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Figure 3-13. You can manage the bridged adapters from the General tab of the Network Bridge Properties dialog box.
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The reaction shown above was used in an electrolytic cell. The voltage measured for the cell was not equal to the calculated E for the cell. This discrepancy could be caused by which of the following (A) Both of the solutions were at 25 C instead of 0 C. (B) The anode and cathode were different sizes. (C) The anion in the anode compartment was chloride instead of nitrate, as in the cathode compartment. (D) One or more of the ion concentrations was not 1 M. (E) The solution in the salt bridge was Na2SO4 instead of KNO3. Questions 79 80 are concerned with the following half-reaction in an electrolytic cell:
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9. A This is an acid-dissociation constant, thus the solution must he acidic (pH < 7). The pH of a 0.010 M strong acid would be 2.0. This is not a strong acid, so the pH must be above 2. 10. A A is the salt of a strong acid and a weak base; it is acidic. B and E are salts of a strong acid and a strong base; they are neutral. C and D are salts of a weak acid and a strong base; they are basic. The lowest pH would be the acidic choice. 11. E Sodium nitrite is a salt of a weak acid and a + strong base. Ions from strong bases, Na in this case, do not undergo hydrolysis, and do not affect the pH. Ions from weak acids, NO2 in this case, undergo hydrolysis to produce basic solutions. 12. B The presence of a strong acid, HNO3, would make this the most acidic (lowest pH). 13. E The weak acid and the weak base partially cancel each other to give a nearly neutral solution. 14. C Both A and C are buffers, because they have conjugate acid base pairs of either a weak acid (A) or a weak base (C). The weak acid buffer would have a pH below 7, and the weak base buffer would have a pH above 7. 15. A See the answer to question 14. 16. A The equilibrium constant expression is: Kb = 10 4.0 10 = [OH ][C6H5NH+ ]/[C6H5NH2]. 3 This expression becomes: (x)(x)/(1.0 x) = 4.0 10 2 10 10 , which simplifies to: x /1.0 = 4.0 10 . Taking the square root of each side gives: x = 5 2.0 10 = [OH ]. Since you can estimate the answer, no actual calculations need be done. 17. E The solubility-product constant expression 2+ 6 2 is: Ksp = [Zn ][IO3 ] = 4 10 . This may be 6 2+ 2 rearranged to: [IO3 ] = 4 10 /[Zn ]. Inserting the desired zinc ion concentration 6 6 2 gives: [IO3 ] = 4 10 /(1 10 ) = 4. Taking the square root of each side leaves a desired IO3 concentration of 2 M. 2 mol of KIO3 must be added to 1.00 L of solution to produce this concentration. Since you can estimate the answer, no actual calculations need be done.
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Address translation (routing) Admission control Minimal bandwidth control request processing Zone management Optional gatekeeper functions include the following:
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Figure 3-20 Output of MyFirstWindowsApplication
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