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Calculate the pressure exerted by \(1.55 \mathrm{~g}\) Xe gas at \(20 .{ }^{\circ} \mathrm{C}\) in a sealed \(560-\mathrm{mL}\) flask.

Short Answer

Expert verified
The pressure is approximately 0.510 atm.

Step by step solution

01

Convert temperature to Kelvin

Convert the given temperature from degrees Celsius to Kelvin using the formula: \[T(K) = T(°C) + 273.15\]Given that, \(T = 20°\)C, so the temperature in Kelvin is:\[T(K) = 20 + 273.15 = 293.15\text{ K}\]
02

Convert mass to moles

Use the molar mass of xenon (Xe), which is approximately 131.29 g/mol, to convert mass to moles using the formula:\[n = \frac{m}{M}\]where \(m\) is the mass and \(M\) is the molar mass. So we have:\[n = \frac{1.55}{131.29} \approx 0.0118 \text{ moles}\]
03

Convert volume from mL to L

Convert the given volume from milliliters to liters using the conversion:\[1 \text{ L} = 1000 \text{ mL}\]Thus, the volume is:\[V = \frac{560}{1000} = 0.560 \text{ L}\]
04

Use the Ideal Gas Law to calculate pressure

The Ideal Gas Law is given by:\[PV = nRT\]where:- \(P\) is the pressure,- \(V\) is the volume,- \(n\) is the moles of gas,- \(R\) is the ideal gas constant, approximately \(0.0821\text{ L atm K}^{-1}\text{ mol}^{-1}\),- \(T\) is the temperature in Kelvin.Rearranging for pressure \(P\), we get:\[P = \frac{nRT}{V}\]Plugging in the known values:\[P = \frac{0.0118 \times 0.0821 \times 293.15}{0.560} \approx 0.510 \text{ atm}\]
05

Check the calculation

Double-check the calculations to ensure all conversions and operations were performed correctly:1. Temperature: 293.15 K (converted correctly)2. Moles: 0.0118 moles (using \(131.29\) g/mol for Xe)3. Volume: 0.56 L4. Ideal Gas Law: \[P = \frac{(0.0118 \text{ mol}) \times (0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}) \times (293.15 \text{ K})}{0.56 \text{ L}} = 0.510 \text{ atm}\]Ensure all steps and inputs align, confirming that the final answer is consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Pressure calculation is an essential step in many chemistry problems, especially when dealing with gases. The pressure of a gas is typically measured in units like atmospheres (atm), which reflects the force that the gas particles exert when they collide with the walls of their container. To find the pressure of a gas using the Ideal Gas Law, you use the formula: \[P = \frac{nRT}{V}\]where:- **\(P\)** is the pressure,- **\(n\)** is the number of moles of the gas,- **\(R\)** is the ideal gas constant, - **\(T\)** is the temperature in Kelvin,- **\(V\)** is the volume of the gas in liters. For example, if you have a known amount of gas in a specific volume at a given temperature, you can calculate the pressure exerted by the gas in a sealed container. This is done by rearranging the Ideal Gas Law formula to solve for pressure, \(P\). Always remember to:- Ensure all units are consistent (e.g., convert volumes to liters and temperatures to Kelvin).- Double check your calculations for accuracy.By following these steps, you can successfully determine the pressure in various chemical contexts.
Temperature Conversion
Temperature conversion is crucial when working with gas laws. Most of the equations, including the Ideal Gas Law, require temperatures to be in Kelvin. The Kelvin scale is an absolute temperature scale commonly used in science because it starts at absolute zero, the lowest possible temperature. To convert a temperature from degrees Celsius to Kelvin, use the formula:\[T(\text{K}) = T(\text{°C}) + 273.15\]This simple addition accounts for the difference between the Celsius scale and the Kelvin scale. For example, to convert \(20 \text{°C}\) to Kelvin:- Add 273.15 to get \(293.15 \text{ K}\).
  • This step is necessary for accurate pressure and volume calculations in gas law problems.
  • Failing to convert temperatures to Kelvin can lead to significant errors in your calculations.
By remembering this conversion, you can avoid common pitfalls in temperature-related calculations.
Moles Calculation
Calculating moles is a key step when working with reactions and equations in chemistry. The number of moles represents a specific quantity of a substance and allows you to relate the mass of a substance to its molecular characteristics. To calculate moles from mass, use the formula:\[n = \frac{m}{M}\]where:- **\(n\)** is the number of moles,- **\(m\)** is the mass in grams,- **\(M\)** is the molar mass in grams per mole.For instance, if you have \(1.55 \text{ g}\) of xenon with a molar mass of approximately \(131.29 \text{ g/mol}\), the calculation is:\[n = \frac{1.55}{131.29} \approx 0.0118 \text{ moles}\]
  • Make sure to use the correct molar mass for the substance you're working with.
  • Convert the substance's mass to moles to enable further calculations, such as pressure or volume estimations.
Understanding how to perform moles calculations ensures successful problem-solving in gas laws and stoichiometry.

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Most popular questions from this chapter

Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O},\) is a volatile organic compound that is sometimes released from insulation used in home construction, and it can be trapped and build up inside the home. When this happens, people exposed to the formaldehyde can suffer adverse health effects. The U. S. National Institute of Occupational Health and Safety (NIOSH) guideline for the maximum allowable concentration of formaldehyde in air in the workplace is \(16 \mathrm{ppb}\) (parts per billion) for an eight-hour average exposure. (a) Determine the partial pressure of formaldehyde at the maximum allowable level of \(16 \mathrm{ppb}\). (b) Calculate how many molecules of formaldehyde are present in each cubic centimeter of air when formaldehyde is present at \(16 \mathrm{ppb}\). (c) Calculate how many total molecules of formaldehyde are present in a room: \(15.0 \mathrm{ft}\) long \(\times 10.0 \mathrm{ft}\) wide \(X\) \(8.00 \mathrm{ft}\) high (at \(16 \mathrm{ppb}\) ).

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Explain the major roles played by nitrogen in the atmosphere. Do the same for oxygen.

A sample of gas at \(30 .{ }^{\circ} \mathrm{C}\) has a pressure of \(2.0 \mathrm{~atm}\) in a sealed 1.0 - \(\mathrm{L}\) container. Calculate the pressure it will exert in a 4.0 -L container. The temperature does not change.

At high temperatures and low pressures, gases behave ideally, but as the pressure is increased the product \(P V\) becomes greater than the product \(n R T\). Give a molecularlevel explanation of this fact.

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