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Chapter 3: Problem 27

If you had a mole of U.S. dollar bills and equally distributed the money to all of the people of the world, how rich would every person be? Assume a world population of 7 billion.

Short Answer

Expert verified
If a mole of U.S. dollar bills (6.022 x 10^23 dollars) were equally distributed among the world population of 7 billion people (7 x 10^9 people), each person would be approximately 8.6 x 10^13 dollars rich.

Step by step solution

01

Convert a mole of U.S. dollar bills into a number

To convert a mole of U.S. dollar bills into a numerical amount, we can use Avogadro's number, which is approximately 6.022 x 10^23. So, a mole of U.S. dollar bills would be equal to: 6.022 x 10^23 dollars.
02

Calculate the total amount to be distributed among the world population

We know that the world population is given as 7 billion, or 7 x 10^9 people. So, to find out how much money would be distributed among the world population, we can use the following formula: Total amount to be distributed = Total money (A mole of U.S. dollar bills) / World population
03

Solve for the total amount to be distributed

In this step, we will plug in the values calculated in the previous steps into our formula: Total amount to be distributed = (6.022 x 10^23 dollars) / (7 x 10^9 people) To simplify, divide the numerator by the denominator: Total amount to be distributed ≈ 8.6 x 10^13 dollars per person
04

Final answer

Therefore, if a mole of U.S. dollar bills were equally distributed among the world population of 7 billion people, each person would be approximately 8.6 x 10^13 dollars rich.

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Most popular questions from this chapter

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$ 2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If $15.0 \mathrm{g} \mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\( and 5.00 \)\mathrm{g} \mathrm{NH}_{3}$ are reacted, what mass of acrylonitrile can be produced, assuming 100\(\%\) yield?

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A gas contains a mixture of \(\mathrm{NH}_{3}(g)\) and $\mathrm{N}_{2} \mathrm{H}_{4}(g),\( both of which react with \)\mathrm{O}_{2}(g)$ to form \(\mathrm{NO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) The gaseous mixture (with an initial mass of 61.00 \(\mathrm{g}\) ) is reacted with 10.00 \(\mathrm{moles}\) \(\mathrm{O}_{2},\) and after the reaction is complete, 4.062 moles of \(\mathrm{O}_{2}\) remains. Calculate the mass percent of \(\mathrm{N}_{2} \mathrm{H}_{4}(g)\) in the original gaseous mixture.
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