12.07.2022 - 04:56

The EPA’s secondary standards for contaminants that may cause cosmetic or aesthetic effects in drinking water suggest an upper limit of 0.300 mg/L for iron. If 2.96 x 104 liters of water in a storage tank contains 26.8 grams of Fe, what is the contaminant

Question:

The EPA’s secondary standards for contaminants that may cause cosmetic or aesthetic effects in drinking water suggest an upper limit of 0.300 mg/L for iron. If {eq}2.96 times 10^4 {/eq} liters of water in a storage tank contains 26.8 grams of {eq}Fe {/eq}, what is the contaminant level in ppm? What about in ppb? Is this level acceptable based on EPA guidelines?

Answers (1)
  • Delma
    April 14, 2023 в 00:54
    The first step is to convert the given amount of iron (26.8 grams) to milligrams, since the EPA guideline for iron is given in milligrams per liter (mg/L): 26.8 grams = 26,800 milligrams Then, we can use the given volume of water in the storage tank and the amount of iron to calculate the contaminant level: contaminant level = (26,800 mg) / (2.96 x 10^4 L) contaminant level = 0.905 mg/L To convert this to parts per million (ppm), we can multiply by 1000: contaminant level = 0.905 ppm To convert this to parts per billion (ppb), we can multiply by 1000 again: contaminant level = 905 ppb Based on EPA guidelines, the upper limit for iron in drinking water is 0.300 mg/L or 0.300 ppm. The calculated contaminant level of 0.905 ppm is higher than this limit, indicating that the iron content in the water exceeds the EPA guideline. Therefore, this level is not acceptable based on EPA guidelines.
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