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Common aluminum foil for household use is nearly pure aluminum. A box of this product at a local supermarket is advertised as giving 75 ft'of material (in a roll 304 mm wide by 22.8 m long). If the foil is 0.5 mil (12.7 µm) thick, calculate the number of atoms of aluminum in the roll. Given: Density of Aluminum: p = 2.7 Mg/m Molar mass of Aluminum: MAI= 26.98 /mol Avogadro’s Number: AN = 0.6203 (10)´

Common aluminum foil for household use is nearly pure aluminum. A box of this product at a local supermarket is advertised as giving 75 ft'of material (in a roll 304 mm wide by 22.8 m long). If the foil is 0.5 mil (12.7 µm) thick, calculate the number of atoms of aluminum in the roll. Given: Density of Aluminum: p = 2.7 Mg/m Molar mass of Aluminum: MAI= 26.98 /mol Avogadro’s Number: AN = 0.6203 (10)´

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Question 1 (Problem 2.2) Common aluminum foil for household use is nearly pure aluminum. A box of this product at a local supermarket is advertised as giving 75 ft'of material (in a roll 304 mm wide by 22.8 m long). If the foil is 0.5 mil (12.7 µm) thick, calculate the number of atoms of aluminum in the roll. Given: Density of Aluminum: p= 2.7 Mg/m³ Molar mass of Aluminum: MAI= 26.98 g/mol Avogadro's Number: AN = 0.6203 (10)² Solution:
Transcribed Image Text:Question 1 (Problem 2.2) Common aluminum foil for household use is nearly pure aluminum. A box of this product at a local supermarket is advertised as giving 75 ft'of material (in a roll 304 mm wide by 22.8 m long). If the foil is 0.5 mil (12.7 µm) thick, calculate the number of atoms of aluminum in the roll. Given: Density of Aluminum: p= 2.7 Mg/m³ Molar mass of Aluminum: MAI= 26.98 g/mol Avogadro's Number: AN = 0.6203 (10)² Solution:
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