There are 50 billion gold atoms in every drop of sea water. 1 drop = .05ml. is it feasible to “mine” the ocean for gold? in other words, how many gallons of sea water would have to be stripped of its gold to produce 1 troy ounce of the precious metal?

In order to answer this question, set up dimensional analysis, you have to cancel out the number of grams while keeping the amount of sea water per troy oz. 
So now you have the density of sea water in terms of gold (g/mL) flip it upside down mL/g. 
(.05mL seawater/1.635*10^-11 grams of gold) * (31.3 grams/troy oz) = mL sea water needed to make 1 troy oz (9.5107x10^10 mL/troy oz) 
then convert mL to gallons ( (9.5107x10^10 mL/troy oz) x (2.64x10^-4 gallons/1mL) ) 
= 25108256.88 gallons which when you round this off, you’ll have 30000000.

1.from liquid state to gaseous state is evaporation 3.solid to liquid is melting 5. solid to gaseous state is sublimation 6. gas to solid state is decomposition 7. a change of state is known as the states of matters

257 grams h2o i think