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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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Answer:

You must add 48.97 mL of water to make the 0.200 M diluted solution.

Explanation:

In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.

In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.

The expression for the dilution calculations is:

Cinitial* Vinitial = Cfinal* Vfinal

In this case:

  • Cinitial= 12 M
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Replacing:

12 M*0.830 mL= 0.200 M*Vfinal

Solving:

Vfinal=\frac{12 M*0.830 mL}{0.200 M}

Vfinal= 49.8 mL

Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:

49.8 mL - 0.830 mL= 48.97 mL

<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>

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A three-carbon chain has a straight line extending from the center carbon.

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