voltages across the battery and capacitor will be equal within the limits of measurement.
