By Maria E. Gamboa-Adelco, Robert J. Gale (auth.)
It has been regularly an incentive for college kids to discover no matter if his/her efforts to unravel routines provide right effects, or to discover suggestions for difficulties that he/she unearths more challenging. those are the most purposes for the looks of the current publication. As a part of the textbook ModernElectrochemistry 1: Ionics, A advisor to difficulties in ModernElectrochemistry: half 1: Ionics compiles a few of the suggestions to the routines and difficulties offered within the textual content, in addition to many new problems.
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Additional info for A Guide to Problems in Modern Electrochemistry: 1: Ionics
114) -I and the enthalpy of solution, from Eq. 115) The heat of solvation from Eq. 6 Since L1Gsoln > 0, then the three salts are insoluble, with AgI being the most insoluble salt. In the three cases i1Hsoln > 0, and the entropy change, L1Ssoln is not positive enough to make the process happen (see table above). The stable lattice of these salts --marked by the large negative value of i1Hlallice- makes it difficult for the solvent to grab the ions away. 37 The densities of aqueous NaCI solutions at 25°C are given as a function of NaCI molality in the table below.
00158kgdm -3 ) and Eq. 02142 amoINaclkg",/v bkg",,'/dm3"',. 0179 'dm 31mol (av m,2/ On 2) D. 2 vs. n2 curve. Plotting Vm•2 against n2 (or m) gives the graph in Fig. 4. 4. The apparent NaCI molar volume vs. 126) n1 Now we have the parameters to fmd V:1' Thus, substituting the corresponding values into Eq. 01781dm 3-1 mol (b) Once the values of the partial-molar volume of the solute are known, the partial-molar volume of water, VI' can be calculated from Eq. ::::.. 129) -1 The corresponding values of V2 and VI at different molalities of NaCI are given in the previous table.
113) 48 CHAPTER 2 Thus, for AgCI, the free energy of solution from Eq. 114) -I and the enthalpy of solution, from Eq. 115) The heat of solvation from Eq. 6 Since L1Gsoln > 0, then the three salts are insoluble, with AgI being the most insoluble salt. In the three cases i1Hsoln > 0, and the entropy change, L1Ssoln is not positive enough to make the process happen (see table above). The stable lattice of these salts --marked by the large negative value of i1Hlallice- makes it difficult for the solvent to grab the ions away.