Water is flowing through a fire hose at a rate of #450# cubic centimetres per second. What is the total negative charge being carried by the electrons in the water, and what current is that flow of charge equivalent to?

#450# #cm^3s^-1# = #450# #mLs^-1# = #0.45# #Ls^-1#

Water has a density of #1.00# #kgL^-1#, so #0.45# #L# = #0.45# #kg#. It has a molar mass of #18# #g mol^-1#, so divide the rate of flow by #18# #gmol^-1# = #0.018# #kgmol^-1# and the flow rate is #25# #mols^-1#.

Now comes the tricky bit: do they mean only the negative charge in the valence shells that is involved in forming the bonds, or do they mean all negative charge? I'm going to assume they mean the latter, but the calculation is the same with small tweaks for the former.

In a #H_2O# molecule there are 2 hydrogen atoms each bringing 1 electron and 1 oxygen atom bringing 8, for a total of 10 electrons per molecule. There are Avogadro's Number #-# #6.02 xx 10^23# #-# molecules per mole.

That means there are #25*6.02xx10^23*10# electrons passing any given point in the hose per second. That's #1.5 xx 10^26# electrons per second. We could leave it at that and say that's how many electron charges pass by, but let's put in the charge on one electron #-# #1.6 xx 10^-19# #C# #-# to find the flow rate in #Cs^-1#.

#1.5 xx 10^26*1.6 xx 10^-19 = 24,080,000# #Cs^-1#

Some of you are probably already screaming at me, though: one #Cs^-1# is one ampere #(A)#.

So the flow of negative charge in the hose is #24,080,000# #A#.

Which is HUGE!

Of course, that's not the net negative charge: there are as many protons in the nuclei of the hydrogen and oxygen atoms as there are electrons orbiting the nuclei, so the net flow of charge is ... #0#.