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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
The number of moles of Co atoms in \(6.00 \times 10^{9}\) Co atoms is \(9.97 \times 10^{-15}\) moles.

Step by step solution

01

Identify Given and Needed Quantities

Identify the given number of atoms, which is \(6.00 \times 10^{9}\) Co atoms. The task is to find out how many moles this number of atoms corresponds to.
02

Apply Avogadro's Number

Use Avogadro's number, \(6.022\times 10^{23}\) atoms per mole, to convert number of atoms into number of moles. Do this by dividing the number of atoms by Avogadro's number.
03

Calculation

The formula to calculate the number of moles (\(n\)) from number of atoms (\(N\)) is: \(n = \frac{N}{6.022 \times 10^{23}}\). Hence, substituting \(N\) with \(6.00 \times 10^{9}\), the calculation becomes: \(6.00 \times 10^{9} \div 6.022 \times 10^{23} = 9.97 \times 10^{-15}\) moles.

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Key Concepts

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

Avogadro's Number
Avogadro's number is a fundamental concept in chemistry that connects the macroscopic world with the microscopic world of atoms and molecules. It is defined as the number of particles, usually atoms or molecules, in one mole of a substance and is a constant with a value of approximately \(6.022 \times 10^{23}\). This large number helps chemists quantify and convert between atoms and moles, making complex calculations more manageable.

When you're given a certain number of atoms, Avogadro's number allows you to determine how many moles those atoms make up. For example, \(6.00 \times 10^{9}\) cobalt atoms can be converted into moles by dividing by Avogadro's number, as shown in the original problem. This gives insight into how many cobalt atoms collectively form approximately one mole, helping us understand the sheer quantity involved even in small amounts of matter.

Understanding Avogadro's number is essential for anyone studying chemistry, as it is crucial for stoichiometry, molecular weight conversion, and chemical reactions.
Cobalt Atoms
Cobalt is a chemical element with the symbol Co and atomic number 27. It is a transition metal known for its hard, lustrous appearance and is often used in alloys and batteries. On an atomic level, cobalt atoms are the building blocks of this element, and understanding their quantity helps in various chemical calculations.

In exercises like the one given, where you need to determine the number of moles from a specific number of atoms, recognizing the individual element, such as cobalt, is important. This context helps apply the calculations specifically to cobalt, acknowledging its unique properties and how they factor into chemical equations. Considering that \(6.00 \times 10^{9}\) atoms is a small portion of one mole, it illustrates how chemistry works at the atomic scale, providing a perspective on how densely packed even the smallest samples of metal can be with atoms.

Cobalt's role in chemistry is significant, not just for its use in various applications, but also in highlighting the universal principles of atomic structure and chemical interactions.
Chemical Calculations
Chemical calculations involve converting and manipulating numerical data regarding chemical substances. These calculations often include converting between different units, such as atoms, moles, and grams, using various constants like Avogadro's number.

In the context of our exercise, the calculation starts with the number of cobalt atoms, \(6.00 \times 10^{9}\), and involves converting these atoms into moles. This is achieved through the equation \(n = \frac{N}{6.022 \times 10^{23}}\), where \(n\) represents the number of moles, and \(N\) is the given number of atoms. By dividing the total number of atoms by Avogadro’s number, we arrive at the number of moles: \(9.97 \times 10^{-15}\).

This calculation is crucial for practical laboratory work and theoretical chemistry tasks, as understanding the relationship between atoms and moles allows chemists to accurately predict the outcomes of reactions and the quantities of materials needed. Mastery of these calculations is fundamental to success in chemistry, ensuring precise and meaningful results in both experiments and industrial applications.

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Most popular questions from this chapter

Consider the reaction $$ \mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ If 0.86 mole of \(\mathrm{MnO}_{2}\) and \(48.2 \mathrm{~g}\) of \(\mathrm{HCl}\) react, which reagent will be used up first? How many grams of \(\mathrm{Cl}_{2}\) will be produced?

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2} $$ Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of Fe are obtained from a \(2.62 \times 10^{3}-\mathrm{kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can 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 many 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 \(6.55 \mathrm{~g}\).

Which of the following has more atoms: \(1.10 \mathrm{~g}\) of hydrogen atoms or \(14.7 \mathrm{~g}\) of chromium atoms?

Define limiting reagent and excess reagent. What is the significance of the limiting reagent in predicting the amount of the product obtained in a reaction? Can there be a limiting reagent if only one reactant is present?

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