# Comment: 100% contamination of what

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### 100% contamination of what

100% contamination of what mass? What concentration? What is the isotope? What is the uptake? What in the sea water is contaminated?

Your example of 1 drop in 1 gallon in each gallon of the ocean is reversed, you should be dividing by 75,508 since it is being diluted more, excluding concentration again since you gave none. Your example is spreading 1 droplet of contaminated 'substance' throughout every gallon in the Pacific.

Say for 964 days 300 tons of contaminated water (mass, not concentration since none was given, of mystery substance) is dumped into the pacific ocean.

That would be 80,700 gallons/day x 964 days = 77,794,800 gallons in 964 days total.

I'll assume 75,708 drops/gallon is a valid statement.

1 / 75,708 drops/gallon = 0.0000132 gallons/ per drop

So lets see if we have enough drops to be spread evenly over the entire Pacific ocean.

We have 174,353,554,600,000,000 gallons in the Pacific ocean.
We have 77,794,800 gallons of contaminated water to spread.
There are 75,708 drops of unknown concentration of substance per gallon of water. Since no concentration was given this refers only to volume.

75,708 drops x 77,794,800 gallons of total contaminated water = 5,889,688,718,400 drops of unknown concentration contaminated water to spread.

We don't even have enough drops of water to spread evenly through every individual gallon of the Pacific
174,353,554,600,000,000 - 5,889,688,718,400 = 174,347,664,911,281,600 Gallons short that did not recieve drops.

Southern Agrarian