You should invest time in reading up on battery safety and proper care and maintenance of batteries before proceeding any further. I'm not the right person to ask about this; I don't know that much about batteries.
I basically need this explained to me like I was 5.
Just to start out, I plan on an 85 watt panel, 2, 12v, 100 amp hour deep cycle batteries, an 1100 watt inverter, and one 7 watt LED bulb.
Units:
A joule is a unit of energy. This is a small unit of energy derived from the definition of mechanical work in classical physis. 1 Joule is about the energy released when a small apple drops to the floor from 1 meter of height in Earth's gravity.
A watt is a unit of power. Power is a rate of energy consumption. 1 watt is equal to consuming 1 joule every second[1 J/s]. e.g. your LED light consumes 7 joules per second.
A watt second is again a unit of energy. 1 W*s = 1 (J/s)*s = 1 J.
A watt hour is a slightly larger unit of energy. 1 W*h = 1(J/s)* (3600 s) = 3600 J. You can of course also go the other way by dividing both sides with 3600: 1 J = 1/3600 Wh.
A kWh is 1000 times larger than a Wh as the kilo prefix would suggest.
1 coloumb is a unit of charge.
1 volt is a unit of electrical potential. When a coloumb moves across an electrical potential of 1 V a joule of energy is release. 1 V = 1 J/C.
1 ampere is a unit of current. 1 A = 1 coloumb/s. An ampere-hour is a unit of electrical charge, in analogy to watt-hours. 1 Ah = 1 (C/s)*(3600 s) = 3600 coloumbs.
A physical quantity is the product of a numerical value and a unit. 0.77 is a numerical value; 1 volt is a unit; 0.77 V is a quantity.
When you perform arithmetic on physical quantities you perform the same arithmetic on their units. So for instance, if you multiply 12 V by 100 Ah you get (12*100) * ( V*Ah) = 1200 VAh = 1200 * ( 1/3600 Wh/C) * (3600 C) = 1200 Wh = 1.2 kWh.
If you're unsure of yourself, keep track of the units and not just the numerical values during a calculation. If the units of the result are nonsense, you've made some mistake.
1.) How long would this bulb stay lit running off of the fully charged batteries?
Deep cycle batteries are made to be discharged between 50% and 80%(check your manufacturer's info) without noticable impact on lifespan.
The amount of energy stored by the batteries 12 V * 2 * 100 Ah = 2400 VAh = 2400 Wh.
The amount of time it takes to fully discharge the batteries with a 7 W load: 2400 Wh / 7 W = 343 h.
If you discharge only to 50-80% as recommended you will get 172 to 274 hours of life on one charge.
2.) Not accounting for inefficiencies, how long would it take to charge the 2 batteries with 6 hours of full sun/day? (no bulb attached)
That might be the wrong question, depending on how fault tolerant you are. At mid US lattitudes you get approximately twice as much solar insolation on a typical summer day as on a typical winter day, not accounting for wheather.
Assuming the panel puts out the full 85 watt for 6 hours/day, which was the question asked, it will take 2400 Wh/(85 W * 6 hours/day * 24 h/day) = 113 hours to fully charge the batteries.
If you are only discharging the batteries by 50%-80% of their full capacity as recommended it will take 57-90 hours to recharge the batteries.
In terms of hours of full sunlight instead of hours of average sunlight it is much less; 14-23 hours not counting inefficiencies and niggles that will affect you in the real world.
Lead-acid batteries have a ~75-80% round-trip efficiency, so this is bit better than the performance you can expect when accounting for inefficiencies. They will discharge slightly quicker and charge slightly slower in the real world.
Batteries also have self-discharge, but it is quite negligible for lead acid batteries at room temperature or slightly above.
Deep cycle lead-acid batteries need proper care to minimize degradation through sulfation and corrosion; consult your manufacturer for detailed instructions on how the battery should be cycled, stored and topped up with sulfuric acid for best performance.
An 1100 W inverter seems quite large. Batteries have internal resistance; if you discharge them too quickly they become very hot. They can withstand short bursts of very high power, but not sustained high power.
If the batteries cannot withstand 1100 W sustained(consult your manufacturer) I believe you will want two fuses. A fast fuse for your inverter and a slow-blow fuse for your batteries so that your batteries can put out up to 1100 W for a short period of time, but not long enough to harm your batteries(again, consult your manufacturer).
