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 solid metal sphere has a volume of \(4.2 \mathrm{ft}^{3}\). The mass of the sphere is 155 lb. Find the density of the metal sphere in grams per cubic centimeter.

Short Answer

Expert verified
The density of the metal sphere is approximately \(0.5910\ \frac{g}{cm^{3}}\).

Step by step solution

01

Convert volume to cubic centimeters

Given the volume of the sphere is 4.2 ft³. We need to convert this volume to cubic centimeters (cm³). We know that 1 ft = 30.48 cm. Therefore, 1 ft³ = (30.48 cm)³ = 28316.8466 cm³. So, the volume of the sphere in cubic centimeters is: \(4.2\ ft^{3} \times 28316.8466\frac{cm^{3}}{ft^{3}} = 118929.1557\ cm^{3}\)
02

Convert mass to grams

Given the mass of the sphere is 155 lb. We need to convert this mass to grams (g). We know that 1 lb = 453.59237 g. So, the mass of the sphere in grams is: \(155\ lb \times 453.59237\frac{g}{lb} = 70305.8172\ g\)
03

Calculate density

We have the volume and mass in the appropriate units (cm³ and g). Now, we can calculate the density using the formula: Density (ρ) = \( \frac{Mass}{Volume} \) \( \rho = \frac{70305.8172 g}{118929.1557 cm^{3}} = 0.5910\ \frac{g}{cm^{3}}\) The density of the metal sphere is approximately 0.5910 grams per cubic centimeter.

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

Convert the following Fahrenheit temperatures to Celsius degrees. a. a chilly morning in early autumn, \(45^{\circ} \mathrm{F}\) b. a hot, dry day in the Arizona desert, \(115^{\circ} \mathrm{F}\) c. the temperature in winter when my car won't start, \(-10^{\circ} \mathrm{F}\) d. the surface of a star, \(10,000^{\circ} \mathrm{F}\)

Write each of the following numbers in standard scientific notation. a. \(102.3 \times 10^{-5}\) b. \(32.03 \times 10^{-3}\) c. \(59933 \times 10^{2}\) d. \(599.33 \times 10^{4}\) e. \(5993.3 \times 10^{3}\) f. \(2054 \times 10^{-1}\) g. \(32,000,000 \times 10^{-6}\) h. \(59.933 \times 10^{5}\)

Apples cost \(\$ 0.79\) per pound. What conversion factor is appropriate to express the cost of 5.3 lb of apples?

Perform each of the following conversions, being sure to set up clearly the appropriate conversion factor in each case. The inside cover of this book provides equivalence statements in addition to those contained in this chapter. a. 17.3 L to cubic feet b. 17.3 L to milliliters c. 8.75 L to gallons d. 762 g to ounces e. \(1.00 \mathrm{g}\) to atomic mass units f. 1.00 L to pints g. \(64.5 \mathrm{g}\) to kilograms h. 72.1 mL to liters

Without actually performing the calculations indicated, tell to how many significant digits the answer to the calculation should be expressed. a. \(\frac{(9.7871)(2)}{(0.00182)(43.21)}\) b. \((67.41+0.32+1.98) /(18.225)\) c. \(\left(2.001 \times 10^{-3}\right)\left(4.7 \times 10^{-6}\right)\left(68.224 \times 10^{-2}\right)\) d. (72.15)(63.9)\([1.98+4.8981]\)

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