How is powering an LED with a 9 volt battery possible without burning the LED?
Here is how this works -----







simulate this circuit &ndash; Schematic created using CircuitLab

I picked a 180 ohm resistor, because at 20 ohms/volt we are guaranteed a
50 milliamp intersection at the left axis. Thus we scale 20 ohms/volt by 9 volts, finding a 180 ohm resistor provides that current.

Note other LEDs will have lower forward voltage, moving solution-intersection
to the left and raising the solution-current closer to 50mA.