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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.

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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.

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Most popular questions from this chapter

The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are 99.99 percent \({ }_{1}^{1} \mathrm{H}\) and 0.01 percent \({ }_{1}^{2} \mathrm{H}\). Assume that water exists as either \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{D}_{2} \mathrm{O} .\) Calculate the number of \(\mathrm{D}_{2} \mathrm{O}\) molecules in exactly \(400 \mathrm{~mL}\) of water \((\) density \(1.00 \mathrm{~g} / \mathrm{mL})\).

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

Limestone \(\left(\mathrm{CaCO}_{3}\right)\) is decomposed by heating to quicklime \((\mathrm{CaO})\) and carbon dioxide. Calculate how many grams of quicklime can be produced from \(1.0 \mathrm{~kg}\) of limestone.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

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 \(4 \pi r^{2}\).]

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