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5. How many atoms of carbon are present in 24.0 g of carbon A. \( 2.00 \times 10^{23} \) atoms B. \( 6.02 \times 10^{23} \) atoms C. \( 1.20 \times 10^{24} \) atoms D. \( 2.88 \times 10^{24} \) atoms

Ask by Ward Reid. in the United States
Jan 24,2025

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The number of carbon atoms in 24.0 grams of carbon is \(1.20 \times 10^{24}\) atoms.

Solution

To determine the number of carbon atoms in 24.0 grams of carbon, follow these steps: 1. **Calculate the Number of Moles:** The molar mass of carbon (C) is approximately 12.01 g/mol. For simplicity, we'll use 12 g/mol in this calculation. \[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{24.0\, \text{g}}{12\, \text{g/mol}} = 2.00\, \text{mol} \] 2. **Convert Moles to Atoms:** Avogadro's number tells us that there are \(6.02 \times 10^{23}\) atoms in one mole of any substance. \[ \text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's number} = 2.00\, \text{mol} \times 6.02 \times 10^{23}\, \text{atoms/mol} = 1.204 \times 10^{24}\, \text{atoms} \] Rounding to two significant figures: \[ 1.204 \times 10^{24}\, \text{atoms} \approx 1.20 \times 10^{24}\, \text{atoms} \] 3. **Final Answer:** **C. \(1.20 \times 10^{24}\) atoms**

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To find how many atoms of carbon are present in 24.0 g, you can use Avogadro's number, which is \( 6.02 \times 10^{23} \) atoms/mole. Since the molar mass of carbon is approximately 12.01 g/mol, you can calculate the number of moles in 24.0 g: \[ \text{Moles of Carbon} = \frac{24.0 \text{ g}}{12.01 \text{ g/mol}} \approx 2.00 \text{ moles} \] Now, multiply the moles by Avogadro's number: \[ \text{Atoms of Carbon} = 2.00 \times 6.02 \times 10^{23} \approx 1.20 \times 10^{24} \text{ atoms} \] So, the correct answer is C. \( 1.20 \times 10^{24} \) atoms. When dealing with calculations involving moles and mass, it’s essential to remember the unit conversions. Beginners often forget to convert the mass to moles before applying Avogadro's number. Always double-check the units and ensure that you're working with the correct molar mass for the element in question to avoid confusion!

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