**Topic: Calculating the Atomic Radius of Nickel** **Problem Statement:** Nickel has a face-centered cubic (FCC) unit cell. The density of nickel is \(6.84 \, \text{g/cm}^3\). Calculate a value for the atomic radius of nickel. **Calculation Steps:** 1. **Formula Noted:** \[ l = 2.828 \times r \] - **Note:** This is needed to calculate the radius. 2. **Given Data:** - For FCC structure: 4 atoms per unit cell - Density, \(d = 6.84 \, \text{g/cm}^3\) - Volume, \(V = l^3\) 3. **Rearranging the Formula:** \[ r = \frac{l}{2.828} \] 4. **Volume Calculation:** \[ V = \frac{m}{d} = \frac{3.898 \times 10^{-22} \, \text{g}}{6.84 \, \text{g/cm}^3} = 5.83 \times 10^{-23} \, \text{cm}^3 \] 5. **Mass of Nickel Calculation:** - Convert to mass of a single nickel atom: \[ \text{Atoms of Ni} \times \left( \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{atoms}} \right) \times \left( \frac{58.69 \, \text{g/mol}}{1 \, \text{mol}} \right) = 3.898 \times 10^{-22} \, \text{g} \] 6. **Final Answer:** - The values from the textbook are: - \(r = 136 \, \text{pm}\) - \(l = 3.85 \times 10^{-8} \, \text{cm}\) **Summary:** This solution demonstrates the process to find the atomic radius of nickel given its density and cubic unit structure. The calculation involves using known formulas for FCC unit cells and performing conversions between atomic mass and molar quantities.
**Topic: Calculating the Atomic Radius of Nickel** **Problem Statement:** Nickel has a face-centered cubic (FCC) unit cell. The density of nickel is \(6.84 \, \text{g/cm}^3\). Calculate a value for the atomic radius of nickel. **Calculation Steps:** 1. **Formula Noted:** \[ l = 2.828 \times r \] - **Note:** This is needed to calculate the radius. 2. **Given Data:** - For FCC structure: 4 atoms per unit cell - Density, \(d = 6.84 \, \text{g/cm}^3\) - Volume, \(V = l^3\) 3. **Rearranging the Formula:** \[ r = \frac{l}{2.828} \] 4. **Volume Calculation:** \[ V = \frac{m}{d} = \frac{3.898 \times 10^{-22} \, \text{g}}{6.84 \, \text{g/cm}^3} = 5.83 \times 10^{-23} \, \text{cm}^3 \] 5. **Mass of Nickel Calculation:** - Convert to mass of a single nickel atom: \[ \text{Atoms of Ni} \times \left( \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{atoms}} \right) \times \left( \frac{58.69 \, \text{g/mol}}{1 \, \text{mol}} \right) = 3.898 \times 10^{-22} \, \text{g} \] 6. **Final Answer:** - The values from the textbook are: - \(r = 136 \, \text{pm}\) - \(l = 3.85 \times 10^{-8} \, \text{cm}\) **Summary:** This solution demonstrates the process to find the atomic radius of nickel given its density and cubic unit structure. The calculation involves using known formulas for FCC unit cells and performing conversions between atomic mass and molar quantities.
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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Hello I need help with this problem. I tried solving it, but i couldnt figure it out, i provided the correct answers. the problem is that the formula im being asked to use for radius is the one i added below, so i have to get that answer using that, can u plz explain it. thanks much!
![**Topic: Calculating the Atomic Radius of Nickel**
**Problem Statement:**
Nickel has a face-centered cubic (FCC) unit cell. The density of nickel is \(6.84 \, \text{g/cm}^3\). Calculate a value for the atomic radius of nickel.
**Calculation Steps:**
1. **Formula Noted:**
\[
l = 2.828 \times r
\]
- **Note:** This is needed to calculate the radius.
2. **Given Data:**
- For FCC structure: 4 atoms per unit cell
- Density, \(d = 6.84 \, \text{g/cm}^3\)
- Volume, \(V = l^3\)
3. **Rearranging the Formula:**
\[
r = \frac{l}{2.828}
\]
4. **Volume Calculation:**
\[
V = \frac{m}{d} = \frac{3.898 \times 10^{-22} \, \text{g}}{6.84 \, \text{g/cm}^3} = 5.83 \times 10^{-23} \, \text{cm}^3
\]
5. **Mass of Nickel Calculation:**
- Convert to mass of a single nickel atom:
\[
\text{Atoms of Ni} \times \left( \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{atoms}} \right) \times \left( \frac{58.69 \, \text{g/mol}}{1 \, \text{mol}} \right) = 3.898 \times 10^{-22} \, \text{g}
\]
6. **Final Answer:**
- The values from the textbook are:
- \(r = 136 \, \text{pm}\)
- \(l = 3.85 \times 10^{-8} \, \text{cm}\)
**Summary:**
This solution demonstrates the process to find the atomic radius of nickel given its density and cubic unit structure. The calculation involves using known formulas for FCC unit cells and performing conversions between atomic mass and molar quantities.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08bb2a94-da17-43f2-90ab-4ad444df37b3%2Fbc439c20-2add-4277-a274-980777be3194%2Flj2fg2_processed.png&w=3840&q=75)
Transcribed Image Text:**Topic: Calculating the Atomic Radius of Nickel**
**Problem Statement:**
Nickel has a face-centered cubic (FCC) unit cell. The density of nickel is \(6.84 \, \text{g/cm}^3\). Calculate a value for the atomic radius of nickel.
**Calculation Steps:**
1. **Formula Noted:**
\[
l = 2.828 \times r
\]
- **Note:** This is needed to calculate the radius.
2. **Given Data:**
- For FCC structure: 4 atoms per unit cell
- Density, \(d = 6.84 \, \text{g/cm}^3\)
- Volume, \(V = l^3\)
3. **Rearranging the Formula:**
\[
r = \frac{l}{2.828}
\]
4. **Volume Calculation:**
\[
V = \frac{m}{d} = \frac{3.898 \times 10^{-22} \, \text{g}}{6.84 \, \text{g/cm}^3} = 5.83 \times 10^{-23} \, \text{cm}^3
\]
5. **Mass of Nickel Calculation:**
- Convert to mass of a single nickel atom:
\[
\text{Atoms of Ni} \times \left( \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{atoms}} \right) \times \left( \frac{58.69 \, \text{g/mol}}{1 \, \text{mol}} \right) = 3.898 \times 10^{-22} \, \text{g}
\]
6. **Final Answer:**
- The values from the textbook are:
- \(r = 136 \, \text{pm}\)
- \(l = 3.85 \times 10^{-8} \, \text{cm}\)
**Summary:**
This solution demonstrates the process to find the atomic radius of nickel given its density and cubic unit structure. The calculation involves using known formulas for FCC unit cells and performing conversions between atomic mass and molar quantities.
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