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

You are asked to help set up a historical display in the park by stacking some cannonballs next to a Revolutionary War cannon. You are told to stack them by starting with a triangle in which each side is composed of four touching cannonballs. You are to continue stacking them until you have a single ball on the top centered over the middle of the triangular base. a. How many cannonballs do you need? b. What type of closest packing is displayed by the cannonballs? c. The four corners of the pyramid of cannonballs form the corners of what type of regular geometric solid?

Short Answer

Expert verified
a. You will need 20 cannonballs to form the pyramid. b. The type of closest packing displayed by the cannonballs is face-centered cubic (fcc) packing. c. The four corners of the pyramid of cannonballs form the corners of a regular tetrahedron.

Step by step solution

01

a. Calculate the number of cannonballs needed

To find out the number of cannonballs needed, let's analyze the structure of the pyramid. Each layer of the pyramid will have the shape of an equilateral triangle. Starting from the base and moving upwards: - Layer 1 (base layer) - An equilateral triangle with 4 cannonballs on each side. - Layer 2 - An equilateral triangle with 3 cannonballs on each side. - Layer 3 - An equilateral triangle with 2 cannonballs on each side. - Layer 4 (top layer) - A single cannonball. To find the number of cannonballs in each layer, we can calculate the number of cannonballs in an equilateral triangular shape: Layer 1: \(1 + 2 + 3 + 4 = 10\) cannonballs Layer 2: \(1 + 2 + 3 = 6\) cannonballs Layer 3: \(1 + 2 = 3\) cannonballs Layer 4: \(1\) cannonball Now, sum up the number of cannonballs in all layers: \(10 + 6 + 3 + 1 = 20\). So, you will need 20 cannonballs.
02

b. Type of closest packing displayed by the cannonballs

The closest packing displayed by the cannonballs is called face-centered cubic (fcc) packing. In this type of packing, the centers of the cannonballs form a 3D lattice with a regular tetrahedron and an octahedron in each unit cell. The fcc packing is one of the most efficient ways to pack spheres as it allows them to fill up approximately 74% of the available space.
03

c. Regular geometric solid formed by the four corners of the pyramid

The four corners of the pyramid of cannonballs form the corners of a regular tetrahedron. A tetrahedron is a polyhedron with four triangular faces, six edges, and four vertices. In a regular tetrahedron, all faces are equilateral triangles, and all edges and angles are congruent.

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!

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 common prank on college campuses is to switch the salt and sugar on dining hall tables, which is usually easy because the substances look so much alike. Yet, despite the similarity in their appearance, these two substances differ greatly in their properties, since one is a molecular solid and the other is an ionic solid. How do the properties differ and why?

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime \((\mathrm{CaO})\) at \(600 .{ }^{\circ} \mathrm{C}\) followed by condensation of the mercury vapor. Given the heat of vaporization of mercury \((296 \mathrm{~J} / \mathrm{g})\) and the vapor pressure of mercury at \(25.0^{\circ} \mathrm{C}\) \(\left(2.56 \times 10^{-3}\right.\) torr), what is the vapor pressure of the condensed mercury at \(300 .{ }^{\circ} \mathrm{C} ?\) How many atoms of mercury are present in the mercury vapor at \(300 .{ }^{\circ} \mathrm{C}\) if the reaction is conducted in a closed 15.0-L container?

A substance, \(X\), has the following properties: Sketch a heating curve for substance \(\mathrm{X}\) starting at \(-50 .{ }^{\circ} \mathrm{C}\).

A \(0.250-\mathrm{g}\) chunk of sodium metal is cautiously dropped into a mixture of \(50.0 \mathrm{~g}\) water and \(50.0 \mathrm{~g}\) ice, both at \(0{ }^{\circ} \mathrm{C}\). The reaction is \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) \quad \Delta H=-368 \mathrm{~kJ}\) Will the ice melt? Assuming the final mixture has a specific heat capacity of \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), calculate the final temperature. The enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

A \(20.0-\mathrm{g}\) sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with \(100.0 \mathrm{~g}\) water at \(80.0^{\circ} \mathrm{C}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(2.03\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

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