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Ch 18: Thermal Properties of Matter
Chapter 18, Problem 18

A 20.0-L tank contains 4.86 * 10^-4 kg of helium at 18.0°C. The molar mass of helium is 4.00 g/mol. (a) How many moles of helium are in the tank?

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1
Convert the mass of helium from kilograms to grams. Since 1 kg equals 1000 grams, multiply the mass of helium in kilograms by 1000 to get the mass in grams.
Calculate the number of moles of helium using the formula: number of moles = mass of substance (g) / molar mass (g/mol).
Substitute the mass of helium in grams and the molar mass of helium into the formula to find the number of moles.
Ensure the units are consistent when performing the calculations to avoid any errors in the final result.
Interpret the result, which gives the amount of helium in the tank in terms of moles, a standard unit in chemistry for quantifying the amount of substance.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moles and Molar Mass

A mole is a unit in chemistry that represents a specific quantity of particles, typically atoms or molecules, equivalent to Avogadro's number (approximately 6.022 x 10^23). Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). For helium, the molar mass is 4.00 g/mol, meaning one mole of helium weighs 4.00 grams.
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Conversion of Mass to Moles

To find the number of moles from a given mass, the formula used is: moles = mass (g) / molar mass (g/mol). This relationship allows us to convert the mass of a substance into moles, which is essential for stoichiometric calculations in chemistry. In this case, the mass of helium must be converted from kilograms to grams before applying the formula.
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Ideal Gas Law Context

While the question specifically asks for moles, understanding the Ideal Gas Law (PV = nRT) is beneficial in gas-related problems. This law relates pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas, providing a framework for understanding gas behavior under various conditions. Although not directly needed for this calculation, it contextualizes the properties of gases.
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