Answered: A 0.100 L Solution Is Prepared With…

Skip to main contentHomework Help is Here – Start Your Trial Now!SEARCHHomework help starts here!ASK AN EXPERTASKScienceChemistryA 0.100 L solution is prepared with initial concentrations of 4.0 × 10−3 M iodine I2 , 8.0×10−3 M iodide I– , and 5.0×10−3 M ascorbic acid C6H8O6 . After the second reaction goes to completion, what will the molar concentrations of iodide and ascorbic acid in the solution be?A 0.100 L solution is prepared with initial concentrations of 4.0 × 10−3 M iodine I2 , 8.0×10−3 M iodide I– , and 5.0×10−3 M ascorbic acid C6H8O6 . After the second reaction goes to completion, what will the molar concentrations of iodide and ascorbic acid in the solution be?ReportChemistryBUYChemistry 10th EditionISBN: 9781305957404Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher: Cengage Learning1 Chemical Foundations2 Atoms, Molecules, And Ions3 Stoichiometry4 Types Of Chemical Reactions And Solution Stoichiometry5 Gases6 Thermochemistry7 Atomic Structure And Periodicity8 Bonding: General Concepts9 Covalent Bonding: Orbitals10 Liquids And Solids11 Properties Of Solutions12 Chemical Kinetics13 Chemical Equilibrium14 Acids And Bases15 Acid-base Equilibria16 Solubility And Complex Ion Equilibria17 Spontaneity, Entropy, And Free Energy18 Electrochemistry19 The Nucleus: A Chemist's View20 The Representative Elements21 Transition Metals And Coordination Chemistry22 Organic And Biological MoleculesChapter QuestionsProblem 1RQProblem 2RQProblem 3RQProblem 4RQProblem 5RQProblem 6RQProblem 7RQProblem 8RQProblem 9RQProblem 10RQProblem 1ALQProblem 2ALQProblem 3ALQProblem 4ALQProblem 5ALQProblem 6ALQProblem 7ALQProblem 8ALQProblem 9ALQProblem 10ALQProblem 11ALQProblem 14ALQProblem 15ALQProblem 16ALQProblem 17ALQProblem 18ALQProblem 19QProblem 20QProblem 21QProblem 22QProblem 23QProblem 24QProblem 25QProblem 26QProblem 28QProblem 29QProblem 30QProblem 31EProblem 32EProblem 33EProblem 34EProblem 35EProblem 36EProblem 37EProblem 38EProblem 39EProblem 40EProblem 41EProblem 42EProblem 43EProblem 44EProblem 45EProblem 46EProblem 47EProblem 48EProblem 49EProblem 50EProblem 51EProblem 52EProblem 53EProblem 54EProblem 55EProblem 56EProblem 57EProblem 58EProblem 59EProblem 60EProblem 61EProblem 62EProblem 63EProblem 64EProblem 65EProblem 66EProblem 67EProblem 68EProblem 69EProblem 70EProblem 71EProblem 72EProblem 73EProblem 74EProblem 75EProblem 76EProblem 77EProblem 78EProblem 79EProblem 80EProblem 81EProblem 82EProblem 83EProblem 84EProblem 85EProblem 86EProblem 87EProblem 88EProblem 89EProblem 90EProblem 91EProblem 92EProblem 93AEProblem 94AEProblem 95AEProblem 96AEProblem 97AEProblem 98AEProblem 99AEProblem 100AEProblem 101AEProblem 102AEProblem 103AEProblem 104AEProblem 105AEProblem 106AEProblem 107AEProblem 108AEProblem 109AEProblem 110AEProblem 111CWPProblem 112CWPProblem 113CWPProblem 114CWPProblem 115CWPProblem 116CWPProblem 117CPProblem 118CPProblem 119CPProblem 120CPProblem 121CPProblem 122CPProblem 123CPProblem 124CPProblem 125CPSee similar textbooksBartleby Related Questions Icon

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bartlebyIonic EquilibriumChemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibri…Arrhenius AcidArrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.Bronsted Lowry Base In Inorganic ChemistryBronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.Question

2I– (aq) + H2O2 (aq) + 2H3O + (aq) → I2 (aq) + 4H2O (l) (slow)

C6H8O6 (aq) + 2H2O (l) + I2 (aq) → C6H6O6 (aq) + 2H3O + (aq) + 2I– (aq) (very fast)

I2 (aq) + I– (aq) ⇌ I3- (aq)

I3- (aq) + starch → blue I3- · starch complex (aq) (fast)

A 0.100 L solution is prepared with initial concentrations of 4.0 × 10−3 M iodine I2 , 8.0×10−3 M iodide I– , and 5.0×10−3 M ascorbic acid C6H8O6 . After the second reaction goes to completion, what will the molar concentrations of iodide and ascorbic acid in the solution be?

Expert SolutionCheck MarkStep 1

Volume of Solution = 0.100 L

[I2]initial = 4.0 * 10-3 M

[I-]initial = 8.0 * 10-3 M

[C6H8O6]initial = 5.0 * 10-3M

I2 (aq) + I-(aq)--------> I3- (aq)

Therefore, I3- formed = 4.0 * 10-3M

(0.004 M iodine combines with 0.004 moles iodide and form 0.004 moles triiodide ion)

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  • (ii) At elevated temperatures, nitrogen dioxide decomposes to nitric oxide and molecular oxygen: 2NO2(g) – → 2NO(g) + O2(g) The change in concentration of NO2 with time at 300 °C is given below: t (s) 100 200 300 400 500 [NO2]/103 (mol dm-3) 8.0 5.6 4.3 3.5 2.9 2.5 By appropriate plotting of the data and use of this graph: (a) verify that this reaction is second order with respect to NO2; (b) determine the rate constant for this reaction.Use this data to determine the value of the rate constant. Ex #           [A] (M)          [B] (M)             Rate (M/hr) 1                  .240              .120                 2.00 2                  .120              .120                  .500 3                 .240               .0600              1.00 Question 4 options:   289 per hour   579 per hour   8.3 per hour   0.0144 per hour   69.4 per hour   .00346 per hour   0.060 per hour   0.0173 per hour   17 per hour   0.12 per hourConsider this initial rate data at a certain temperature in the table for the reaction OH (aq) OCI (aq) + I (aq) OI (aq) + Cl(aq) Determine the rate law. rate k [OC] (1) Answer Bank (1-1² [OH Trial 1 2 3 4 [I Jo [OCI ]o (M) [OH lo (M) (M) 0.00171 0.00171 0.510 0.00171 0.00299 0.510 0.00279 0.00171 0.643 0.00171 0.00299 0.826 [OCI"] [OH-] Initial rates (M/s) 0.000359 0.000629 0.000465 0.000388
  • 10. CHALLENGE. Calculate AHrxn. N2H4(1) + O2(g) → N2(g) + 2 H₂O(1) given: 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(D) N2O(g) + 3 H2(g) → N2H4(D) + H₂O(1) ->> 2 NH3(g) + 12 O2(g) → N2H4(1) + H₂O(1) H₂(g) + O2(g) → H₂O(n) AH = -1013 kJ ΔΗ = -317 kJ AH = -142.9 kJ ΔΗ = -285.8 kJIf I am given the attached information how would I go about finding the rate order for Iodine?Use the following equations (picture) to determine the deltaH ° for the target reaction. Target: N2H2(l)+2H2O2(l)——> N2(g)+4H2O(l)
  • 1:41 1 .5Gc NH, (aq) B) H+ (aq) + OH¯ (aq) → H₂O (1) C) NH3 (aq) + HI (aq) → NHÂI (aq) D) NH₂ (aq) → NH3 (aq) + H+ (aq) Tap here or pull up for additional resourcesWhat is the reaction constant and the rate law of the reaction in the table?Calculate AH for the reaction below
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