The maximum contamination level of arsenic ion in a water system is 0.050 parts per million. if the arsenic is present as ascl3, how many grams of arsenic chloride could be present in a system that contains 8.2 x 10^5 liters?

mass AsCl₃ in 8.2x10⁵ L = 99.2gAsCl₃

Step-by-step explanation:

Let's remember that 0.05 ppm (parts per million) would be write terms of mg/L. We have 0.05 mg/L, that is a concentration unit.

First of all we need to find the g/L of AsCl₃ present in water.

Now, we just need to multiply it by 8.2 x 10⁵ L to determine the mass of AsCl₃ present in that volume.

Finally we have:      

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Have a nice day!

question is incomplete.

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13