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Chapter 2: Problem 108

A 5.585-kg sample of iron (Fe) contains (a) \(10.0 \mathrm{mol} \mathrm{Fe}\) (b) twice as many atoms as does \(600.6 \mathrm{g} \mathrm{C}\) (c) 10 times as many atoms as does \(52.00 \mathrm{g} \mathrm{Cr}\) (d) \(6.022 \times 10^{24}\) atoms

Short Answer

Expert verified
To know if the statements are correct, the calculations explained in steps 1 through 4 have to be done. That is, conversion of kg to moles, and then moles to number of atoms. Then compare these calculated values with the given quantities.

Step by step solution

01

Convert mass to moles

First, start by converting the mass of iron to moles. The molar mass of iron is 55.85 g/mol. But, the mass we have is in kilograms, so we first need to convert it to grams. Then, the moles of iron can be calculated by dividing the mass of iron by its molar mass. This will give us the number of moles present in the given sample of iron.
02

Compare moles

For part a, simply equate the moles of iron calculated in step 1 with 10.0 moles as specified in the exercise.
03

Compare atoms

For part b and c, calculate the number of atoms by multiplying the moles of iron (from step 1) by Avogadro's number (\(6.022 \times 10^{23} \) atoms/mol), then compare this with the number of atoms in the given amount of carbon and chromium. For carbon, the number of atoms can be calculated by dividing its given mass by its molar mass (12.01 g/mol) and then multiplying by Avogadro's number. For chromium, make a similar calculation using its molar mass (52.00 g/mol).
04

Compare atoms

In part d, do a direct comparison of the total number of atoms calculated for the iron sample with \(6.022 \times 10^{24}\) atoms, as given in the problem.

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Most popular questions from this chapter

Use fundamental definitions and statements from Chapters 1 and 2 to establish the fact that \(6.022 \times 10^{23} \mathrm{u}=1.000 \mathrm{g}\)

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