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Close Problem Tutored Practice Problem 13.1.3 COUNTS TOWAR CA Use ppm and ppb concentration units. The EPAs secondary standards for contaminants that may cause cosmetic or aesthetic effects in drinking water suggest an upper limit of 5.00 102 mgL for manganese. If 4.60 104 liters of water in a storage tank contains 0.837 grams of Mn, what is the contaminant level in ppm? in ppb? Is this level acceptable based on EPA guidelines? ppm ppb Check & Submit Answer Show Approach ype here to search Desktop 5 627-2493

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1000 mg -837 mg mass of the manganese = 0.837 g × volume of soluti on -4.60x10 L 837 mg 4.60×10 L Concentrati on of manganese a mg/L X1Ppm(: 1 ppm -1 mg/L) 1 mg/L = 1.82×10-2 ppm Since the contamination level (1.82× 10-2 mg/L ) İs less than 5.00× 10-2 mg/L this level is acceptable 14g_ = .37×105 μg 8 mass of the manganese -0.837g x volume of solution-4.60×10 L Concentrati on of manganese - 10° g 831×105 ㎍ 4.60x104 L = 18.2 μg/L × 1㎍L -18.2 ppb Since the contamination level (182 ㎍/L ) is less than 5.00×10-2 mgL this level is acceptable

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