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15.0 gPb(NO₂)₂ X STARTING AMOUNT 15.0 g Pb(NO₂)₂ × 1 mol Pb(NO₂) 1 299.22 malPb(NO₂) aPb(NO₂)₂ 2 atoms N Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO₂)2 X 6.022 x 102 atoms N 2 1 atoms N malPb(NO₂) = 6.04 x 10²² atoms N X 6.022 x 1023 1 atoms N mol N 6.04 x 1022 atoms N

15.0 gPb(NO₂)₂ X STARTING AMOUNT 15.0 g Pb(NO₂)₂ × 1 mol Pb(NO₂) 1 299.22 malPb(NO₂) aPb(NO₂)₂ 2 atoms N Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO₂)2 X 6.022 x 102 atoms N 2 1 atoms N malPb(NO₂) = 6.04 x 10²² atoms N X 6.022 x 1023 1 atoms N mol N 6.04 x 1022 atoms N

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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what am I doing wrong how do I correctly follow with converstions

**Transcription and Explanation of Calculation** **Problem:** Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO₃)₂. **Calculation Steps:** 1. **Starting Amount:** - 15.0 grams of Pb(NO₃)₂ 2. **Conversion Factor 1: Molar Mass** - Multiply by \( \frac{1 \text{ mol Pb(NO₃)₂}}{299.22 \text{ g Pb(NO₃)₂}} \) 3. **Conversion Factor 2: Mole Ratio** - Multiply by \( \frac{2 \text{ atoms N}}{1 \text{ mol Pb(NO₃)₂}} \) 4. **Conversion Factor 3: Avogadro’s Number** - Multiply by \( \frac{6.022 \times 10^{23} \text{ atoms N}}{1 \text{ mol N}} \) 5. **Result:** - Equal to \( 6.04 \times 10^{22} \text{ atoms N} \) **Feedback:** - **Incorrect Result Message:** - "Incorrect, 4 attempts remaining." - **Hint:** - "Your ratio between atoms N and mol Pb(NO₃)₂ in your second conversion factor is incorrect. Reconsider the mole ratio that exists between N and Pb(NO₃)₂ and try again." **Explanation of Error:** The conversion step involving the mole ratio should accurately reflect the number of nitrogen atoms per formula unit of Pb(NO₃)₂. Adjusting this step will correct the calculation. **Conclusion:** Recalculate with the correct mole ratio to find the precise number of nitrogen atoms in the given mass of Pb(NO₃)₂.
Transcribed Image Text:**Transcription and Explanation of Calculation** **Problem:** Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO₃)₂. **Calculation Steps:** 1. **Starting Amount:** - 15.0 grams of Pb(NO₃)₂ 2. **Conversion Factor 1: Molar Mass** - Multiply by \( \frac{1 \text{ mol Pb(NO₃)₂}}{299.22 \text{ g Pb(NO₃)₂}} \) 3. **Conversion Factor 2: Mole Ratio** - Multiply by \( \frac{2 \text{ atoms N}}{1 \text{ mol Pb(NO₃)₂}} \) 4. **Conversion Factor 3: Avogadro’s Number** - Multiply by \( \frac{6.022 \times 10^{23} \text{ atoms N}}{1 \text{ mol N}} \) 5. **Result:** - Equal to \( 6.04 \times 10^{22} \text{ atoms N} \) **Feedback:** - **Incorrect Result Message:** - "Incorrect, 4 attempts remaining." - **Hint:** - "Your ratio between atoms N and mol Pb(NO₃)₂ in your second conversion factor is incorrect. Reconsider the mole ratio that exists between N and Pb(NO₃)₂ and try again." **Explanation of Error:** The conversion step involving the mole ratio should accurately reflect the number of nitrogen atoms per formula unit of Pb(NO₃)₂. Adjusting this step will correct the calculation. **Conclusion:** Recalculate with the correct mole ratio to find the precise number of nitrogen atoms in the given mass of Pb(NO₃)₂.
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