Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An atom of rhodium (Rh) has a diameter of about \(2.7 \times 10^{-8} \mathrm{~cm}\). (a) What is the radius of a rhodium atom in angstroms \((\AA)\) and in meters \((\mathrm{m}) ?(\mathbf{b})\) How many \(\mathrm{Rh}\) atoms would have to be placed side by side to span a distance of \(6.0 \mu \mathrm{m} ?\) (c) If you assume that the \(\mathrm{Rh}\) atom is a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single atom?

Short Answer

Expert verified
(a) The radius of a rhodium atom in angstroms is approximately \(6.75 \times 10^{1} \AA\) and in meters is approximately \(6.75 \times 10^{-11} \mathrm{m}\). (b) About 2.22 $\times$ 10^3 Rh atoms would need to be placed side by side to span a distance of \(6.0 \mu \mathrm{m}\). (c) Assuming the Rh atom is a sphere, the volume of a single atom is approximately \(1.29 \times 10^{-30} \mathrm{m}^3\).

Step by step solution

01

- Conversion of Diameter to Radius

We are given that the diameter of a Rhodium atom is \(2.7 \times 10^{-8} \mathrm{cm}\). To convert diameter into radius, we must simply divide the diameter by 2. Radius = diameter / 2
02

- Conversion of Radius to Angstroms and Meters

First, convert the radius in centimeters to angstroms(\(\AA\)) and meters. 1 cm = \(1 \times 10^8\) \(\AA\) 1 cm = 0.01 meters Now, multiply the radius in centimeters with the equivalent conversion factors.
03

- Calculate the Number of Rhodium Atoms

We are given a distance of \(6.0 \mu \mathrm{m}\) that needs to be spanned by placing Rh atoms side by side. To determine the number of atoms required, we need to divide the given distance by the diameter of one Rh atom. First, we need to convert the given distance to centimeters. 1 \(\mu \mathrm{m}\) = \(1 \times 10^{-4} \mathrm{cm}\) Now, divide the distance in cm by the diameter of one Rh atom.
04

- Calculate the Volume of a Rhodium Atom

To calculate the volume of a Rh atom considering it as a sphere, we will use the formula for the volume of a sphere: Sphere Volume = \(\dfrac{4}{3}\pi r^{3}\) Use the radius in meters calculated from step 2 and plug it into the formula to calculate the volume of a single Rh atom in \(m^3\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conversion of Units in Chemistry
When studying chemistry, it's essential to understand how to convert between different units, as this allows chemists to communicate information clearly and perform calculations accurately. Important to many areas, including atomic measurements, this skill is fundamental to recognizing the incredibly small scales at which chemical processes occur.

For example, converting the radius of a rhodium atom from centimeters to angstroms and meters involves multiplying by appropriate conversion factors. Since 1 cm equals to both \(1 \times 10^8\) angstroms and 0.01 meters, understanding how to apply these factors is crucial. It's similar to translating words between languages; each unit represents a different 'language' for measuring the same concept.

Students often struggle with converting units due to the variety of systems in use. A helpful tip is to keep a reference of common conversions handy and to practice by solving problems involving unit changes, just like converting the atomic radius from centimeters to angstroms and meters in the rhodium atom example.
Atomic Radius Measurement
The atomic radius is a measure of the size of an atom, typically its size when bonded with another atom. Measuring the atomic radius is not as straightforward as it sounds, because electrons in atoms are not stationary, and the outermost electrons can be found in different probabilities at various distances from the nucleus.

In practice, scientists use different methods to estimate the radius, such as X-ray crystallography or van der Waals radii. The rhodium atom in our example has a straightforward measurement provided, but remember these are often approximate. After obtaining this measurement, the next step is usually to convert it into a more useful unit, depending on the context of the calculation or the requirements of the formula it will be used in.

To visualize atomic radii better, it might help to compare different elements on the periodic table: atoms with larger atomic numbers typically have larger atomic radii, while those closer to the top-right corner generally have smaller radii, due to increased nuclear charge pulling electrons closer.
Volume of Atoms Calculation
The volume of an atom can provide insightful information regarding properties like density and packing in materials. To calculate the volume, we often assume that atoms are perfectly spherical, which simplifies the use of the volume formula for a sphere \(\dfrac{4}{3}\pi r^{3}\), where \(r\) is the radius.

Using this assumption for the rhodium atom, once we have the atomic radius in the correct unit (meters), we can simply substitute it into the formula to determine the volume. The concept of volume at the atomic level might be difficult to grasp since atoms are incredibly tiny, but this mathematical representation allows scientists to make practical predictions about the substance's behavior.

A common confusion arises regarding the size of atoms and their volume, as it's challenging to picture the three-dimensional aspect based on radius alone. However, understanding the relationship between the radius and volume reinforces the concept of atoms occupying three-dimensional space, which is key in advanced topics like molecular geometry and crystal structure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hydrogen sulfide is composed of two elements: hydrogen and sulfur. In an experiment, \(6.500 \mathrm{~g}\) of hydrogen sulfide is fully decomposed into its elements. (a) If \(0.384 \mathrm{~g}\) of hydrogen is obtained in this experiment, how many grams of sulfur must be obtained? (b) What fundamental law does this experiment demonstrate? (c) How is this law explained by Dalton's atomic theory?

Answer the following questions without referring to Table 2.1: (a) What are the main subatomic particles that make up the atom? (b) What is the relative charge (in multiples of the electronic charge) of each of the particles? (c) Which of the particles is the most massive? (d) Which is the least massive?

Predict the chemical formula for the ionic compound formed by (a) \(\mathrm{Ca}^{2+}\) and \(\mathrm{Br}^{-},\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{CO}_{3}^{2-},\) (c) \(\mathrm{Al}^{3+}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) (d) \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\), (e) \(\mathrm{Mg}^{2+}\) and \(\mathrm{PO}_{4}^{3-}\).

The explosion of an atomic bomb releases many radioactive isotopes, including strontium-90. Considering the location of strontium in the periodic table, suggest a reason for the fact that this isotope is particularly harmful to humans.

Because many ions and compounds have very similar names, there is great potential for confusing them. Write the correct chemical formulas to distinguish between (a) calcium sulfide and calcium hydrogen sulfide, (b) hydrobromic acid and bromic acid, (c) aluminum nitride and aluminum nitrite, (d) iron(II) oxide and iron(III) oxide, (e) ammonia and ammonium ion, (f) potassium sulfite and potassium bisulfite, (g) mercurous chloride and mercuric chloride, (h) chloric acid and perchloric acid.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free