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In class, students were given the following problem: "Aluminum satellite dishes resist corrosion because aluminum reacts with oxygen gas (0₂) to form a coating of aluminum oxide (Al2O3) according to the following chemical equation: 4Al(s)+302(g) → 2Al2O3(s) What is the mass of 9.45 moles of Al 20 3? After correctly solving for the molar mass of Al2O3, Jacob set up the following conversion: 9.45 moles Al2O3* 102.2 grams of Al, 0, 1 mole Al, 0₁ Is this answer correct? = 0.0926 grams of Al2O3

In class, students were given the following problem: "Aluminum satellite dishes resist corrosion because aluminum reacts with oxygen gas (0₂) to form a coating of aluminum oxide (Al2O3) according to the following chemical equation: 4Al(s)+302(g) → 2Al2O3(s) What is the mass of 9.45 moles of Al 20 3? After correctly solving for the molar mass of Al2O3, Jacob set up the following conversion: 9.45 moles Al2O3* 102.2 grams of Al, 0, 1 mole Al, 0₁ Is this answer correct? = 0.0926 grams of Al2O3

Chemistry
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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In class, students were given the following problem: "Aluminum satellite dishes resist corrosion because aluminum reacts with oxygen gas (0₂) to form a coating of aluminum oxide (Al2O3) according to the following chemical equation: 4Al(s) + 302(g) →2A/203(s) What is the mass of 9.45 moles of Al 203? After correctly solving for the molar mass of Al2O3, Jacob set up the following conversion: 1 mole AL, Q 9.45 moles Al₂O3× 102.2 grams of Al, 0, Is this answer correct? = 0.0926 grams of Al2O3
Transcribed Image Text:In class, students were given the following problem: "Aluminum satellite dishes resist corrosion because aluminum reacts with oxygen gas (0₂) to form a coating of aluminum oxide (Al2O3) according to the following chemical equation: 4Al(s) + 302(g) →2A/203(s) What is the mass of 9.45 moles of Al 203? After correctly solving for the molar mass of Al2O3, Jacob set up the following conversion: 1 mole AL, Q 9.45 moles Al₂O3× 102.2 grams of Al, 0, Is this answer correct? = 0.0926 grams of Al2O3
A 1 mole = 6.02 x 10²³ particles ✓ N 6.02 x 10 particles 1 mole C 6.02 x 10¹ particles B 1 mole = molar mass* (grams) "S determined by summing the ma 1 of all elements in the formula 1 mole = 22.4 L of gas Inde 7241 22.4 L
Transcribed Image Text:A 1 mole = 6.02 x 10²³ particles ✓ N 6.02 x 10 particles 1 mole C 6.02 x 10¹ particles B 1 mole = molar mass* (grams) "S determined by summing the ma 1 of all elements in the formula 1 mole = 22.4 L of gas Inde 7241 22.4 L
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