×
Log in to StudySoup
Get Full Access to Chemistry - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Chemistry - Textbook Survival Guide

The average concentration of gold in seawater is 100 fM

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 111 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

4 5 0 320 Reviews
30
2
Problem 111

The average concentration of gold in seawater is 100 fM (femtomolar). Given that the price of gold is $1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect $5000 worth of gold, assuming your processing technique captures only 50% of the gold present in the samples?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the volume of seawater needed to process to collect $5000 worth of gold. Step1: Given, cost of 1 troy ounce of gold = $1764.20 i.e., For $1764.20, quantity of gold collected = 1 troy ounce Therefore, for $5000, quantity of gold collected = 1 X 5000 [(troy ounce / $) X $] 1764.20 = 2.834 troy ounce = 2.834 x 31.103 g [1 troy ounce = 31.103 g] = 88.146 g Step2: Since, the processing technique captures only 50% of the gold present in the sample,...

Step 2 of 3

Chapter 4, Problem 111 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

This full solution covers the following key subjects: gold, seawater, Troy, ounce, assuming. This expansive textbook survival guide covers 82 chapters, and 9464 solutions. Since the solution to 111 from 4 chapter was answered, more than 357 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 111 from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. The answer to “The average concentration of gold in seawater is 100 fM (femtomolar). Given that the price of gold is $1764.20 per troy ounce (1 troy ounce = 31.103 g), how many liters of seawater would you need to process to collect $5000 worth of gold, assuming your processing technique captures only 50% of the gold present in the samples?” is broken down into a number of easy to follow steps, and 58 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247.

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

The average concentration of gold in seawater is 100 fM

×
Log in to StudySoup
Get Full Access to Chemistry - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Chemistry - Textbook Survival Guide
×
Reset your password