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

Earth's population is about 7.0 billion. Suppose that every person on Earth participates in a process of counting identical particles at the rate of two particles per second. How many years would it take to count \(6.0 \times 10^{23}\) particles? Assume that there are 365 days in a vear.

Short Answer

Expert verified
It takes about 1.36 million years.

Step by step solution

01

Understand the Problem

We are asked to find out how long it would take for 7 billion people to count a total of \(6.0 \times 10^{23}\) particles if each person can count two particles per second. We need to determine the duration in years.
02

Calculate Total Counting Rate

Each person counts 2 particles per second. Therefore, with 7 billion people counting, the total rate of counting is \[ 2 \times 7 \times 10^9 = 1.4 \times 10^{10} \text{ particles per second.} \]
03

Calculate Total Seconds Needed

We need to divide the total number of particles, \(6.0 \times 10^{23}\), by the total counting rate:\[ \frac{6.0 \times 10^{23}}{1.4 \times 10^{10}} = 4.2857 \times 10^{13} \text{ seconds.} \]
04

Convert Seconds to Years

First, convert seconds to minutes by dividing by 60, then to hours, then to days, and finally to years. - Seconds to minutes: \( \frac{4.2857 \times 10^{13}}{60} = 7.1428 \times 10^{11} \text{ minutes} \)- Minutes to hours: \( \frac{7.1428 \times 10^{11}}{60} = 1.1905 \times 10^{10} \text{ hours} \)- Hours to days: \( \frac{1.1905 \times 10^{10}}{24} = 4.9604 \times 10^8 \text{ days} \)- Days to years: \( \frac{4.9604 \times 10^8}{365} = 1.3598 \times 10^6 \text{ years} \)
05

Final Answer

It would take approximately \(1.36 \times 10^6\) years for the entire Earth's population to count \(6.0 \times 10^{23}\) particles at the rate of two particles per second per person.

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.

Avogadro's number
Avogadro's number plays a crucial role in chemistry by linking the macroscopic world to the atomic scale. The number, which is approximately equal to \( 6.022 imes 10^{23} \), represents the number of atoms, ions, or molecules in one mole of a substance. This vast number is a constant in calculations that involve the mole concept.

Understanding Avogadro's number can help make sense of extremely large quantities you could never count by hand, like the number of molecules in a glass of water. Avogadro's number is used not only in theoretical chemistry but also in real-world calculations, such as determining the number of particles in a sample. In our exercise, it serves as a point of reference for how many particles the world's population would need to count.
  • Represents the size of an atom or molecule population in a mole.
  • Crucial for converting between a substance's atomic scale and macroscopic measurements.
  • Enables calculations that involve stoichiometry and chemical reactions.
Mole concept
The mole concept is a fundamental aspect of chemistry that provides a method for counting atoms, molecules, or particles in a given amount of substance. A mole is defined as the amount of a substance that contains as many elementary entities as there are atoms in 12 grams of carbon-12.

This concept allows chemists to count particles using moles rather than numbers that are astronomically large, like those used in Avogadro's number. In the context of our exercise, it highlights how counting trillions upon trillions of particles can be managed using simpler calculations based on moles.
  • Essential for quantifying substances in chemical equations and reactions.
  • Makes it feasible to perform stoichiometric calculations efficiently.
  • Provides a bridge between the atomic world and laboratory-scale chemistry.
Mathematical calculations
Mathematical calculations in stoichiometry can often seem complex or abstract, but they boil down to applying simple arithmetic and algebra to solve chemical problems. Understanding these calculations helps to view chemistry problems as logical puzzles.

In this exercise, we apply mathematical calculations to determine how long it would take for a vast number of particles to be counted. First, by calculating the total counting rate, then the total time in seconds, and finally converting that time into units of years.
  • Enhances problem-solving skills involved in chemical calculations.
  • Utilizes unit conversion to switch between seconds, minutes, hours, days, and years.
  • Applies exponential notation to simplify large-number arithmetic.
Population chemistry problem
A population chemistry problem involves applying chemical concepts and calculations to real-world or hypothetical scenarios involving large groups of people or populations. In our problem, we explore how chemists use population-based calculations to extend the understanding of counting atoms or molecules.

By thinking about the Earth's population attempting to count particles, this exercise demonstrates the vastness of Avogadro's number and the mole concept in stark, relatable terms. It turns an abstract chemical principle into a tangible scenario that highlights the scale of stoichiometric calculations.
  • Connects abstract chemical concepts to everyday life scenarios.
  • Highlights the magnitude of Avogadro's number in practical terms.
  • Demonstrates the power of collective effort in comprehending large-scale chemical processes.

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

Aspirin or acetylsalicylic acid is synthesized by combining salicylic acid with acetic anhydride: $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ \(\begin{array}{l}\text { salicylic acid acetic anhydride } \\ \text { aspirin } & \text { acetic acid }\end{array}\) (a) How much salicylic acid is required to produce \(0.400 \mathrm{~g}\) of aspirin (about the content in a tablet), assuming acetic anhydride is present in excess? (b) Calculate the amount of salicylic acid needed if only 74.9 percent of salicylic is converted to aspirin. (c) In one experiment, \(9.26 \mathrm{~g}\) of salicylic acid reacts with \(8.54 \mathrm{~g}\) of acetic anhydride. Calculate the theoretical yield of aspirin an the percent yield if only \(10.9 \mathrm{~g}\) of aspirin is produced.

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride \((\mathrm{KCl})\) and potassium sulfate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right) .\) Potash production is often reported as the potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) equivalent or the amount of \(\mathrm{K}_{2} \mathrm{O}\) that could be made from a given mineral. (a) If \(\mathrm{KCl}\) costs \(\$ 0.55\) per \(\mathrm{kg},\) for what price (dollar per kg) must \(\mathrm{K}_{2} \mathrm{SO}_{4}\) be sold to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of \(\mathrm{K}_{2} \mathrm{O}\) contains the same number of moles of \(\mathrm{K}\) atoms as \(1.00 \mathrm{~kg}\) of \(\mathrm{KCl}\) ?

Disulfide dichloride \(\left(\mathrm{S}_{2} \mathrm{Cl}_{2}\right)\) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine: $$ \mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{~S}_{2} \mathrm{Cl}_{2}(l) $$ What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams when \(4.06 \mathrm{~g}\) of \(\mathrm{S}_{8}\) is heated with \(6.24 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\) ? If the actual yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is \(6.55 \mathrm{~g},\) what is the percent yield?

Determine the empirical formulas of the compounds with the following compositions: (a) 40.1 percent \(\mathrm{C}\), 6.6 percent \(\mathrm{H}, 53.3\) percent \(\mathrm{O} ;\) (b) 18.4 percent \(\mathrm{C}\), 21.5 percent \(\mathrm{N}, 60.1\) percent \(\mathrm{K}\)

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