Im wiring my first lm7805 and I want to make sure I got it right:
Here you can see that on a breadboard Ive got the + from my 7.4V pack into the + leg of one cap and its - leg to the - from my 7.4V pack. Kinda like this:
+(7.4)-(7.4)
+CAP1____________-CAP1____________
________-CAP2+CAP2
IN__________________GND______________OUT
And I will take my 5V from the OUT and GND, right?
Yes.
You're looking at the back of the part.
The red wire is 7.4V and the black wire is Gnd?
Yes that is the back of the lm7805, so the left pin on that picture would be its pin 1, which is IN.
Yes the red wire is 7.4V and the black is the ground from the battery pack.
Im just a bit scared of setting something on fire or having it explode 
Kindly refer to this diagram.
Once you have it connected properly, there is is the issue of power disapation. Without a heat sink, don't exceed ~200mw, give or take. Since power is a function of Vin-Vout*I and you're as low as you can go on Vin so the only consideration is load. To stay in that ~200mw range puts load at about 100ma max.
Since the 7805 has internal thermal protection, it should shut down by dropping the output to zero volts if it becomes too hot - which can confuse newbies that don't understand the behavior. Just know that when it shuts down, the metal side is hot enough to give you a nasty burn.
Thanks guys. One last question related to the caps. I know what they do in theory but what would the difference be in using 10uF vs 100uF caps?
I know what they do in theory but what would the difference be in using 10uF vs 100uF caps?
100uF on the output of the regulator will be better, that is less ripple on the output voltage.
