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Chapter 3: Problem 17

What is the mass in grams of a single atom of each of the following elements? (a) \(\mathrm{Hg},\) (b) Ne.

Short Answer

Expert verified
The mass of a single atom of Hg (Mercury) is approximately \(3.33 \times 10^{-22} \, \text{grams}\) and the mass of a single atom of Ne (Neon) is approximately \(3.35 \times 10^{-23} \, \text{grams}\).

Step by step solution

01

Find Atomic Mass of elements

Look up the atomic masses of the elements from the periodic table. The atomic masses of Hg (Mercury) and Ne (Neon) are approximately \(200.59 \, \text{amu}\) and \(20.18 \, \text{amu}\) respectively.
02

Apply Avogadro's number

Avogadro's number (\(6.022 \times 10^{23}\)) is the number of units in one mole of any substance. This implies that one atom of an element weighs the atomic mass in grams per mole. Therefore, the mass of an atom can be calculated by dividing the atomic mass in grams per mole by Avogadro's number.
03

Calculate for Hg

The atomic mass of Hg is 200.59 amu. Therefore, its mass in grams is \( \frac{200.59}{6.022 \times 10^{23}} \approx 3.33 \times 10^{-22} \, \text{grams}\).
04

Calculate for Ne

The atomic mass of Ne is 20.18 amu. Therefore, its mass in grams is \( \frac{20.18}{6.022 \times 10^{23}} \approx 3.35 \times 10^{-23} \, \text{grams}\).

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) is also called "laughing gas." \(\mathrm{It}\) can be prepared by the thermal decomposition of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\). The other product is \(\mathrm{H}_{2} \mathrm{O} .\) (a) Write a balanced equation for this reaction. (b) How many grams of \(\mathrm{N}_{2} \mathrm{O}\) are formed if 0.46 mole of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is used in the reaction?

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