95011&OpenVellumHMAC=712fec996d2e4be570362447e9826797#100.. Part C Cl₂(g) + NaOH(aq) → NaCl(aq) + NaClO3(aq) + H₂O(1) Express your answers as integers separated by commas. Submit Request Answer Part D MgCO3(aq) + HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g) Express your answers as integers separated by commas. Submit Request Answer Provide Feedback © Pearson

95011&OpenVellumHMAC=712fec996d2e4be570362447e9826797#100.. Part C Cl₂(g) + NaOH(aq) → NaCl(aq) + NaClO3(aq) + H₂O(1) Express your answers as integers separated by commas. Submit Request Answer Part D MgCO3(aq) + HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g) Express your answers as integers separated by commas. Submit Request Answer Provide Feedback © Pearson

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter3: Measurement And Chemical Calculations

Section: Chapter Questions

Problem 128E: In Active Example 3-29 you calculated that you would have to work six weeks to earn enough money to...

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Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
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