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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

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.

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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.

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Most popular questions from this chapter

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 \(6.55 \mathrm{~g}\).

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