Ch.12 - SolutionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A 2.90 L sample of water contains 217.5 μg of Lead (II) ions. Calculate the concentration of lead in ppm.A) 7.5 X 10‐1 ppmB) 7.5 X 10‐2 ppmC) 7.5 X 10‐3 ppmD) 7.5 X 10‐4 ppmE) 7.5 X 10+4 ppm

Problem

A 2.90 L sample of water contains 217.5 μg of Lead (II) ions. Calculate the concentration of lead in ppm.

A) 7.5 X 10‐1 ppm
B) 7.5 X 10‐2 ppm
C) 7.5 X 10‐3 ppm
D) 7.5 X 10‐4 ppm
E) 7.5 X 10+4 ppm