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Chapter 1: Problem 74

A \(40-\mathrm{lb}\) container of peat moss measures \(14 \times 20 \times 30 \mathrm{in}\). A \(40-\mathrm{lb}\) container of topsoil has a volume of 1.9 gal. (a) Calculate the average densities of peat moss and topsoil in units of \(\mathrm{g} / \mathrm{cm}^{3}\). Would it be correct to say that peat moss is "lighter" than topsoil? Explain. (b) How many bags of peat moss are needed to cover an area measuring \(15.0 \mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of 3.0 in.?

Short Answer

Expert verified
The average densities of peat moss and topsoil are 0.0851 g/cm鲁 and 0.2514 g/cm鲁, respectively. Since the density of peat moss is less than the density of topsoil, it is correct to say that peat moss is "lighter" than topsoil. To cover an area of 15.0 ft 脳 20.0 ft to a depth of 3.0 inches, approximately 4 bags of peat moss are required.

Step by step solution

01

Convert units to match

We need to convert the volumes of peat moss and topsoil to the same unit to find their densities. Let's convert the volume of the peat moss container from inches to centimeters and the volume of the topsoil container from gallons to cubic centimeters. 1 inch = 2.54 cm 1 gallon = 3,785.41 cm鲁
02

Calculate the volume of the peat moss container

Using the given dimensions for the peat moss container, we can find the volume by multiplying the length, width, and height. Volume = Length 脳 Width 脳 Height Volume = \(14\:\text{in}\) 脳 \(20\:\text{in}\) 脳 \(30\:\text{in}\) Volume = \(14\:\text{in}\) 脳 \(20\:\text{in}\) 脳 \(30\:\text{in}\) 脳 \((2.54\:\mathrm{cm/in})^3\) Volume = 21282.24 cm鲁
03

Calculate the volume of the topsoil container

We will now convert the given volume of the topsoil container to cubic centimeters. Volume = 1.9 gal 脳 3,785.41 cm鲁/gal Volume = 7192.279 cm鲁
04

Calculate the densities of peat moss and topsoil

To calculate the densities of peat moss and topsoil, we will use the formula: Density = Mass / Volume (a) Peat moss: Density = \(40\:\text{lb}\) 脳 453.592 g/lb 梅 21282.24 cm鲁 Density = 0.0851 g/cm鲁 (b) Topsoil: Density = \(40\:\text{lb}\) 脳 453.592 g/lb 梅 7192.279 cm鲁 Density = 0.2514 g/cm鲁
05

Compare the densities

Now, we can compare the densities and determine if it is correct to say that peat moss is lighter than topsoil: Density of peat moss = 0.0851 g/cm鲁 Density of topsoil = 0.2514 g/cm鲁 Since the density of peat moss is less than the density of topsoil, it is correct to say that peat moss is "lighter" than topsoil.
06

Calculate the number of bags of peat moss required

To find the number of bags of peat moss needed to cover the given area to a certain depth, we first need to calculate the volume of peat moss needed. Area = Length 脳 Width = \(15.0\:\text{ft}\) 脳 \(20.0\:\text{ft} = 300\:\text{ft}^2\) Depth = 3.0 in = 3.0 in 脳 2.54 cm/in = 7.62 cm Volume_needed = Area 脳 Depth Volume_needed = 300 ft虏 脳 7.62 cm 脳 \((30.48\:\mathrm{cm/ft})^2\) Volume_needed = 65776.32 cm鲁 Now, we will divide the volume needed by the volume of one bag of peat moss to obtain the number of bags required. Number_of_bags = Volume_needed / Volume_of_one_bag Number_of_bags = 65776.32 cm鲁 梅 21282.24 cm鲁 Number_of_bags 鈮� 3.09 As we can't use a fraction of a bag, we will round up to find that 4 bags of peat moss are required to cover the specified area to a depth of 3.0 inches.

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Key Concepts

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

Unit Conversion in Chemistry
In chemistry, unit conversion is a fundamental skill that enables scientists and students to compare measurements and perform calculations with different units. It often involves converting units of mass, volume, and density to a common system so that they can be compared accurately. For instance, the exercise presented required converting the volume of peat moss from inches to centimeters and the volume of topsoil from gallons to cubic centimeters.

To make such conversions accurate and easy to understand, it's essential to know the conversion factors, such as:
  • 1 inch = 2.54 cm
  • 1 gallon = 3,785.41 cm鲁
Additionally, remembering that conversions can be performed by multiplying the initial measurement by the conversion factor helps ensure that units will be consistent across different measurements, allowing for correct density calculations.
Volume Calculation
Volume is the amount of space that a substance or object occupies, and it is a crucial parameter in density calculations. To calculate the volume of solid objects, like the container of peat moss, we typically use the formula for the volume of a rectangular prism:
Volume = Length 脳 Width 脳 Height
Using the exercise's measurements, the volume was calculated by multiplying the container's length, width, and height. Converting these measurements from inches to the standard unit of centimeters in chemistry (cm鲁) involved raising the conversion factor (2.54 cm/in) to the third power, since volume is a three-dimensional measurement.

For liquids, such as the topsoil in gallons, we use conversion factors to translate gallons into cubic centimeters, as different materials are often compared in a common unit of measure in scientific calculations.
Mass to Volume Ratio
The mass to volume ratio, more commonly known as density, is a characteristic property of a material and a critical concept in many fields of science, including chemistry. Density is defined as the mass of the material divided by its volume, and it is usually expressed in grams per cubic centimeter (g/cm鲁) in the metric system.

In the given exercise, the density of peat moss and topsoil was calculated using the formula:
Density = Mass / Volume
First, the mass of each container was converted to grams (g), and then the volume was calculated in cubic centimeters (cm鲁) as described in the previous sections. These values were then used to find the densities. The comparison of the densities of peat moss and topsoil allowed us to determine that peat moss is 'lighter' since it has a lower density, meaning less mass per unit of volume.

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Most popular questions from this chapter

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0 -mL portion of the liquid had a mass of \(21.95 \mathrm{~g}\). A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) ? (The volume of a sphere is \((4 / 3) \pi r^{3}\) where \(r\) is the radius.)

(a) To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a \(45-\mathrm{mL}\) sample of the substance. She then measured the mass of the sample, finding that it weighed \(38.5 \mathrm{~g}\). She knew that the substance had to be either isopropyl alcohol (density \(0.785 \mathrm{~g} / \mathrm{mL})\) or toluene (density \(0.866 / \mathrm{mL}\) ). What are the calculated density and the probable identity of the substance? (b) An experiment requires \(45.0 \mathrm{~g}\) of ethylene glycol, a liquid whose density is \(1.114 \mathrm{~g} / \mathrm{mL}\). Rather than weigh the sample on a balance, a chemist chooses to dispense the liquid using a graduated cylinder. What volume of the liquid should he use? (c) A cubic piece of metal measures \(5.00 \mathrm{~cm}\) on each edge. If the metal is nickel, whose density is \(8.90 \mathrm{~g} / \mathrm{cm}^{3},\) what is the mass of the cube?

Use appropriate metric prefixes to write the following measurements without use of exponents: (a) \(2.3 \times 10^{-10} \mathrm{~L}\) (b) \(4.7 \times 10^{-6} \mathrm{~g}\), (c) \(1.85 \times 10^{-12} \mathrm{~m}\) (d) \(16.7 \times 10^{6} \mathrm{~s}\); (e) \(15.7 \times 10^{3} \mathrm{~g}\) (f) \(1.34 \times 10^{-3} \mathrm{~m},(\mathrm{~g}) 1.84 \times 10^{2} \mathrm{~cm}\)

Silicon for computer chips is grown in large cylinders called "boules" that are \(300 \mathrm{~mm}\) in diameter and \(2 \mathrm{~m}\) in height. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3}\). Silicon wafers for making integrated circuits are sliced from a \(2.0 \mathrm{~m}\) boule and are typically \(0.75 \mathrm{~mm}\) thick and \(300 \mathrm{~mm}\) in diameter. (a) How many wafers can be cut from a single boule? (b) What is the mass of a silicon wafer? (The volume of a cylinder is given by \(\pi r^{2} h\), where \(r\) is the radius and \(h\) is its height.)

When you convert units, how do you decide which part of the conversion factor is in the numerator and which is in the denominator? [Section 1.6]
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