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Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion. 4AI( s) + 30 2( g) 2Al 20 3( s) Calculate AG° for this reaction, given that AG° f of aluminum oxide is -1576.4 kJ/mol. 2-3152.8 kJ/mol O -1576.4 kJ/mol O 3152.8 kJ/mol -788.2 kJ/mol 1576.4 kJ/mol

Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion. 4AI( s) + 30 2( g) 2Al 20 3( s) Calculate AG° for this reaction, given that AG° f of aluminum oxide is -1576.4 kJ/mol. 2-3152.8 kJ/mol O -1576.4 kJ/mol O 3152.8 kJ/mol -788.2 kJ/mol 1576.4 kJ/mol

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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Question 16

**Question 16** Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion. \[4\text{Al}(s) + 3\text{O}_2(g) \rightarrow 2\text{Al}_2\text{O}_3(s)\] Calculate \(\Delta G^\circ\) for this reaction, given that \(\Delta G^\circ_f\) of aluminum oxide is \(-1576.4 \text{ kJ/mol}\). - \(-3152.8 \text{ kJ/mol}\) - \(-1576.4 \text{ kJ/mol}\) - \(3152.8 \text{ kJ/mol}\) - \(-788.2 \text{ kJ/mol}\) - \(1576.4 \text{ kJ/mol}\) **Question 17** Calculate the pH at the equivalence point for the titration of \(0.20 \text{ M HCl}\) with \(0.20 \text{ M NH}_3\) (\(K_b = 1.8 \times 10^{-5}\)). - \(5.12\) - \(7.00\) - \(2.87\) - \(11.12\) - \(4.98\)
Transcribed Image Text:**Question 16** Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion. \[4\text{Al}(s) + 3\text{O}_2(g) \rightarrow 2\text{Al}_2\text{O}_3(s)\] Calculate \(\Delta G^\circ\) for this reaction, given that \(\Delta G^\circ_f\) of aluminum oxide is \(-1576.4 \text{ kJ/mol}\). - \(-3152.8 \text{ kJ/mol}\) - \(-1576.4 \text{ kJ/mol}\) - \(3152.8 \text{ kJ/mol}\) - \(-788.2 \text{ kJ/mol}\) - \(1576.4 \text{ kJ/mol}\) **Question 17** Calculate the pH at the equivalence point for the titration of \(0.20 \text{ M HCl}\) with \(0.20 \text{ M NH}_3\) (\(K_b = 1.8 \times 10^{-5}\)). - \(5.12\) - \(7.00\) - \(2.87\) - \(11.12\) - \(4.98\)
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