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What is the weight of 1,000 cubic feet of air at STP?

Short Answer

Expert verified
The weight of 1,000 cubic feet of air at STP is approximately 36.53 kg.

Step by step solution

01

Convert volume to cubic meters

Given volume is 1,000 cubic feet. To convert cubic feet to cubic meters, we use the following conversion factor: 1 cubic foot = 0.0283168 cubic meter. So, 1,000 cubic feet = 1,000 × 0.0283168 m³.
02

Calculate volume of air in cubic meters

Calculate the volume in cubic meters: 1,000 cubic feet * 0.0283168 (m³/ft³) = 28.3168 m³.
03

Use density to find the mass of air

We are given the density of air at STP: 1.29 kg/m³. To find the mass of the air, we use the formula mass = density × volume: mass = 1.29 kg/m³ × 28.3168 m³.
04

Calculate mass of air

Calculate the mass of air: mass = 1.29 kg/m³ × 28.3168 m³ = 36.5288 kg. So, the weight of 1,000 cubic feet of air at STP is approximately 36.53 kg.

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Key Concepts

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

Conversion of Volume Units
One of the fundamental skills in chemistry and related fields is being able to convert between various units of measurement. Volume, which is the amount of space an object or substance occupies, can be measured in different units such as liters, cubic meters, and cubic feet. A common conversion that students encounter is changing cubic feet to cubic meters as many scientific calculations are done using the metric system.

For example, to convert the volume from cubic feet to cubic meters, you use the conversion factor 0.0283168 cubic meters per cubic foot. This factor is derived from the definition of a cubic foot, based on the International foot, which is exactly 0.3048 meters long. Therefore, a cubic foot is \(0.3048^3\) cubic meters. In practice, to convert volume, you multiply the number of cubic feet by 0.0283168 to obtain the volume in cubic meters. Understanding this conversion is crucial when working with standard temperature and pressure (STP) conditions in chemistry, as it ensures accuracy in calculations.
Density and Mass Relationship
Density is a measure of how much mass is contained in a given volume and is usually expressed in kilograms per cubic meter (kg/m³) in the International System of Units (SI). In chemistry and physics, understanding the relationship between density and mass is essential when trying to calculate the mass of a substance if its volume is known, or vice versa.

The formula connecting these two properties is as simple as it is profound: \( \text{mass} = \text{density} \times \text{volume} \). This implies that if you know the density of a substance and you have a volume of it, you can determine the mass simply by multiplication. Conversely, if the mass and the volume are known, the density can be calculated by dividing the mass by the volume. The direct proportionality between density and mass makes it a straightforward calculation, but one must never forget to use consistent units to avoid errors in the results.
Calculating Mass of Air
Air, like all substances, has mass and density. The density of air can vary depending on temperature, pressure, and humidity, but at STP — which is defined as 0°C temperature and 1 atmosphere pressure — the density of air is typically about 1.29 kg/m³. When it comes to calculating the mass of air, this standard density value is extremely useful.

Applying the density and mass relationship, the mass of a sample of air can be calculated if its volume is known. As demonstrated in the given exercise, the mass of air (in kilograms) can be calculated by multiplying the volume of air (in cubic meters) by the density of air at STP. This allows for the determination of the weight of a particular amount of air under specified conditions, which is a common task in various scientific calculations involving gases. This straightforward calculation is beneficial for solving practical problems in environmental science, engineering, and other disciplines where the properties of air are relevant.

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Most popular questions from this chapter

Balance the equations: (a) \(\mathrm{Ag}_{2} \mathrm{O} \rightarrow \mathrm{Ag}+\mathrm{O}_{2}\) (b) \(\mathrm{Zn}+\mathrm{HCl}+\mathrm{ZnCl}_{2}+\mathrm{H}_{2}\); (c) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\).

Heating of \(\mathrm{NaNO}_{3}\) decomposed it to \(\mathrm{NaNO}_{2}\) and \(\mathrm{O}_{2}\). How much \(\mathrm{NaNO}_{3}\) would you have to decompose to produce \(1.50 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\)

Baking powder consists of a mixture of cream of tartar (potassium hydrogen tartrate, \(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), molecular weight \(=188 \mathrm{~g} /\) mole \()\) and baking soda (sodium bicarbonate, \(\mathrm{NaHC} 0_{3}\), molecular weight \(=84 \mathrm{~g} /\) mole \()\). These two components react according to the equation \(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}+\mathrm{NaHCO}_{3} \rightarrow \mathrm{KNaC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) How much baking soda must be added to \(8.0 \mathrm{~g}\) of cream of tartar for both materials to react completely?

A silicious rock contains the mineral \(\mathrm{ZnS}\). To analyze for \(\mathrm{Zn}\), a sample of the rock is pulverized and treated with \(\mathrm{HCl}\) to dissolve the ZnS (silicious matter is insoluable). Zinc is precipitated from solution by the addition of potassium ferrocyanide \(\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6}\). After filtering, the precipitate is dried and weighed. The reactions which occur are $$ \begin{gathered} \mathrm{ZnS}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \mathrm{~S} \\ 2 \mathrm{ZnCl}_{2}+\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6} \rightarrow \mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}+4 \mathrm{KCl} \end{gathered} $$ If a 2 gram sample of rock yields \(0.969\) gram of \(\mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}\), what is the percentage of \(\mathrm{Zn}\) in the sample? Atomic weight \(\mathrm{Zn}=65.4\), molecular weight \(\mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}=342.6\).

Lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right.\), molecular weight \(=30 \mathrm{~g} / \mathrm{mole}\) ) reacts with water \(\left(\mathrm{H}_{2} \mathrm{O}\right.\), molecular weight \(=18 \mathrm{~g} /\) mole, density \(=1.0\) \(\mathrm{g} / \mathrm{cm}^{3}\) ) to produce lithium hydroxide ( \(\mathrm{LiOH}\) ) according to the following reaction: \(\mathrm{Li}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{LiOH}\). What mass of \(\mathrm{Li}_{2} \mathrm{O}\) is required to completely react with 24 liters of \(\mathrm{H}_{2} \mathrm{O} ?\)

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