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mr Goodwill [35]
3 years ago
11

A low-pressure weather system comes into the city of Denver. The atmospheric pressure is 693 mmHg.693 mmHg. If 78.0%78.0% of dry

air is nitrogen, what is the partial pressure of nitrogen in this low-pressure system?
Chemistry
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

540.54 mmHg

Explanation:

We know that the partial pressure of a substance is defined as; Mole fraction * total pressure.

If the total amount of gases in the atmosphere is 100%, the mole fraction of nitrogen gas is now

78/100 = 0.78

Thus, partial pressure of nitrogen gas = 0.78 * 693 = 540.54 mmHg

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An occluded front forms when a warm air mass is caught between two cooler air masses.

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Which of the following is an indication of a chemical reaction?
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How many grams of Fe will be produced if 705.0g of H20 are produced?
fredd [130]
<h3>Answer:</h3>

2265 g Fe₃O₄

<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
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

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

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

<u>Step 3: Convert</u>

  1. Set up stoich:                              \displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})
  2. Multiply/Divide/Cancel units:                                                                           \displaystyle 2264.74 \ g \ Fe_3O_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

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