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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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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

<u>Step 2: Identify Conversions</u>

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Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})
  2. Multiply:                                                                                                             \displaystyle 1003.77 \ g \ CCl_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1003.77 g CCl₄ ≈ 1000 g CCl₄

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