How many grams of Co are there in a sample of Co that contains 8.40×1023 atoms? grams

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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**Question:**

How many grams of Co are there in a sample of Co that contains \(8.40 \times 10^{23}\) atoms? [Text box] grams

**Explanation:**

This question asks for the mass in grams of a sample of cobalt (Co) which contains a specified number of atoms. To solve this, you can use Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol, and the molar mass of cobalt. The calculation involves converting the number of atoms to moles and then to grams. 

**Steps for Calculation:**

1. **Calculate Moles:**
   \[
   \text{Moles of Co} = \frac{8.40 \times 10^{23} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}}
   \]

2. **Convert to Grams:**
   \[
   \text{Mass in grams} = \text{Moles of Co} \times \text{Molar mass of Co (approximately 58.93 g/mol)}
   \]

Provide the calculated result in the text box.
Transcribed Image Text:**Question:** How many grams of Co are there in a sample of Co that contains \(8.40 \times 10^{23}\) atoms? [Text box] grams **Explanation:** This question asks for the mass in grams of a sample of cobalt (Co) which contains a specified number of atoms. To solve this, you can use Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol, and the molar mass of cobalt. The calculation involves converting the number of atoms to moles and then to grams. **Steps for Calculation:** 1. **Calculate Moles:** \[ \text{Moles of Co} = \frac{8.40 \times 10^{23} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} \] 2. **Convert to Grams:** \[ \text{Mass in grams} = \text{Moles of Co} \times \text{Molar mass of Co (approximately 58.93 g/mol)} \] Provide the calculated result in the text box.
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