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

How many moles of cobalt (Co) atoms are there in \(6.00 \times 10^{9}(6\) billion) Co atoms?

Short Answer

Expert verified
Approximately \(9.97 \times 10^{-15}\) moles of cobalt.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Understanding the Problem

We are asked to find out how many moles of cobalt atoms there are in a given number of cobalt atoms. The number of atoms given is \(6.00 \times 10^{9}\), which is 6 billion atoms.
02

Using Avogadro's Number

Avogadro's number tells us that 1 mole of any substance contains \(6.022 \times 10^{23}\) atoms. This number allows us to convert between atoms and moles.
03

Setting Up the Conversion

To find the number of moles, we use the formula: \(\text{number of moles} = \frac{\text{number of atoms}}{\text{Avogadro's number}}\). Here, number of atoms is \(6.00 \times 10^{9}\) and Avogadro's number is \(6.022 \times 10^{23}\).
04

Calculating the Moles

Substitute the values into the formula: \[ \frac{6.00 \times 10^{9}}{6.022 \times 10^{23}} \]. Divide the numbers: \(6.00 \div 6.022\), which will give a value close to \(0.997\) and adjust the exponent \(10^{9} - 10^{23} = 10^{-14}\).
05

Presenting the Final Answer

The final result of the division gives us the number of moles of cobalt, which is approximately \(9.97 \times 10^{-15}\) moles.

Key Concepts

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

Avogadro's number
Named after the scientist Amedeo Avogadro, Avogadro's number is a fundamental constant in chemistry. It represents the number of atoms, molecules, or particles contained in one mole of a substance. Specifically, Avogadro's number is valued at approximately \(6.022 \times 10^{23}\). This constant provides chemists with a bridge between the atomic scale and the laboratory scale, allowing for conversions between the two.

When dealing with atoms, using Avogadro's number enables us to express quantities in moles, a more practical unit that relates to the macroscopic quantities we often handle in labs. If you think of moles like a "chemist's dozen," where instead of 12 items, we count approximately \(6.022 \times 10^{23}\) items, it all starts to make sense. This mass-to-quantity conversion is what makes Avogadro's number vital for calculations involving chemical reactions and physical substances.
cobalt atoms
Cobalt (Co) is a shiny, bluish-grey metal, categorized under transitional metals on the periodic table. Cobalt atoms are the smallest constituents of cobalt elements. They embody all the properties that make cobalt unique, such as ferromagnetism and its vibrant blue color when used in pigments.

In chemistry, understanding the number of cobalt atoms in a sample is crucial for determining the substance's mass, reacting proportions, and potential applications. When we say we have "6 billion cobalt atoms," we are specifying a very tiny amount from a macroscopic point of view, but quite substantial when talking about atomic-level quantities. Remember, due to Avogadro's immense number, \(6.00 \times 10^{9}\) atoms equate to just a tiny fraction of a mole, showcasing how vast an Avogadro number truly is.
conversion formula
The conversion formula is an essential tool in chemistry, enabling us to shift from counting individual atoms to dealing with easily manageable moles. It is given by the relation:
  • \(\text{number of moles} = \frac{\text{number of atoms}}{\text{Avogadro's number}}\)
This equation simplifies the process of converting a vast number of atoms into comparable units that chemists can use in equations and formulations.

Let's break down how it works: If you have \(6.00 \times 10^{9}\) cobalt atoms and wish to find the equivalent number of moles, you place the number of atoms over Avogadro's number in the formula. By performing the division \(\frac{6.00 \times 10^{9}}{6.022 \times 10^{23}}\), you go from counting individual atoms to expressing the quantity as approximately \(9.97 \times 10^{-15}\) moles.

This systematic approach goes beyond cobalt atoms, applying broadly to any substance you deal with on an atomic level. It's fundamental to mastering chemical calculations.

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

Nickel carbonyl can be prepared by the direct combination of nickel metal with carbon monoxide gas according to the following chemical equation: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \longrightarrow \mathrm{Ni}(\mathrm{CO})_{4}(s) $$ Determine the mass of nickel carbonyl that can be produced by the combination of \(50.03 \mathrm{~g} \mathrm{Ni}(s)\) with \(78.25 \mathrm{~g} \mathrm{CO}(g)\). Which reactant is consumed completely? How much of the other reactant remains when the reaction is complete?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent C \(, 4.77\) percent \(\mathrm{H}, 37.85\) percent \(\mathrm{O}, 8.29\) percent \(\mathrm{N},\) and 13.60 percent Na. What is its molecular formula if its molar mass is about \(169 \mathrm{~g}\) ?

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride \((\mathrm{KCl})\) and potassium sulfate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right) .\) Potash production is often reported as the potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) equivalent or the amount of \(\mathrm{K}_{2} \mathrm{O}\) that could be made from a given mineral. (a) If \(\mathrm{KCl}\) costs \(\$ 0.55\) per \(\mathrm{kg},\) for what price (dollar per kg) must \(\mathrm{K}_{2} \mathrm{SO}_{4}\) be sold to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of \(\mathrm{K}_{2} \mathrm{O}\) contains the same number of moles of \(\mathrm{K}\) atoms as \(1.00 \mathrm{~kg}\) of \(\mathrm{KCl}\) ?

Consider the combustion of carbon monoxide (CO) in oxygen gas: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Starting with 3.60 moles of \(\mathrm{CO},\) calculate the number of moles of \(\mathrm{CO}_{2}\) produced if there is enough oxygen gas to react with all the CO.

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