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A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of 1.78 torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h\).)

Short Answer

Expert verified
The neon sign contains approximately \(0.0069\,\mathrm{g}\) of neon.

Step by step solution

01

Find the volume of the glass tubing

The volume of a cylinder can be calculated using the formula: \[V = \pi r^2h\] where V is the volume, r is the radius of the tube, and h is the length of the tube. Given that the inside diameter of the tube is \(2.5\,\mathrm{cm}\), we need to find the radius. The radius (r) can be calculated by dividing the diameter by 2: \[r = \frac{2.5}{2} \,\mathrm{cm} = 1.25\,\mathrm{cm}\] Now, convert the length of the tube and the radius from centimeters to meters: \[ r = 1.25 \,\mathrm{cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.0125 \mathrm{~m} \] \[ h = 5.5 \,\mathrm{m} \] Now, we can calculate the volume (V) of the tube using the given formula: \[ V = \pi (0.0125 \,\mathrm{m})^{2}(5.5\,\mathrm{m}) \approx 0.0027 \,\mathrm{m^3} \]
02

Apply the Ideal Gas Law to find the moles of neon

The Ideal Gas Law equation is given as: \[PV = nRT\] Where P is the pressure, V is the volume, n is the moles of the gas, R is the universal gas constant, and T is the temperature in Kelvins. We are given the pressure in torr, and it should be convered to atmosphere (atm): \[ P = 1.78\,\mathrm{torr} \times \frac{1\, \mathrm{atm}}{760\, \mathrm{torr}} \approx 0.00234\, \mathrm{atm} \] Next, we need to convert the temperature from Celsius to Kelvin: \[ T = 35^\circ\text{C} + 273.15\text{K} = 308.15\text{K} \] Now, we can rearrange the Ideal Gas Law equation to find the moles of neon (n): \[ n = \frac{PV}{RT} \] We will use the universal gas constant R (R = 0.0821 L atm/mol K) in our calculations: \[ n = \frac{(0.00234\, \mathrm{atm})(0.0027\,\mathrm{m^3} \times 1000\,\mathrm{L/m^3})}{(0.0821\, \mathrm{L\, atm/mol\, K})(308.15\,\mathrm{K})} \approx 3.42 \times 10^{-4}\, \mathrm{mol} \]
03

Convert moles of neon to grams

To convert moles of neon into grams, we can use the molar mass of neon, which is 20.18 g/mol. We can use the equation: Mass = moles × molar mass \[ \text{Mass}_\text{Neon} = (3.42 \times 10^{-4} \,\mathrm{mol}) (20.18\,\mathrm{g/mol}) \approx 0.0069\,\mathrm{g} \] Therefore, there are approximately 0.0069 grams of neon inside the glass tubing.

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Key Concepts

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

Gas Law Calculations
Understanding gas law calculations is crucial for solving problems that involve the behavior of gases under different conditions. This is exemplified by the task of determining the mass of neon in a lit sign, which is a practical application of the Ideal Gas Law. The Ideal Gas Law is an equation that relates the pressure (P), volume (V), temperature (T), and amount of substance in moles (n) for an ideal gas.

The mathematical form of the Ideal Gas Law is expressed as: \[PV = nRT\]Here, R represents the ideal, or universal, gas constant, which has a value of 0.0821 L atm/mol K. This constant allows us to use standard pressure and temperature units when working with gases. By mastering gas law calculations, one can predict how changing one variable, like pressure or temperature, will affect the others, assuming the gas acts ideally. This is pivotal in fields ranging from meteorology to engineering.

To solve our particular problem, it was necessary to first determine the volume of the neon gas, which was contained in a cylindrical tube, and then apply the Ideal Gas Law to find the number of moles of neon. With each step, conversions were appropriately made to ensure that the units matched those needed for the constant R. This exercise wonderfully illustrates the interconnectedness of the gas properties and the importance of unit consistency in calculations.
Mole Concept
The mole concept is a fundamental principle in chemistry that allows chemists to work with the submicroscopic world of atoms and molecules on a macroscopic level. One mole is defined as exactly 6.02214076×10²³ elementary entities (this number is known as Avogadro's number) and is used to express the amount of a chemical substance.

By using the mole concept, one can relate the mass of a substance to the number of particles it contains. In the case of the neon in the sign, we used the molar mass of neon (20.18 g/mol) to convert moles to grams:\[\text{Mass}_\text{Neon} = (n_\text{moles}) (M_\text{molar})\]This formula is essential as it allows for the mass-to-mole conversion, which is a critical step in relating the chemical behavior of a substance to its mass. Knowing how to apply the mole concept helps bridge the gap between the tangible and intangible aspects of chemical substances.
Chemical Quantities
Chemical quantities are vital to describe the quantitative relationships in chemical reactions and processes. These include mass, volume, concentration, and the amount of substance, which in the context of the mole concept, is measured in moles. Accurate measurements and conversions of these quantities are essential in stoichiometry, the part of chemistry that involves calculating the quantities of reactants and products in a chemical reaction.

In the exercise provided, after finding the number of moles of neon, we needed to accurately convert this quantity into a mass using neon's molar mass. Molar mass links the mass of a substance to the amount in moles, and is expressed in grams per mole (g/mol). This step demonstrates how a seemingly abstract number of particles can translate into a tangible mass that can be measured. Understanding chemical quantities within these calculations is not only critical in laboratory and industrial settings but it is also a key component in environmental science, pharmaceuticals, and material design.

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Most popular questions from this chapter

A piece of dry ice (solid carbon dioxide) with a mass of \(5.50 \mathrm{~g}\) is placed in a 10.0 - \(\mathrm{L}\) vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24{ }^{\circ} \mathrm{C} ?\)

Perform the following conversions: (a) 0.912 atm to torr, (b) 0.685 bar to kilopascals, (c) \(655 \mathrm{~mm}\) Hg to atmospheres, (d) \(1.323 \times 10^{5}\) Pa to atmospheres, (e) 2.50 atm to psi.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164{ }^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23{ }^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in \(\mathrm{kPa},\) if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

A 4.00 -g sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO},\) forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3} .\) When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

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