단위환산. 오염된 호수에 수은 0.4 μg/mL가 존재

단위환산. 오염된 호수에 수은 0.4 μg/mL가 존재

 

 

오염된 호수에 수은 0.4 μg/mL가 존재한다면 그 호수에 있는 총 수은의 양은 몇 kg인가? (호수의 표면적은 100 mi^2이고, 평균 깊이는 20 ft이다.)

 

Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is 0.4 μg Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi^2 and an average depth of 20 ft.)

 

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1 mi = 1609.344 m

1 mi^2 = (1609.344 m)^2

 

1 ft = 0.3048 m

 

 

100 mi^2 * [(1609.344 m)^2 / 1 mi^2] * 20 ft * (0.3048 m / 1 ft)

= 1578856752 m^3

= 1.5789*10^15 mL

( 1 m3 = 1000 L )

( 1 L = 1000 mL )

 

 

1.5789*10^15 mL * (0.4 μg/mL) * (1 kg / 10^9 μg)

= 6.3156*10^5 kg

 

 

답: 6*10^5 kg

( 7*10^5 kg은 오답입니다. )