Which of the following will have the lowest average kinetic energy? OA) H₂ at 400 °C O B) O₂ at 300 °C OC) H₂O at 300 °C OD) He at 200 °C O E) CH4 at 400 °C

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Question:** Which of the following will have the lowest average kinetic energy?

- **A)** H₂ at 400 °C
- **B)** O₂ at 300 °C
- **C)** H₂O at 300 °C
- **D)** He at 200 °C
- **E)** CH₄ at 400 °C

**Correct Answer:** A) H₂ at 400 °C

### Explanation:

The average kinetic energy of a gas is directly related to its temperature, regardless of the type of gas. Since all options list different temperatures, the gas with the lowest temperature will have the lowest average kinetic energy. In this case:

- He at 200 °C (option D) would have the lowest average kinetic energy, not option A as initially indicated.

It seems there might be an error if option A is marked as correct. Option D should be correct based on the understanding of kinetic energy in relation to temperature.
Transcribed Image Text:**Question:** Which of the following will have the lowest average kinetic energy? - **A)** H₂ at 400 °C - **B)** O₂ at 300 °C - **C)** H₂O at 300 °C - **D)** He at 200 °C - **E)** CH₄ at 400 °C **Correct Answer:** A) H₂ at 400 °C ### Explanation: The average kinetic energy of a gas is directly related to its temperature, regardless of the type of gas. Since all options list different temperatures, the gas with the lowest temperature will have the lowest average kinetic energy. In this case: - He at 200 °C (option D) would have the lowest average kinetic energy, not option A as initially indicated. It seems there might be an error if option A is marked as correct. Option D should be correct based on the understanding of kinetic energy in relation to temperature.
**Question:**

Combustion of an unknown compound containing only carbon and hydrogen produces 37.65 g of CO₂ and 15.42 g of H₂O. What is the empirical formula of the compound? Show your work.

**Options:**

A) C₂H₃  
B) C₃H₁₀  
C) C₄H₈  
D) CH₂  
E) CH  

**Explanation:**

To solve this problem, first, calculate the moles of carbon and hydrogen in the produced CO₂ and H₂O.

1. **Find moles of carbon:**

   - Molar mass of CO₂ is approximately 44.01 g/mol.
   - Moles of CO₂ = 37.65 g / 44.01 g/mol = 0.855 moles of CO₂.
   - Since each molecule of CO₂ has one carbon atom, moles of carbon = 0.855.

2. **Find moles of hydrogen:**

   - Molar mass of H₂O is approximately 18.02 g/mol.
   - Moles of H₂O = 15.42 g / 18.02 g/mol = 0.856 moles of H₂O.
   - Each molecule of H₂O contains two hydrogen atoms, so moles of hydrogen = 2 * 0.856 = 1.712.

3. **Determine the empirical formula:**

   - Calculate the mole ratio:
     - Ratio of C:H = 0.855:1.712 ≈ 1:2.
   - Thus, the empirical formula is CH₂.

So, the answer is **D) CH₂**.
Transcribed Image Text:**Question:** Combustion of an unknown compound containing only carbon and hydrogen produces 37.65 g of CO₂ and 15.42 g of H₂O. What is the empirical formula of the compound? Show your work. **Options:** A) C₂H₃ B) C₃H₁₀ C) C₄H₈ D) CH₂ E) CH **Explanation:** To solve this problem, first, calculate the moles of carbon and hydrogen in the produced CO₂ and H₂O. 1. **Find moles of carbon:** - Molar mass of CO₂ is approximately 44.01 g/mol. - Moles of CO₂ = 37.65 g / 44.01 g/mol = 0.855 moles of CO₂. - Since each molecule of CO₂ has one carbon atom, moles of carbon = 0.855. 2. **Find moles of hydrogen:** - Molar mass of H₂O is approximately 18.02 g/mol. - Moles of H₂O = 15.42 g / 18.02 g/mol = 0.856 moles of H₂O. - Each molecule of H₂O contains two hydrogen atoms, so moles of hydrogen = 2 * 0.856 = 1.712. 3. **Determine the empirical formula:** - Calculate the mole ratio: - Ratio of C:H = 0.855:1.712 ≈ 1:2. - Thus, the empirical formula is CH₂. So, the answer is **D) CH₂**.
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