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Chapter 2: Problem 57

Determine (a) the number of moles of \(\mathrm{Zn}\) in a \(415.0 \mathrm{g}\) sample of zinc metal (b) the number of \(\mathrm{Cr}\) atoms in \(147.4 \mathrm{kg}\) chromium (c) the mass of a one-trillion-atom \(\left(1.0 \times 10^{12}\right)\) sample of metallic gold (d) the mass of one fluorine atom

Short Answer

Expert verified
For (a), the number of moles in \(415.0\) g of Zn is given in Step 1. For (b), the number of Cr atoms in \(147.4 kg\) of Cr is given in Step 2. For (c), the mass of a \(1.0 * 10^{12}\)-atom sample of Au is provided in Step 3. Finally, for (d), the mass of one F atom is computed in Step 4.

Step by step solution

01

Conversion of Zn grams into moles

The number of Zn moles is determined using the formula: moles = mass / molar mass. The molar mass of Zn is 65.38 g/mol. Therefore, Moles of Zn = \(415.0/65.38\)
02

Conversion of Cr kilograms into atoms

Firstly, convert kg to g: 1kg = 1000g. So Cr mass = \(147.4 * 1000g\). The number of moles can then be determined using the formula: moles = mass / molar mass. The molar mass of Cr is 52.00 g/mol. Therefore, Moles of Cr = \(147.4 * 1000 / 52.00\). Lastly, convert moles to atoms using Avogadro's number (\(6.02 * 10^{23}\)): Number of Cr atoms = Moles of Cr * Avogadro's number.
03

Calculation of Au sample mass

The number of moles can be determined using the formula: moles = number of atoms / Avogadro's number. Therefore, Moles of Au = \(1.0 * 10^{12} / 6.02 * 10^{23}\). Finally, mass can be determined using the formula: mass = moles * molar mass. The molar mass of Au is 197.0 g/mol. Total mass of Au = Moles of Au * molar mass.
04

Computation of F atom mass

The mass of a single atom is determined using the formula: mass = molar mass / Avogadro's number. The molar mass of F is 18.998 g/mol. Therefore, the mass of one F atom = molar mass / Avogadro's number.

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Most popular questions from this chapter

Before \(1961,\) the standard for atomic masses was the isotope \(^{16} \mathrm{O},\) to which physicists assigned a value of exactly \(16 .\) At the same time, chemists assigned a value of exactly 16 to the naturally occurring mixture of the isotopes \(^{16} \mathrm{O},^{17} \mathrm{O},\) and \(^{18} \mathrm{O}\). Would you expect atomic masses listed in a 60 -year-old text to be the same, generally higher, or generally lower than in this text? Explain.

Determine the only possible isotope (E) for which the following conditions are met: \(\bullet\)The mass number of \(\mathrm{E}\) is 2.50 times its atomic number. \(\bullet\)The atomic number of \(\mathrm{E}\) is equal to the mass number of another isotope (Y). In turn, isotope Y has a neutron number that is 1.33 times the atomic number of \(Y\) and equal to the neutron number of selenium- 82 .

In Example \(2-1,\) we established that the mass ratio of magnesium to magnesium oxide is 0.455 g magnesium/ \(0.755 \mathrm{g}\) magnesium oxide. (a) What is the ratio of oxygen to magnesium oxide, by mass? (b) What is the mass ratio of oxygen to magnesium in magnesium oxide? (c) What is the percent by mass of magnesium in magnesium oxide?

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

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