Explanation Given: Galena contains 86.6 % lead by mass; an ore contains 68.5 % galena by mass. The efficiency of extraction of lead is 92.5 % and the radius of the lead sphere is 5.00 cm . Volume of sphere ( V ) is given as follows: V = 4 3 π r 3 …… (1) Here, r is the radius of lead sphere. Substitute π as 3.14 and r as 5.00 cm in equation (1) as follows: V = 4 3 × 3.14 × 5.00 cm × 5.00 cm × 5.00 cm =523 .34 cm 3 Density of lead ( ρ ) is 11.34 g/cm 3 . Calculate the mass of lead ( M ) as follows: M = ρ × V Substitute V as 523 .34 cm 3 and ρ as 11.34 g/cm 3 in the above expression as follows: M = 11.34 g/cm 3 × 523 .34 cm 3 = 5934.68 g Assume the total mass to be m g

Chemistry: A Molecular Approach, Books a la Carte Edition; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card -- for Chemistry: A Molecular Approach, 4/e

1st Edition

ISBN: 9780134465654

Author: Tro

Publisher: Pearson Education

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Chapter 1, Problem 130E

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Interpretation: The mass of ore required to make a lead sphere is to be determined.

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Chapter 1, Problem 129E

Chapter 1, Problem 131E

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Chemistry: A Molecular Approach, Books a la Carte Edition; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card -- for Chemistry: A Molecular Approach, 4/e

Ch. 1 - A chemist mixes sodium with water and witnesses a...Ch. 1 - Q2. This image represents a particulate view of a...Ch. 1 - Which change is a physical change? a) Wood burning...Ch. 1 - Q4. Which property of rubbing alcohol is a...Ch. 1 - Convert 85F to K. a) 181.1 K b) 358 K c) 29.4 K d)...Ch. 1 - Express the quantity 33.2 10?4 m in mm. a) 33.2...Ch. 1 - Q7. What is the mass of a 1.75 L sample of a...Ch. 1 - Perform the calculation to the correct number of...Ch. 1 - Q9. Perform the calculation to the correct number...Ch. 1 - Q10. Convert 1285 cm2 to m2. a) 1.285 × 107...

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