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worty [1.4K]
3 years ago
7

Air is comprised predominantly of nitrogen (N2) and oxygen gas. The ambient air in this room contains about 0.28g of oxygen (O2)

gas per liter of air. A typical human at rest inhales about 19.3 fluid ounces (fl.oz.) of air per breath. And at rest takes about 20 breaths per minute. How many grams of oxygen gas does a human at rest inhale in an hour? Show all calculations with the correct units for full credit.
Chemistry
1 answer:
viktelen [127]3 years ago
7 0

Answer:

190 g

Explanation:

Knowing that:

  • O₂ gas in air: 0.28 g O₂ / L of air
  • Air inhaled at rest: 19.3 fl.oz. air / breath
  • Human takes 20 breaths / min

The grams of oxygen gas (O₂) that a human at rest inhales in 1 hour (60 minutes) will be calculated as:

grams of O₂ = breaths in 1 hour * air inhaled per breath * amount of O₂ in air

grams O_{2} = 60 min* 20 \frac{breath}{min} * 19.3 \frac{fl.oz. air}{breath} * \frac{1 L}{33.814 fl.oz.} * 0.28 \frac{g O_{2} }{L}

grams of O₂ = 191.78 g

190 g of O₂ are inhaled in 1 hour (written in correct number of significant figures)

*In the calculation the amount of air inhaled per breath was converted from fl.oz. to Liters knowing that <em>1 L = 33.814 fl.oz.</em>

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<h3>Answer:</h3>

690 g AgCl

<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
  5. Division
  6. Addition
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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂

↓

[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

[Given] 2.4 mol ZnCl₂

[Solve] <em>x</em> g AgCl

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

[RxN] 1 mol ZnCl₂ → 2 mol AgCl

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 2.4 \ mol \ ZnCl_2(\frac{2 \ mol \ AgCl}{1 \ mol \ ZnCl_2})(\frac{143.32 \ g \ AgCl}{1 \ mol \ AgCl})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 687.936 \ g \ AgCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

687.936 g AgCl ≈ 690 g AgCl

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