Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A \(40-1 b\) container of peat moss measures \(14 \times 20 \times 30\) in. A 40-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? (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
(a) The average density of peat moss is approximately 0.081 g/cm³, and the average density of topsoil is approximately 0.951 g/cm³. 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. (b) To cover an area measuring \(15.0 \mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of 3.0 in., we will need approximately 30 bags of peat moss.

Step by step solution

01

Convert given measurements

First, we need to convert all given measurements into appropriate metric units. We will convert inches to centimeters, pounds to grams, and gallons to cubic centimeters using the conversion factors: 1 inch = 2.54 cm, 1 lb = 453.6 g, and 1 gal = 3785.41 cm³. For the peat moss container (dimensions in inches): \(14 \times 20 \times 30\) in. In centimeters: \(14 \times 2.54 \times 20 \times 2.54 \times 30 \times 2.54\) cm³ For the peat moss mass (in pounds): 40.1 lb In grams: \(40.1 \times 453.6\) g For the topsoil container (volume in gallons): 1.9 gal In cubic centimeters: \(1.9 \times 3785.41\) cm³ For the topsoil mass (in pounds): 40 lb In grams: \(40 \times 453.6\) g
02

Calculate densities of peat moss and topsoil

Density is defined as mass per unit volume. Thus, we can calculate the density of both peat moss and topsoil using their mass and volume (in grams and cm³, respectively). Density of peat moss = \(\frac{\text{Mass of peat moss}}{\text{Volume of peat moss container}} = \frac{40.1 \times 453.6}{14 \times 2.54 \times 20 \times 2.54 \times 30 \times 2.54}\) g/cm³. Density of topsoil = \(\frac{\text{Mass of topsoil}}{\text{Volume of topsoil container}} = \frac{40 \times 453.6}{1.9 \times 3785.41}\) g/cm³.
03

Compare peat moss and topsoil densities

Now that we have calculated the average densities of peat moss and topsoil, we can compare them to determine if peat moss is lighter than topsoil: If the average density of peat moss is less than that of topsoil, then it is correct to say that peat moss is lighter than topsoil.
04

Calculate the number of bags required

To calculate the number of bags of peat moss required to cover an area of \(15.0 \mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of 3.0 in., we need to first convert the dimensions to appropriate metric units: 1 ft = 30.48 cm and 1 in = 2.54 cm Area = \(15.0 \times 30.48 \times 20.0 \times 30.48\) cm². Depth = \(3.0 \times 2.54\) cm. Now, we will multiply the area by the depth to find the volume required and then divide this volume by the volume of one bag of peat moss to find the number of bags: Volume required = Area × Depth Number of bags = \(\frac{\text{Volume required}}{\text{Volume of one bag of peat moss}}\). Now we can calculate the number of bags required to cover the given area at the specified depth.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Peat Moss
Peat moss is an organic material derived from decomposed plant matter, primarily sphagnum moss, which accumulates in damp, acidic bogs over thousands of years. It's typically used in gardening and agriculture due to its excellent moisture retention abilities and its capability to improve soil structure.

**Characteristics of Peat Moss**
Peat moss is renowned for its light and fluffy texture, which helps increase soil aeration and drainage. This can encourage healthier root growth in plants. It has an acidic nature, with a pH usually between 3.0 and 4.5, making it ideal for acid-loving plants such as blueberries and azaleas. Furthermore, peat moss is relatively sterile, preventing the introduction of harmful pathogens or weed seeds into garden soil.

**Usage of Peat Moss in Gardening**
  • Improves soil structure by adding organic matter.
  • Acts as a growing medium for germinating seeds.
  • Retains moisture, reducing the frequency of watering.
  • Acidifies alkaline soils for acid-loving plants.
Understanding its properties and measuring its density accurately, as in the given exercise, is crucial for predicting how much is needed for specific gardening tasks.
Topsoil
Topsoil is the uppermost layer of soil, rich in nutrients and organic matter, which supports plant life. Comprising around the top 5 to 10 inches of the soil profile, it is crucial for plant growth due to its nutrient content and its role in retaining moisture.

**Topsoil Characteristics**
Topsoil composition can vary widely depending on location, but it generally contains a mix of sand, silt, and clay, along with organic material that provides nutrients essential for plants. It's typically darker in color than the subsoil due to its organic matter content, including decomposed leaves and plants.

**Significance of Topsoil in Landscaping**
  • Provides necessary nutrients and minerals for plants.
  • Supports root systems in holding plants securely.
  • Facilitates good drainage while retaining beneficial moisture.
In the context of density calculations, like in the exercise, understanding the mass and volume of topsoil is key to determining its ability to support plant growth effectively.
Unit Conversion
Unit conversion is a fundamental skill in mathematics and science, allowing you to change measurements from one unit to another, ensuring consistency and accuracy within calculations. It involves the use of conversion factors, which are precise multipliers that relate one set of units to another.

**Common Unit Conversions in Volume and Mass**
In the given exercise, we see the transformation of measurements between imperial and metric systems. Here are some essential conversions used:
  • Inches to centimeters: 1 inch = 2.54 cm
  • Pounds to grams: 1 lb = 453.6 g
  • Gallons to cubic centimeters: 1 gallon = 3785.41 cm³
Being proficient in unit conversions is essential for ensuring that measurements are correctly interpreted across different systems.

**Importance in Practical Calculations**
In scientific calculations, even a slight inaccuracy can lead to significant errors, emphasizing the necessity of proper conversion. The given exercise illustrates how every element, from the peat moss and topsoil density calculations to area coverage, hinges on precise unit conversions to maintain consistency and accuracy in wildlife management, agriculture, and construction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) A sample of tetrachloroethylene, a liquid used in dry cleaning that is being phased out because of its potential to cause cancer, has a mass of \(40.55 \mathrm{~g}\) and a volume of \(25.0 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\). What is its density at this temperature? Will tetrachloroethylene float on water? (Materials that are less dense than water will float.) (b) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "supercritical fluid" that is a much safer dry-cleaning agent than tetrachloroethylene. At a certain pressure, the density of supercritical \(\mathrm{CO}_{2}\) is \(0.469 \mathrm{~g} / \mathrm{cm}^{3}\). What is the mass of a \(25.0-\mathrm{mL}\) sample of supercritical \(\mathrm{CO}_{2}\) at this pressure?

Make the following conversions: (a) \(83^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) (b) \(29^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\) (c) \(294^{\circ} \mathrm{C}\) to \(\mathrm{K}\) (d) \(832 \mathrm{~K}\) to \({ }^{\circ} \mathrm{C}\) (f) \(35^{\circ} \mathrm{F}\) to \(\mathrm{K}\). (e) \(721 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\)

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 metric prefix.

Using your knowledge of metric units, English units, and the information on the back inside cover, write down the conversion factors needed to convert (a) \(\mathrm{km} / \mathrm{hr}\) to \(\mathrm{m} / \mathrm{s}\) (b) \(\mathrm{mL}\) to \(\mu \mathrm{L}(\mathbf{c}) \mathrm{ps}\) to \(\mathrm{s}(\mathbf{d}) \mathrm{m}^{3}\) to gal.

A solid white substance \(A\) is heated strongly in the absence of air. It decomposes to form a new white substance \(\mathrm{B}\) and a gas C. The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids A and \(\mathrm{B}\) and gas \(\mathrm{C}\) are elements or compounds?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free