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Salsk061 [2.6K]
3 years ago
13

In 1993 the Minnesota Department of Health set a health risk limit for chloroform in groundwater of 60.0 g/L Suppose an analytic

al chemist receives a sample of groundwater with a measured volume of 79.0 mL. Calculate the maximum mass in milligrams of chloroform which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits. mg
Chemistry
1 answer:
tangare [24]3 years ago
6 0

Answer:

4.74 × 10³ mg

Explanation:

Given data

  • Health risk limit for chloroform in groundwater: 60.0 g/L
  • Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L

The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:

79.0 × 10⁻³ L × 60.0 g/L = 4.74 g

1 gram is equal to 10³ milligrams. Then,

4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg

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The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver
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Answer: Partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

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Also,   P_{N_{2}} = x_{N_{2}}P

where,    P_{N_{2}} = partial pressure of N_{2}

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Putting the given values into the above formula as follows.

      P_{N_{2}} = x_{N_{2}}P

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Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of N_{2} is as follows.

         P_{N_{2}} = x_{N_{2}}P

                  = 0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}

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Therefore, we can conclude that partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

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