Please help me answer part VI Part VI. Balance the following oxidation reduction reactions by oxidation number method.1. NO2 + H2 - H2O + NH32. K+ KNO3- N2 + K203. N2H4 + H2O2 - HNO3 + H2O4. H2SO4 + HI → H2S + 12+ H2O5. KMNO4 + HCI - MnCl2 + Cl2 + KCI + H2O6. Zn + NO3 → Zn?2* + NHa" (in acidic solution)7. MnOs + H2S - Mn2* + S (in acidic solution)8. Cu + NO3 - Cu2* + NO (in acidic solution)9. Al + OH AIO2- + H2 (in basic solution)10. MnO4 + C2O42- CO2 + MnO2 (in basic solution)Part VII. Write the balanced half reactions of the following reactions.1. NIO2 + 2 H2O + Fe Ni(OH)2 + Fe(OH)2 (in basic solution)2. 2 H* + H2O2 + 2 Fe2* 2 Fe* + 2 H2O (in acidic solution)3. CO2 + 2 NH2OH CO + N2 + 3 H20 (in basic solution)4. H* + 2 H20 + 2 MnO4 + 5 SO2 2 Mn2 + 5 HSO4 (in acidic solution)

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Answered: Part VI. Balance the following…

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter3: Chemical Reactions

Section: Chapter Questions

Problem 47PS

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In balancing oxidation-reduction equations, why is it permissible to add water to either side of the equation?
Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting Cr3+ to Cr2O72 with the peroxydisulfate ion: S2O82(aq) + Cr3+(aq) + H2O(l) Cr2O72(aq) + SO42(aq) + H+(aq) (Unbalanced) After removal of unreacted S2O82 an excess of ferrous ammonium sulfate [Fe(NH4)2(SO4)26H2O] is added, reacting with Cr2O72 produced from the first reaction. The unreacted Fe2+ from the excess ferrous ammonium sulfate is titrated with a separate K2Cr2O7 solution. The reaction is: H+(aq) + Fe2+(aq) + Cr2O72(aq) Fe3+(aq) + Cr3+(aq) + H2O(l) (Unbalanced) a. Write balanced chemical equations for the two reactions. b. In one analysis, a 40.0-cm2 sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess S2O82, 3.000 g of Fe(NH4)2(SO4)26H2O was added. It took 8.58 mL of 0.0520 M K2Cr2O7 solution to completely react with the excess Fe2+. Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 g/cm3)
Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen. 4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O() 4 NaAu(CN)2(aq) + 4 NaOH(aq) (a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced? (b) If you have exactly one metric ton (1 metric ton = 1000 kg) of gold-bearing rock, what volume of 0.075 M NaCN, in liters, do you need to extract the gold if the rock is 0.019% gold?
In theory, most metals should easily corrode in air. Why? A group of metals called the noble metals are relatively difficult to corrode in air. Some noble metals include gold, platinum, and silver. Reference Table 17-1 to come up with a possible reason why the noble metals are relatively difficult to corrode.
In balancing oxidation-reduction reactions, why is it permissible to add water to either side of the equation?
You have two 500.0-mL aqueous solutions. Solution A is a solution of a metal nitrate that is 8.246% nitrogen by mass. The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of + 6 and there are 2 potassiums and 1 chromium in the formula. The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction bas gone to completion, you dry the solid and find that it has a mass of 331.8 g. a. Identify the ionic compounds in solution A and solution B. b. Identify the blood-red precipitate. c. Calculate the concentration (molarity) of all ions in the original solutions. d. Calculate the concentration (molarity) of all ions in the final solution.

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