Respuesta :
A) 10.75 is the concentration of arsenic in the sample in parts per billion .
B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.
C)It will take 1.37 years to remove all of the arsenic from the lake.
Explanation:
A) Mass of arsenic in lake water sample = 164.5 ng
The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.
To calculate the ppm of oxygen in sea water, we use the equation:
[tex]\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9[/tex]
Both the masses are in grams.
We are given:
Mass of arsenic = 164.5 ng = [tex]164.5\times 10^{-9} g[/tex]
[tex]1 ng=10^{-9} g[/tex]
Volume of the sample = V = [tex]15.3 cm^3[/tex]
Density of the lake water sample ,d= [tex]1.00 g/cm^3[/tex]
Mass of sample = M = [tex]d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g[/tex]
[tex]ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75[/tex]
10.75 is the concentration of arsenic in the sample in parts per billion.
B)
Mass of arsenic in [tex]1 cm^3[/tex] of lake water = [tex]\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g[/tex]
Mass of arsenic in [tex]0.710 km^3[/tex] lake water be m.
[tex]1 km^3=10^{15} cm^3[/tex]
Mass of arsenic in [tex]0.710\times 10^{15} cm^3[/tex] lake water :
[tex]m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g[/tex]
1 g = 0.001 kg
7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg
7,633.66 kg the total mass of arsenic in the lake that the company have to remove.
C)
Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.
Days required to remove 1 kilogram of arsenic from the lake water :
[tex]\frac{2.74}{41.90} days[/tex]
Then days required to remove 7,633.66 kg of arsenic from the lake water :
[tex]=7,633.66\times \frac{2.74}{41.90} days=499.19 days[/tex]
1 year = 365 days
499.19 days = [tex]\frac{499.19}{365} years = 1.367 years\approx 1.37 years[/tex]
C)
Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.
Days required to remove 1 kilogram of arsenic from the lake water :
[tex]\frac{2.74}{41.90} days[/tex]
Then days required to remove 7,633.66 kg of arsenic from the lake water :
[tex]=7,633.66\times \frac{2.74}{41.90} days=499.19 days[/tex]
1 year = 365 days
499.19 days = [tex]\frac{499.19}{365} years = 1.367 years\approx 1.37 years[/tex]
It will take 1.37 years to remove all of the arsenic from the lake.
