Nickel Percentage B8C3C0

1. **State the problem:** We have a mixture of nickel and lead spheres with total mass $4.02$ kg and total volume $423$ mL. We want to find the percentage of the spheres that are nickel by mass. 2. **Known data:** - Total mass, $m_{total} = 4.02$ kg - Total volume, $V_{total} = 423$ mL = $0.423$ L (since $1000$ mL = $1$ L) - Density of nickel, $\rho_{Ni} = 8.90$ g/mL = $8900$ kg/m$^3$ - Density of lead, $\rho_{Pb} = 11.34$ g/mL = $11340$ kg/m$^3$ 3. **Define variables:** - Let $m_{Ni}$ = mass of nickel spheres (kg) - Let $m_{Pb}$ = mass of lead spheres (kg) 4. **Write equations:** - Total mass: $$m_{Ni} + m_{Pb} = 4.02$$ - Total volume: $$V_{Ni} + V_{Pb} = 0.423$$ L 5. **Express volumes in terms of mass and density:** $$V_{Ni} = \frac{m_{Ni}}{\rho_{Ni}}$$ $$V_{Pb} = \frac{m_{Pb}}{\rho_{Pb}}$$ 6. **Substitute into volume equation:** $$\frac{m_{Ni}}{8900} + \frac{m_{Pb}}{11340} = 0.423$$ 7. **Use mass equation to express $m_{Pb}$:** $$m_{Pb} = 4.02 - m_{Ni}$$ 8. **Substitute $m_{Pb}$ into volume equation:** $$\frac{m_{Ni}}{8900} + \frac{4.02 - m_{Ni}}{11340} = 0.423$$ 9. **Multiply both sides by $\cancel{1000}$ to convert kg to g for easier calculation:** $$\frac{1000 m_{Ni}}{8900} + \frac{1000 (4.02 - m_{Ni})}{11340} = 423$$ 10. **Simplify fractions:** $$\frac{1000}{8900} = 0.11236, \quad \frac{1000}{11340} = 0.08822$$ 11. **Rewrite equation:** $$0.11236 m_{Ni} + 0.08822 (4.02 - m_{Ni}) = 423$$ 12. **Distribute:** $$0.11236 m_{Ni} + 0.3547 - 0.08822 m_{Ni} = 423$$ 13. **Combine like terms:** $$0.02414 m_{Ni} + 0.3547 = 423$$ 14. **Subtract $0.3547$ from both sides:** $$0.02414 m_{Ni} = 422.6453$$ 15. **Solve for $m_{Ni}$:** $$m_{Ni} = \frac{422.6453}{0.02414} = 17515.5$$ grams 16. **Convert back to kg:** $$m_{Ni} = 17.5155$$ kg (which is impossible since total mass is $4.02$ kg, so we must not convert units incorrectly) **Correction:** Step 9 was incorrect to multiply by 1000 because $m_{Ni}$ is in kg and densities are in kg/m$^3$, volumes in L (1 L = 0.001 m$^3$). Let's redo step 9 correctly. 9. **Convert volume to m$^3$:** $$V_{total} = 423 \text{ mL} = 423 \times 10^{-6} \text{ m}^3 = 0.000423 \text{ m}^3$$ 10. **Rewrite volume equation with consistent units:** $$\frac{m_{Ni}}{8900} + \frac{4.02 - m_{Ni}}{11340} = 0.000423$$ 11. **Multiply both sides by $\cancel{1}$ (no unit change needed):** $$\frac{m_{Ni}}{8900} + \frac{4.02 - m_{Ni}}{11340} = 0.000423$$ 12. **Multiply entire equation by $8900 \times 11340$ to clear denominators:** $$11340 m_{Ni} + 8900 (4.02 - m_{Ni}) = 0.000423 \times 8900 \times 11340$$ 13. **Calculate right side:** $$0.000423 \times 8900 \times 11340 = 0.000423 \times 100926000 = 42700.6$$ 14. **Expand left side:** $$11340 m_{Ni} + 358,980 - 8900 m_{Ni} = 42700.6$$ 15. **Combine like terms:** $$(11340 - 8900) m_{Ni} + 358,980 = 42700.6$$ $$2440 m_{Ni} + 358,980 = 42700.6$$ 16. **Subtract 358,980 from both sides:** $$2440 m_{Ni} = 42700.6 - 358,980 = -316,279.4$$ 17. **Solve for $m_{Ni}$:** $$m_{Ni} = \frac{-316,279.4}{2440} = -129.6$$ kg (negative mass is impossible, so check units again) **Re-examining units:** - Densities are in g/mL, masses in kg, volumes in mL. - Let's convert all to consistent units: mass in g, volume in mL, density in g/mL. 18. **Convert total mass to grams:** $$4.02 \text{ kg} = 4020 \text{ g}$$ 19. **Use densities in g/mL and volumes in mL:** $$\frac{m_{Ni}}{8.90} + \frac{4020 - m_{Ni}}{11.34} = 423$$ 20. **Multiply both sides by $8.90 \times 11.34 = 100.926$:** $$11.34 m_{Ni} + 8.90 (4020 - m_{Ni}) = 423 \times 100.926$$ 21. **Calculate right side:** $$423 \times 100.926 = 42,700.6$$ 22. **Expand left side:** $$11.34 m_{Ni} + 358,980 - 8.90 m_{Ni} = 42,700.6$$ 23. **Combine like terms:** $$(11.34 - 8.90) m_{Ni} + 358,980 = 42,700.6$$ $$2.44 m_{Ni} + 358,980 = 42,700.6$$ 24. **Subtract 358,980 from both sides:** $$2.44 m_{Ni} = 42,700.6 - 358,980 = -316,279.4$$ 25. **Solve for $m_{Ni}$:** $$m_{Ni} = \frac{-316,279.4}{2.44} = -129,600$$ g (impossible again) **This suggests a misinterpretation: the total volume is 423 mL, total mass 4.02 kg (4020 g), but densities are in g/mL. Since 4020 g mass in 423 mL volume implies average density $\approx 9.5$ g/mL, which is between nickel and lead densities, so calculation should be possible.** 26. **Rearrange volume equation:** $$\frac{m_{Ni}}{8.90} + \frac{4020 - m_{Ni}}{11.34} = 423$$ 27. **Multiply both sides by 8.90 and 11.34 to clear denominators:** Multiply entire equation by $8.90 \times 11.34 = 100.926$ $$11.34 m_{Ni} + 8.90 (4020 - m_{Ni}) = 423 \times 100.926$$ 28. **Calculate right side:** $$423 \times 100.926 = 42,700.6$$ 29. **Expand left side:** $$11.34 m_{Ni} + 358,980 - 8.90 m_{Ni} = 42,700.6$$ 30. **Combine like terms:** $$(11.34 - 8.90) m_{Ni} + 358,980 = 42,700.6$$ $$2.44 m_{Ni} + 358,980 = 42,700.6$$ 31. **Subtract 358,980 from both sides:** $$2.44 m_{Ni} = 42,700.6 - 358,980 = -316,279.4$$ 32. **Solve for $m_{Ni}$:** $$m_{Ni} = \frac{-316,279.4}{2.44} = -129,600$$ g (impossible) **This contradiction suggests the problem data or units might be inconsistent or the densities are not correctly applied.** **Assuming densities are in g/cm$^3$ (same as g/mL), mass in g, volume in mL:** 33. **Try solving for volume fraction instead:** Let $x$ = volume of nickel spheres (mL), then volume of lead spheres is $423 - x$ mL. Mass of nickel spheres: $$m_{Ni} = 8.90 x$$ Mass of lead spheres: $$m_{Pb} = 11.34 (423 - x)$$ Total mass: $$8.90 x + 11.34 (423 - x) = 4020$$ 34. **Expand:** $$8.90 x + 4795.82 - 11.34 x = 4020$$ 35. **Combine like terms:** $$-2.44 x + 4795.82 = 4020$$ 36. **Subtract 4795.82 from both sides:** $$-2.44 x = 4020 - 4795.82 = -775.82$$ 37. **Solve for $x$:** $$x = \frac{775.82}{2.44} = 318$$ mL 38. **Calculate mass of nickel:** $$m_{Ni} = 8.90 \times 318 = 2830.2$$ g 39. **Calculate percentage of nickel by mass:** $$\text{Percentage nickel} = \frac{2830.2}{4020} \times 100 = 70.4\%$$ **Final answer:** Approximately $70.4\%$ of the spheres are nickel by mass.