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66conv6 is correct! Gasoline liquid does not burn, the vapors just above the liquid surface is what burns. That is why a carb converts and meters gasoline into a vapor for combustion in each cylinder.
The answer to the first question is the the sending unit has been designed such that it will not be an ignition source to the gasoline vapor. It is operating at 12 volts and uses very little resistance to register liquid levels to the fuel guage. Also, when you have less than a 1/4 of a tank, the sending unit is no longer submerged in liquid.
 

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I have another good one for you to ponder. If the fuel tank has 100% fuel vapors and absolutely no oxygen/air present, it would not be able to ignite or explode anyway. Think about it!
 

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IIRC the resister behind the dash drops the voltage to less than 2 volts going to the sending unit, it also sends pulsed signals to the sending unit not constant voltage. This also keeps the possibility of an explosion to a minimum. If you were to put a test light on your sending unit wire it would barely light up from the lowered voltage...
 

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Discussion Starter #8
I would imagine SOME air would get in....otherwise no fuel would flow out...

never saw the resistor inline, but the pulsing of the regulator is a good point
 

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the pick-up is a sliding contact on the float and is immersed in gasoline usually (bad environment for sparking) and low voltage. biggest danger is some idiot banging wrenches while removing with gas fumes in tank - fumes igniting are what blows your hair off.
 

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Gas has a vapor pressure because of the propanes,butanes and other light ends in it. As it sloshes around like during refueling it forces most of the air out. That puts it above the uel(upper explosive limit). So if the 2 volts that operate the sender could fail so bad to make a spark (like is some one hooks a 110v extension cord to it, it's still is too rich to burn. I guess it pays to be an industrial fire fighter some times.
 
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