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Balance the following using the half-reaction method. Identify oxidation, reduction, reducing agent, oxidizing agent. i. Be(NO3)2 (aq) + Ce(s)--> Ce(NO3)3(aq) + Be(s)

Balance the following using the half-reaction method. Identify oxidation, reduction, reducing agent, oxidizing agent. i. Be(NO3)2 (aq) + Ce(s)--> Ce(NO3)3(aq) + Be(s)

Chemistry
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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Balance the following using the half-reaction method. Identify oxidation, reduction, reducing agent, oxidizing agent.

i. Be(NO3)2 (aq) + Ce(s)--> Ce(NO3)3(aq) + Be(s)

ii. A container contains H2SO3, H2SO4, Pb2+, Pb4+ and NO3- in distilled water. Create a balanced chemical reaction (Net Ionic Equation, including states of matter) for the spontaneous reaction in the container

AH( 1 (K2 In K1 1 [A] = [A]o - kt r=k = (RT)A"K. R (T1 аА + bB cС + dD, AG, = AG? + RT InQ -b + b? – 4ac X = ([A¯] Ereduction pH = pKa+ log |[HA]) cell + Eoxidation 2а pH = -log([H*]) Ecell = Ecel1° (RT ln Q)/nF if T =298.15 K then Ecell = Ecel1° (0.025693V( 1n Q)/n) In K = nF Ecel1°/RT if T = 298.15 K then In K = n AG° = -nF Ecel1° Ecel1/0.025693V Some useful data Element/Ion/Molecu AH?, kJ/mol Element/Ion/Molecu AH?, kJ/mol le le Na*(aq) -240.12 NO, (aq) -205.0 Pb2* (aq) -1.7 PBSO,(s) -919.94 I- (aq) -55.19 CH;COO (aq) -486.0 NH,*(aq) S2- (aq) -132.51 PbS (s) -98.32 +33.1 Element m, g/mol Element т, s/mo1 Ва 137.327 H 1.0079 S 32.065 C1 35.453 12.0107 15.9994 Element/Ion/Molecu pKa Element/Ion/Molecu pKa le le H3PO4 2.12 H,PO, 7.21 HCN 9.31 HPO,2- 12.68 HSO, 1.92 NH,* 9.25 CH3COOH 4.75 HCO3 10.25 HI -10(estimate) H2CO3 6.37 HIO3 0.77 HNO, 3.37 Molecule Ksp Molecule Ksp 2|P age
Transcribed Image Text:AH( 1 (K2 In K1 1 [A] = [A]o - kt r=k = (RT)A"K. R (T1 аА + bB cС + dD, AG, = AG? + RT InQ -b + b? – 4ac X = ([A¯] Ereduction pH = pKa+ log |[HA]) cell + Eoxidation 2а pH = -log([H*]) Ecell = Ecel1° (RT ln Q)/nF if T =298.15 K then Ecell = Ecel1° (0.025693V( 1n Q)/n) In K = nF Ecel1°/RT if T = 298.15 K then In K = n AG° = -nF Ecel1° Ecel1/0.025693V Some useful data Element/Ion/Molecu AH?, kJ/mol Element/Ion/Molecu AH?, kJ/mol le le Na*(aq) -240.12 NO, (aq) -205.0 Pb2* (aq) -1.7 PBSO,(s) -919.94 I- (aq) -55.19 CH;COO (aq) -486.0 NH,*(aq) S2- (aq) -132.51 PbS (s) -98.32 +33.1 Element m, g/mol Element т, s/mo1 Ва 137.327 H 1.0079 S 32.065 C1 35.453 12.0107 15.9994 Element/Ion/Molecu pKa Element/Ion/Molecu pKa le le H3PO4 2.12 H,PO, 7.21 HCN 9.31 HPO,2- 12.68 HSO, 1.92 NH,* 9.25 CH3COOH 4.75 HCO3 10.25 HI -10(estimate) H2CO3 6.37 HIO3 0.77 HNO, 3.37 Molecule Ksp Molecule Ksp 2|P age
BaCO3 8.1 * 10-9 Ba(NO3), 4.64*10-3 BaSO4 1.1*10-10 Be(OH)2 6.92*10-22 PbI2 1.4 *10-8 AgI 8 * 10-17 CuI 5.1* 10-12 Ni(OH)2 2.8 * 10-16 Ni(OH)3 1 * 10-23 Reaction pKa H3C6H507 + H20 5 H30 + + H2C6H507 3.15 H2C6H507 + H20– H30 + + HC6H50?- 4.77 HCGH503- + H20 – H30 + + C6H503- 6.39 Molecule/element Boiling point at 1 Molecule/element Boiling point at 1 atm, °C atm, °C Не -269 Cl2 -34 N2 -196 H20 100 Density of Water Temperature(°C) | density(g/m1) | Temperature(°C) | density(g/m1) 20 0.99819 23 0.99753 21 0.99798 24 0.99730 22 0.99776 25 0.99705
Transcribed Image Text:BaCO3 8.1 * 10-9 Ba(NO3), 4.64*10-3 BaSO4 1.1*10-10 Be(OH)2 6.92*10-22 PbI2 1.4 *10-8 AgI 8 * 10-17 CuI 5.1* 10-12 Ni(OH)2 2.8 * 10-16 Ni(OH)3 1 * 10-23 Reaction pKa H3C6H507 + H20 5 H30 + + H2C6H507 3.15 H2C6H507 + H20– H30 + + HC6H50?- 4.77 HCGH503- + H20 – H30 + + C6H503- 6.39 Molecule/element Boiling point at 1 Molecule/element Boiling point at 1 atm, °C atm, °C Не -269 Cl2 -34 N2 -196 H20 100 Density of Water Temperature(°C) | density(g/m1) | Temperature(°C) | density(g/m1) 20 0.99819 23 0.99753 21 0.99798 24 0.99730 22 0.99776 25 0.99705
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