The waters of the oceans contain many elements in trace amounts. Rubidium, for example, is present at the level of 2.2 nM (1 nM = 10^-9 M).What volume of seawater have to be processed to recover 510 kg of rubidium, assuming that the recovery process was 100% efficient?
2007-02-04 05:29:22 · 2 answers · asked by bollocks 2 in Science & Mathematics ➔ Chemistry
Rubidium has an atomic mass of 85.4678
85.4678 g/mole x 2.2 nM = 188ng of Rb/liter of seawater
1x10^12 ng/kg x 510kg / 188ng/liter = 2.713x10^12 liters seawater
2.713x10^12 liters = 2.713 teraliters of seawater
2007-02-08 13:09:54 · answer #1 · answered by Timothy H 4 · 1⤊ 0⤋
M=moles/L
you know M, and you know what mass you want to have so you can convert that to moles. than it's just plug in numbers and solve for volume.
2007-02-12 05:13:10 · answer #2 · answered by shiara_blade 6 · 0⤊ 0⤋