Linda2016idn
Answered

IDNLearn.com is your go-to resource for finding precise and accurate answers. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.

a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?

Sagot :

Maacastrobridn

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

  • 15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²

Then we convert 27.0 ft to m:

  • 27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 m

Now we calculate the total volume of the lake:

  • 3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³

Converting 3.20x10⁸ m³ to L:

  • 3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ L

Now we calculate the total mass of mercury in the lake, using the given concentration:

  • 0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μg

Finally we convert μg to kg:

  • 9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kg
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.