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O Stas of Matter Using the ideal equation of state A reaction at \( -10.0^{\circ} \mathrm{C} \) evolves \( 490 . \mathrm{mmol} \) of dinitrogen difluoride gas. Calculate the volume of dinitrogen difluoride gas that is collected. You can assume the pressure in the room is exactly 1 atm. Round your answer to 3 significant digits.

Ask by Beck Chen. in the United States
Jan 21,2025

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Answer

The volume of dinitrogen difluoride gas collected is approximately \( 10.6 \, \text{L} \).

Solution

To calculate the volume of dinitrogen difluoride gas collected, we can use the ideal gas equation: \[ PV = nRT \] where: - \( P \) is the pressure in atmospheres (atm), - \( V \) is the volume in liters (L), - \( n \) is the number of moles of gas, - \( R \) is the gas constant (\( 0.08206 \, \text{L atm/mol K} \)), - \( T \) is the temperature in Kelvin (K). Given: - Pressure (\( P \)) = 1 atm - Number of moles (\( n \)) = \( 490 \, \text{mmol} = 0.49 \, \text{mol} \) (converting mmol to mol) - Temperature (\( T \)) = \( -10.0^{\circ} \text{C} = 263.15 \, \text{K} \) (converting Celsius to Kelvin) We can rearrange the ideal gas equation to solve for volume (\( V \)): \[ V = \frac{nRT}{P} \] Substitute the given values into the equation to find the volume of dinitrogen difluoride gas collected. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(0.49\times 0.08206\times 263.15\right)}{1}\) - step1: Remove the parentheses: \(\frac{0.49\times 0.08206\times 263.15}{1}\) - step2: Multiply the terms: \(10.58110361\) The volume of dinitrogen difluoride gas collected is approximately \( 10.581 \, \text{L} \) when rounded to 3 significant digits.

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To calculate the volume of dinitrogen difluoride gas at \( -10.0^{\circ} \mathrm{C} \) using the ideal gas law, we can use the equation: \[ PV = nRT \] Where: - \( P \) = pressure (in atm) - \( V \) = volume (in liters) - \( n \) = number of moles of gas - \( R \) = ideal gas constant \( \approx 0.0821 \, \text{L·atm/(K·mol)} \) - \( T \) = temperature (in Kelvin) First, convert the temperature to Kelvin: \[ T = -10.0^{\circ} \mathrm{C} + 273.15 = 263.15 \, \mathrm{K} \] Next, convert mmol to moles: \[ 490 \, \mathrm{mmol} = 490 \times 10^{-3} \, \mathrm{mol} = 0.490 \, \mathrm{mol} \] Now, plug the values into the ideal gas law equation: \[ V = \frac{nRT}{P} \] \[ V = \frac{(0.490 \, \mathrm{mol}) (0.0821 \, \mathrm{L·atm/(K·mol)}) (263.15 \, \mathrm{K})}{1 \, \mathrm{atm}} \] Calculating this gives: \[ V = \frac{(0.490)(0.0821)(263.15)}{1} \] \[ V \approx \frac{10.0367}{1} \approx 10.04 \, \mathrm{L} \] Rounding to three significant digits: \[ \boxed{10.0 \, \mathrm{L}} \]

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