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Wittaler [7]
3 years ago
6

According to the EPA, the maximum contaminant level (MCL) of antimony in drinking water is 0.0060 mg/L.A farmer recently dug a n

ew well for his property. He sends a 12.0 mL 12.0 mL sample of the well water to the EPA to be tested for the presence of antimony. What is the maximum amount of Sb Sb (in μg μg ) that can be present in this 12.0 mL 12.0 mL sample based on its MCL?
Chemistry
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

The maximum amount of Sb that can be present in this 12 mL sample based on MCL, is 0.072 μg

Explanation:

The maximum contaminant level according to EPA is given as:

Max Level = 0.006 mg/L

Max Level = (0.006 mg/L)(1 L/1000 mL)

Max Level = 0.006 x 10^-3 mg/mL

Now, the sample size is 12 mL. Hence the maximum amount of Antimony (Sb) can be ount out by the use of maximum Contaminant Level.

Max. Amount = (Max. Level)(Sample Size)

Max. Amount = (0.006 x 10^-3 mg/mL) (12 mL)

Max. Amount = 0.072 x 10^-3 mg

<u>Max. Amount = 0.072 μg</u>

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The second-order decomposition of NO2 has a rate constant of 0.255 M-15-1. How much NO2 decomposes in 4.00 s if the initial conc
S_A_V [24]

Answer:

Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol

Explanation:

rate\;constant = 0.255\;M^{-1}s{-1}

Time (t) = 4.00\;s

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Integrated law for second order reaction:

\frac{1}{[A]}=\frac{1}{[A]_0} =kt

Where, [A] = Concentration after time, t

[A]0 = Intitial concentration, k = rate constant, t = time

On substituting values in the above

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[A] = 0.5644 M

Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M

No. of mole = Molarity * volume

                    = 0.7656 * 1

                    = 0.7656 mol 0r 0.77 mol

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