Use the References to access important values if needed for this question. a. How many ATOMS of nitrogen are present in 2.50 moles of nitrogen dioxide? 1.505x10^24 atoms of nitrogen b. How many MOLES of oxygen are present in 1.08 × 1023 molecules of nitrogen dioxide? moles of oxygen 2
Use the References to access important values if needed for this question. a. How many ATOMS of nitrogen are present in 2.50 moles of nitrogen dioxide? 1.505x10^24 atoms of nitrogen b. How many MOLES of oxygen are present in 1.08 × 1023 molecules of nitrogen dioxide? moles of oxygen 2
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:**Question:**
Use the References to access important values if needed for this question.
a. How many ATOMS of nitrogen are present in 2.50 moles of nitrogen dioxide?
**Answer:**
\( 1.505 \times 10^{24} \) atoms of nitrogen
b. How many MOLES of oxygen are present in \(1.08 \times 10^{23}\) molecules of nitrogen dioxide?
**Answer:**
2 moles of oxygen
![### Determining the Sample with the Greatest Number of Atoms
To solve the following problem, you may need to refer to reference materials for important values.
**Problem Statement:**
**Question**: Which sample contains the greatest number of atoms: A sample of Germanium (Ge) that contains \(1.97 \times 10^{24}\) atoms or a 6.28 mole sample of Krypton (Kr)?
**Conclusion**:
The sample of [Blank] contains the greatest number of atoms.
**Explanation:**
- To determine which sample contains the greatest number of atoms, you can compare the given quantity of atoms directly to the quantity converted from moles.
- For Krypton (Kr), we need to convert the amount given in moles to the number of atoms. We use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole):
\[
\text{Number of atoms in 6.28 moles of Kr} = 6.28 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole}
\]
- Calculate this quantity to compare it with the given number of atoms in the Germanium (Ge) sample.
Complete the calculation and fill in the blank accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febc6c9e4-20c2-4aa0-9f79-b773de998cf1%2Fd200a345-bda9-4ef3-b766-a55b7a195743%2Fpbrew5q_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining the Sample with the Greatest Number of Atoms
To solve the following problem, you may need to refer to reference materials for important values.
**Problem Statement:**
**Question**: Which sample contains the greatest number of atoms: A sample of Germanium (Ge) that contains \(1.97 \times 10^{24}\) atoms or a 6.28 mole sample of Krypton (Kr)?
**Conclusion**:
The sample of [Blank] contains the greatest number of atoms.
**Explanation:**
- To determine which sample contains the greatest number of atoms, you can compare the given quantity of atoms directly to the quantity converted from moles.
- For Krypton (Kr), we need to convert the amount given in moles to the number of atoms. We use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole):
\[
\text{Number of atoms in 6.28 moles of Kr} = 6.28 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole}
\]
- Calculate this quantity to compare it with the given number of atoms in the Germanium (Ge) sample.
Complete the calculation and fill in the blank accordingly.
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