For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy Sof the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column.Note for advanced students: you may assume ideal gas and ideal solution behaviour.SystemChangeASO AS < 0The seawater is passed through areverse-osmosis filter, whichAS = 0A liter of seawater at 15°C.separates it into 750. mL ofpureO AS > 0water and 250. mL of brine (verysalty water).not enoughinformationAS < 0300 mL of a solution made from0.5 g of Nal crystallizes out of theO AS = 0sodium iodide (Nal) dissolved insolution, without changing thetemperature.O AS > 0water.not enoughinformationAS < 0A mixture of carbon dioxide (CO,)An additional 2.0 L of pure Kr gasAS = 0gas and krypton (Kr) gas at 3 atmis added to the mixture, with theAS > 0and 41°C.pressure kept constant at 3 atm.not enoughinformation

Initially water was present as a mixture and we know mixture has always higher entropy then it gets…

For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. Note for advanced students: you may assume ideal gas and ideal solution behaviour. System Change AS O AS < 0 The seawater is passed through a reverse-osmosis filter, which O AS = 0 A liter of seawater at 15°C. separates it into 750. mL of pure O AS > 0 water and 250. mL of brine (very salty water). not enough information O AS < 0 300 mL of a solution made from 0.5 g of Nal crystallizes out of the O AS = 0 sodium iodide (Nal) dissolved in solution, without changing the O AS > 0 water. temperature. not enough information O AS < 0 A mixture of carbon dioxide (CO,) O AS = 0 An additional 2.0 L of pure Kr gas is added to the mixture, with the gas and krypton (Kr) gas at 3 atm O AS > 0 pressure kept constant at 3 atm. and 41°C. not enough information O o o O

For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. Note for advanced students: you may assume ideal gas and ideal solution behaviour. System Change AS O AS < 0 The seawater is passed through a reverse-osmosis filter, which O AS = 0 A liter of seawater at 15°C. separates it into 750. mL of pure O AS > 0 water and 250. mL of brine (very salty water). not enough information O AS < 0 300 mL of a solution made from 0.5 g of Nal crystallizes out of the O AS = 0 sodium iodide (Nal) dissolved in solution, without changing the O AS > 0 water. temperature. not enough information O AS < 0 A mixture of carbon dioxide (CO,) O AS = 0 An additional 2.0 L of pure Kr gas is added to the mixture, with the gas and krypton (Kr) gas at 3 atm O AS > 0 pressure kept constant at 3 atm. and 41°C. not enough information O o o O

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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1/5 For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. Note for advanced students: you may assume ideal gas and ideal solution behaviour. System Change AS O AS 0 not enough information O AS 0 and 34°C. pressure kept constant at 4 atm. not enough O. information O AS 0 dissolved in water. not enough information Enplanation Check Terms of Use | Privacy Center | Accessibilit 02022 McGraw Hill LLC. All Rights Reserved. DE O O O O O
the free energy for a reaction can be related to the equilibriumconstant through the formula below.K = e (-ΔG° / RT)Therefore if Kc for a reaction is known, Go can be determined, or vice versa. Furthermore, ifyou have the value for Go at two different temperatures, you can calculate H and S throughthe familiar equation for Gibbs energy below, since you have two unknowns but also twoequations.G = H – T SIn this lab you will be studying the solubility of borax (Na2B4O5(OH)4*8H2O), a slightly solublesodium salt, at two different temperatures. When solid borax is added to water, theequilibrium below is established.Na2B4O5(OH)4*8H2O (s) 2 Na+ (aq) + B4O5(OH)42- (aq) + 8 H2O(l)If you measure the concentrations for those substances that show up in the reaction quotient,then the Kc for the reaction at that temperature can be calculated. In this lab, theconcentration of borate ion (B4O5(OH)42-) in solution will be measured by titration with standardhydrochloric acid according to the…
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The equilibrium constant for a system at chemical equilibrium at 318 oC is 3.4 x 10-3. Calculate the Gibbs free-energy change, deltaGo, for the system at 318 oC.Enter your answer here: kJ/mol
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Discuss the effect on the entropy, enthalpy, heat capacity, and freezing point in a solution of water when a small amount of the following substances are added. Be sure to discuss the chemical reasoning for the observed change. Images are encouraged. Benzene Sodium chloride Sodium dodecyl sulfate
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Can u help me work this problem. I didn’t write notes in how I got the correct answer (2nd option).
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At 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 4 B(g) 2C(g) + 2D(g) were found to be PA = 4.67 atm, På = 5.73 atm, Pc = 5.60 atm, and Pp = 5.84 atm. What is the standard change in Gibbs free energy of this reaction at 25 °C? kJ mol privacy policy contact us help terms of use 6:11 PM about us careers 3/25/202
This question has multiple parts. Work all the parts to get the most points. Use the References to access important values if needed for this question. The Gibbs Free Energy equation is given by the equation: AG=AG + RTln Q - Where: AG = Gibbs Free Energy change AG = Standard Gibbs Free Energy change T= temperature In Q = natural log of the reaction quotient In order to solve for the temperature, T, in two steps you must: Step One Add the same expression to each side of the equation to leave the term that includes the variable by itself on the right-hand side of the expression: (Be sure that the answer field changes from light yellow to dark yellow before releasing your answer.) +AG = +AG + RT In Q Drag and drop your selection from the following list to complete the answer: 1 -AG AG AGⓇ AGⓇ Submit