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Calculate the number of moles in 75.0 g of dinitrogen trioxide.
Sure! Let's calculate the number of moles of dinitrogen trioxide (N₂O₃) given a mass of 75.0 grams.
### Step-by-Step Solution:
1. Understand the Problem: - We are given: - The mass of dinitrogen trioxide (N₂O₃) is 75.0 grams - We need to find the number of moles in this mass.
2. Recall the Formula: - To find the number of moles, we can use the formula: [tex]\[
\text{Number of moles} = \frac{\text{mass (in grams)}}{\text{molar mass (in g/mol)}}
\][/tex]
3. Identify the Molar Mass: - For dinitrogen trioxide (N₂O₃), the molar mass is approximately 76.01 g/mol. This value is typically found using atomic masses from the periodic table: - Nitrogen (N): approximately 14.01 g/mol - Oxygen (O): approximately 16.00 g/mol - For N₂O₃: [tex]\(2 \times 14.01 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 28.02 \text{ g/mol} + 48.00 \text{ g/mol} = 76.02 \text{ g/mol}\)[/tex] - Note: We use the given precise molar mass of 76.01 g/mol for accuracy.
4. Plug in the Values and Solve: - Using the given mass and molar mass: [tex]\[
\text{Number of moles} = \frac{75.0 \text{ grams}}{76.01 \text{ g/mol}}
\][/tex]
5. Compute the Result: - Carry out the division to find the number of moles: [tex]\[
\text{Number of moles} \approx 0.9867
\][/tex]
### Final Answer: The number of moles in 75.0 grams of dinitrogen trioxide is approximately 0.9867 moles.
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