### Problem Statement**Question (6.3):** The volume of a gas is 0.953 L at 30.0°C. If the gas is heated to 60.0°C, what would be the volume of the gas (in L) at this temperature?### Answer Choices- 1.05 L- 0.867 L- 1.91 L- 0.477 L### ExplanationTo solve this problem, one must apply the principles of Charles's Law, which states that at constant pressure, the volume (V) of a gas is directly proportional to its temperature (T) in Kelvin. The formula is given by:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Where:- \( V_1 \) is the initial volume (0.953 L)- \( T_1 \) is the initial temperature in Kelvin (30.0°C = 303.15 K)- \( V_2 \) is the final volume (unknown)- \( T_2 \) is the final temperature in Kelvin (60.0°C = 333.15 K)**Step-by-Step Calculation:**1. Convert both temperatures from Celsius to Kelvin: \[ T_1 = 30.0°C + 273.15 = 303.15 K \] \[ T_2 = 60.0°C + 273.15 = 333.15 K \]2. Use Charles's Law to find the unknown volume \( V_2 \): \[ \frac{0.953 \, \text{L}}{303.15 \, \text{K}} = \frac{V_2}{333.15 \, \text{K}} \]3. Rearrange to solve for \( V_2 \): \[ V_2 = \frac{0.953 \, \text{L} \times 333.15 \, \text{K}}{303.15 \, \text{K}} \]4. Calculate: \[ V_2 = \frac{317.50 \, \text{L K}}{303.15 \, \text{K}} \approx 1.05 \, \text{L} \]### ConclusionThe correct answer is **1

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Answered: (6.3)The volume of a gas is 0.953 L at…

(6.3)The volume of a gas is 0.953 L at 30.0 °C. If the gas is heated to 60.0 °C, what would be the volume of the gas (in L) at this temperature? O 1.05 L O 0.867 L O 1.91 L O 0.477 L

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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