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A die has an edge length of \(1.5 \mathrm{~cm}\). (a) What is the volume of one mole of such dice? (b) Assuming that the mole of dice could be packed in such a way that they were in contact with one another, forming stacking layers covering the entire surface of Earth, calculate the height in meters the layers would extend outward. [The radius \((r)\) of Earth is \(6371 \mathrm{~km}\), and the area of a sphere is \(4 \pi r^{2}\).]

Short Answer

Expert verified
(a) Volume: \(2.03 \times 10^{18} \text{ m}^3\); (b) Height: \(506 \text{ m}\).

Step by step solution

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01

Calculate the volume of a single die

The volume of a cube is given by the formula \( V = a^3 \), where \( a \) is the edge length. Given that the edge length of the die is \( 1.5 \text{ cm} \), the volume is \( V = (1.5)^3 \).
02

Convert the volume to cubic meters

Convert the volume from cubic centimeters to cubic meters by dividing by \( 10^6 \). If the volume in cubic centimeters is \( 3.375 \text{ cm}^3 \), then in cubic meters it is \( 3.375 \times 10^{-6} \text{ m}^3 \).
03

Calculate the volume of one mole of dice

One mole of dice contains Avogadro's number of dice, which is \( 6.022 \times 10^{23} \). Therefore, the volume of one mole of dice is \( 6.022 \times 10^{23} \times 3.375 \times 10^{-6} \text{ m}^3 \).
04

Calculate the surface area of Earth

The surface area of a sphere is given by \( 4 \pi r^2 \). Given \( r = 6371 \text{ km} = 6,371,000 \text{ m} \), the surface area is \( 4 \pi (6371000)^2 \text{ m}^2 \).
05

Calculate the height of the dice stack

To find the height, divide the total volume of the dice by the surface area of Earth. The formula becomes \( \text{Height} = \frac{6.022 \times 10^{23} \times 3.375 \times 10^{-6}}{4 \pi (6371000)^2} \).
06

Compute the final height and volume

Perform the calculations in the prior steps to find the volume of one mole of dice and the height of the dice stack on Earth's surface. The calculated height should be in meters.

Key Concepts

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

Volume Calculation
To understand how to calculate the volume of an object, we need to know its shape. Here, we are dealing with a cube, which is a simple 3D shape.
The volume - is the amount of space inside a three-dimensional object.
- is calculated using measured dimensions like length, width, and height. For a cube, the volume is given by the formula:\[ V = a^3 \]where \( a \) is the length of one edge of the cube. In the case of our exercise, each die has an edge of \(1.5\, \text{cm}\), so:\[ V = (1.5)^3 \text{ cm}^3 = 3.375 \text{ cm}^3 \] This calculation gives us the volume of a single die in cubic centimeters.
Cube Geometry
Understanding the geometry of a cube can simplify calculating its properties like volume and surface area. A cube is a special case of a rectangular prism where all edges are equal.Key Characteristics of a Cube:- All faces are square and identical.- It has 12 edges, 8 vertices, and 6 faces.- Each face intersects at right angles.This symmetry makes the formulas for calculating its attributes quite straightforward, such as the formula for volume mentioned above, \( V = a^3 \). Cubes also facilitate easy stacking because of their regular shape, important in understanding how they could cover an area like Earth’s surface.
Earth's Surface Area
The Earth is roughly spherical, so its surface area is calculated as though it were a perfect sphere. The formula for the surface area of a sphere is:\[ A = 4 \pi r^2 \]Here, \( r \) is the radius of the sphere. For Earth, \( r = 6,371,000 \text{ m} \). Plugging the numbers in gives: \[ A = 4 \pi (6,371,000)^2 \text{ m}^2 \]\ This calculation is vital for determining how high dice stacking layers can reach if a mole of dices covered the Earth's surface. This concept helps in visualizing large-scale distribution of smaller units over a global scale.
Avogadro's Number
Avogadro's number is a foundational concept in chemistry and physics, often denoted as \(6.022 \times 10^{23}\). It represents the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance.Semi-widely Used Applications:- Essential for converting between atomic mass units and grams.- Used in calculating the number of particles in a given mass of substance.In our exercise, it tells us how many dice comprise one mole. This number is then used to calculate the total volume of a mole of dice, which highlights the scope and size of Avogadro's number by relating it to everyday-sized objects rather than atomic scale entities.
Unit Conversion
Unit conversion is the process of converting one unit of measure to another, crucial for ensuring calculations remain consistent. In measurement, especially while dealing with volumes and areas, conversions between metric units like centimeters to meters are common.Key Steps for Unit Conversion in this Exercise:- Converting volume from cubic centimeters (\( \text{cm}^3 \)) to cubic meters (\( \text{m}^3 \)) involves dividing by \(10^6\).For instance, our dice volume:\[ 3.375 \text{ cm}^3 = 3.375 \times 10^{-6} \text{ m}^3 \]This ensures all volumes are in the same unit, making it possible to use them directly in calculations with Earth's surface area in \( \text{m}^2 \). Understanding these conversions ensures precision in scientific calculations and helps bridge different measurement systems.

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Most popular questions from this chapter

Titanium(IV) oxide \(\left(\mathrm{TiO}_{2}\right)\) is a white substance produced by the action of sulfuric acid on the mineral ilmenite \(\left(\mathrm{FeTiO}_{3}\right):\) $$ \mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O} $$ Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, \(8.00 \times\) \(10^{3} \mathrm{~kg}\) of \(\mathrm{FeTiO}_{3}\) yielded \(3.67 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{TiO}_{2}\). What is the percent yield of the reaction?

How many moles of \(\mathrm{O}\) are needed to combine with \(0.212 \mathrm{~mol}\) of \(\mathrm{C}\) to form (a) \(\mathrm{CO}\) and \((\mathrm{b}) \mathrm{CO}_{2} ?\)

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is a minor component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air: $$ \mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ (b) How many grams of carbon dioxide can be produced by burning 3.65 mol of propane? Assume that oxygen is the excess reactant in this reaction.

A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from \(46.0 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\). (The products are \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\).)

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

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