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Chapter 3: Problem 11

Earth's population is about 7.2 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 year.

Short Answer

Expert verified
It would take approximately 1.32 million years for all people on Earth to count \(6.0 \times 10^{23}\) particles at a rate of 2 particles per second, assuming that there are 365 days in a year.

Step by step solution

01

Conversion Time To Number Of Seconds Per Year

The first step is to convert the number of seconds in a year in order to properly calculate the required time. Given that there are 365 days in a year, 24 hours in a day, 60 minutes in an hour and 60 seconds in a minute, the total number of seconds in a year can be calculated by multiplying these values together: \[ 365 \times 24 \times 60 \times 60 = 31,536,000 \] s/year.
02

Calculate the Total Counting Rate

The next step is to calculate how many particles can be counted by the entire human population in one second, which can be done by multiplying the population by the individual counting rate. So, given that there are approximately \(7.2 \times 10^9\) people on Earth, and each counts 2 particles per second: \[ 7.2 \times 10^9 \times 2 = 1.44 \times 10^{10} \] particles/second.
03

Calculate Time

Now that we know how many particles can be counted per second and how many particles there are in total, we can calculate how long it will take to count all the particles. This can be done by dividing the total number of particles by the total counting rate, which gives us the time in seconds. Then, we should convert the time from seconds to years by dividing the resulting number of seconds by the number of seconds in a year. \[ \frac{6.0 \times 10^{23}}{1.44 \times 10^{10}} = 4.17 \times 10^{13} \] seconds, and then \[ \frac{4.17 \times 10^{13}}{31,536,000} = 1.32 \times 10^{6} \] years, or approximately 1.32 million years.

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Most popular questions from this chapter

Consider the combustion of carbon monoxide (CO) in oxygen gas: $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$$Starting with 3.60 moles of \(\mathrm{CO}\), calculate the number of moles of \(\mathrm{CO}_{2}\) produced if there is enough oxygen gas to react with all of the CO.

On what law is stoichiometry based? Why is it essential to use balanced equations in solving stoichiometric problems?

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 many 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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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:$$\mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{~S}_{2} \mathrm{Cl}_{2}(l)$$What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams when \(4.06 \mathrm{~g}\) of \(\mathrm{S}_{8}\) are 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?
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