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The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(4.5 \mathrm{~m} \times 5.0 \mathrm{~m} \times 2.5 \mathrm{~m}\) ?

Short Answer

Expert verified
The mass of the air in the room is 66.9375 kg.

Step by step solution

01

Calculate the Volume of the Room

First, determine the volume of the room by multiplying its length, width, and height. The formula for calculating the volume of a rectangular prism is:\[V = ext{length} imes ext{width} imes ext{height}\]Substitute the given dimensions:\[V = 4.5 \, \text{m} \times 5.0 \, \text{m} \times 2.5 \, \text{m} = 56.25 \, \text{m}^3\]
02

Convert Room Volume to Liters

Convert the volume from cubic meters to liters since the density is given in grams per liter. Remember that:\[1 \, \text{m}^3 = 1000 \, \text{L}\]So:\[56.25 \, \text{m}^3 = 56.25 \, \text{m}^3 \times 1000 \, \text{L/m}^3 = 56250 \, \text{L}\]
03

Use Density to Find Mass in Grams

Using the density of air, calculate the mass of air in the room in grams. The formula to calculate mass from density is:\[\text{mass} = \text{density} \times \text{volume}\]Substitute the given density and the volume in liters:\[\text{mass} = 1.19 \, \text{g/L} \times 56250 \, \text{L} = 66937.5 \, \text{g}\]
04

Convert Mass to Kilograms

Convert the mass from grams to kilograms as the final answer requires kilograms. Knowing:\[1 \, \text{kg} = 1000 \, \text{g}\]Thus:\[66937.5 \, \text{g} = \frac{66937.5}{1000} \, \text{kg} = 66.9375 \, \text{kg}\]

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Key Concepts

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

Volume Conversion
Understanding volume conversion is fundamental when performing calculations involving density. When you are given a volume in cubic meters (m³) and need to work with liters (L), it is crucial to remember the conversion factor between these units. Since density can often be provided in grams per liter (g/L), converting cubic meters to liters allows you to easily use the density in your calculations.
Whenever you encounter a measurement in cubic meters, convert it to liters by multiplying the volume in cubic meters by 1000. Remember that:
  • 1 cubic meter (1 m³) = 1000 liters (L)
For instance, if the volume of a room is 56.25 m³, you convert this to liters by calculating: \[56.25 \, \text{m}^3 \times 1000 \, \text{L/m}^3 = 56250 \, \text{L} \]This step aligns the volume measurement to the density unit, setting the stage for mass calculation in an easily manageable form.
Mass Calculation
The mass of a substance can be easily calculated once you know both its volume and density. After converting your volume into the proper units, the next step is to use the density in the calculation. The formula to find mass when you have volume and density is: \[\text{mass} = \text{density} \times \text{volume}\]For example, if you know the density of air is 1.19 g/L and the volume of the room is 56250 L as derived before, you can calculate the mass of air contained in the room in grams:\[\text{mass} = 1.19 \, \text{g/L} \times 56250 \, \text{L} = 66937.5 \, \text{g}\]After finding the mass in grams, it often needs to be converted into different units such as kilograms for a suitable presentation or further computation.
Unit Conversion
Unit conversion is a crucial skill in problem-solving, especially when different measurement units are involved. Converting units ensures that all parts of the calculation are compatible.
When calculating mass, you might first find the answer in grams, but if the problem requires an answer in kilograms, you need to convert it. The conversion between grams and kilograms is straightforward:
  • 1 kilogram (kg) = 1000 grams (g)
To convert mass from grams to kilograms, divide the mass value in grams by 1000. For example:\[66937.5 \, \text{g} = \frac{66937.5}{1000} \, \text{kg} = 66.9375 \, \text{kg}\]This simple conversion allows you to represent the mass in a more customary unit which might be required by the question or desired for simplification.

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Most popular questions from this chapter

Two students determine the percentage of lead in a sample as a laboratory exercise. The true percentage is \(22.52 \% .\) The students' results for three determinations are as follows: (1) 22.52,22.48,22.54 (2) 22.64,22.58,22.62 (a) Calculate the average percentage for each set of data and state which set is the more accurate based on the average. (b) Precision can be judged by examining the average of the deviations from the average value for that data set. (Calculate the average value for each data set; then calculate the average value of the absolute deviations of each measurement from the average.) Which set is more precise?

Convert the following expressions into exponential notation: (a) 3 terameters (tm) (b) 2.5 femtoseconds (fs) (c) 57 micrometers \((\mu m)\) (d) 8.3 megagrams (mg).

If on a certain year, an estimated amount of 4 million metric tons (1 metric ton \(=1000 \mathrm{~kg}\) ) of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) was emitted worldwide due to agricultural activities, express this mass of \(\mathrm{N}_{2} \mathrm{O}\) in grams without exponential notation, using an appropriate mètric préfix.

Identify each of the following as measurements of length, area, volume, mass, density, time, or temperature: (a) \(25 \mathrm{ps}\), (b) \(374.2 \mathrm{mg}\), (c) \(77 \mathrm{~K}\), (d) \(100,000 \mathrm{~km}^{2}\), (e) \(1.06 \mu \mathrm{m}\), (f) \(16 \mathrm{nm}^{2},(\mathbf{g})-78^{\circ} \mathrm{C},(\mathbf{h}) 2.56 \mathrm{~g} / \mathrm{cm}^{3}\), (i) \(28 \mathrm{~cm}^{3}\).

Label each of the following as either a physical process or a chemical process: (a) crushing a metal can, (b) production of urine in the kidneys, \((\mathbf{c})\) melting a piece of chocolate, \((\mathbf{d})\) burning fossil fuel, (e) discharging a battery.

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