Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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**Chemistry Problem Example: Gas Density and Volume Calculations**

**Problem Statement:**

Gaseous hydrogen has a density of 0.0899 g/L at 0°C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?

**Detailed Solution:**

To find the volume of gas needed, use the formula:

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

For hydrogen:

1. Given:
   - Density of hydrogen (\( \rho_H \)): 0.0899 g/L
   - Mass of hydrogen (\( m_H \)): 1.0078 g

2. Applying the formula:
\[ V_H = \frac{m_H}{\rho_H} \]
\[ V_H = \frac{1.0078\text{ g}}{0.0899\text{ g/L}} \]
\[ V_H \approx 11.21\text{ L} \]

For chlorine:

1. Given:
   - Density of chlorine (\( \rho_{Cl} \)): 3.214 g/L
   - Mass of chlorine (\( m_{Cl} \)): 35.45 g

2. Applying the formula:
\[ V_{Cl} = \frac{m_{Cl}}{\rho_{Cl}} \]
\[ V_{Cl} = \frac{35.45\text{ g}}{3.214\text{ g/L}} \]
\[ V_{Cl} \approx 11.03\text{ L} \]

Hence, you would need approximately 11.21 liters of hydrogen and 11.03 liters of chlorine to meet the required amounts.
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Transcribed Image Text:**Chemistry Problem Example: Gas Density and Volume Calculations** **Problem Statement:** Gaseous hydrogen has a density of 0.0899 g/L at 0°C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine? **Detailed Solution:** To find the volume of gas needed, use the formula: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \] For hydrogen: 1. Given: - Density of hydrogen (\( \rho_H \)): 0.0899 g/L - Mass of hydrogen (\( m_H \)): 1.0078 g 2. Applying the formula: \[ V_H = \frac{m_H}{\rho_H} \] \[ V_H = \frac{1.0078\text{ g}}{0.0899\text{ g/L}} \] \[ V_H \approx 11.21\text{ L} \] For chlorine: 1. Given: - Density of chlorine (\( \rho_{Cl} \)): 3.214 g/L - Mass of chlorine (\( m_{Cl} \)): 35.45 g 2. Applying the formula: \[ V_{Cl} = \frac{m_{Cl}}{\rho_{Cl}} \] \[ V_{Cl} = \frac{35.45\text{ g}}{3.214\text{ g/L}} \] \[ V_{Cl} \approx 11.03\text{ L} \] Hence, you would need approximately 11.21 liters of hydrogen and 11.03 liters of chlorine to meet the required amounts.
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