**Title:** Calculating Lead Concentration in Drinking Water**Educational Content:**---**Problem Statement:**The maximum quantity of lead allowed in drinking water is 0.05 ppm. What mass of lead could be found in 5,000 kg of water?**Solution Requirement:**mass Pb²⁺ = [ ? ] g**Input Box:** (for student responses)```mass Pb²⁺ _____________ [Enter]```---**Explanation:****Step-by-Step Calculation:**1. **Understanding ppm:** ppm stands for parts per million. It is a way of expressing very dilute concentrations of substances. One ppm means one part of substance in one million parts of the solution.2. **Given Data:** - Maximum lead concentration = 0.05 ppm - Total mass of water = 5,000 kg3. **Conversion of water mass to grams:** - 1 kg = 1,000 grams - Therefore, 5,000 kg = 5,000,000 grams4. **Calculating the mass of lead:** - 0.05 ppm indicates 0.05 grams of lead in 1,000,000 grams of water. - Hence, the mass of lead in 5,000,000 grams of water is calculated by setting up a proportion: \[ \frac{0.05 \text{ grams Pb}}{1,000,000 \text{ grams water}} \times 5,000,000 \text{ grams water} = x \text{ grams Pb} \]5. **Performing the Calculation:** - \[ x = \frac{0.05 \times 5,000,000}{1,000,000} \] - \[ x = \frac{250,000}{1,000,000} \] - \[ x = 0.25 \text{ grams Pb} \]Thus, the mass of lead that could be found in 5,000 kg of water is 0.25 grams.---**Interactive Response:**Enter your calculation in the box provided and press "Enter" to check your answer.

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Answered: The maximum quantity of lead allowed in…

The maximum quantity of lead allowed in drinking water is 0.05 ppm. What mass of lead could be found in 5,000 kg of water? mass Pb²+ = [?] g mass Pb2+ Enter

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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