(a)
Interpretation:
The oxidation number for nonmetal X in
Concept introduction:
The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with identical elements have zero oxidation state.
Answer to Problem 1CE
The oxidation number for non metal X in
Explanation of Solution
In
Zero is the oxidation number for non metal X in
(b)
Interpretation:
The oxidation number for non metal X in
Concept introduction:
The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.
Answer to Problem 1CE
The oxidation number for non metal X in
Explanation of Solution
In
Assume the oxidation number of
Calculate the oxidation number of
Therefore, the oxidation number of X in
The oxidation number for non metal X in
(c)
Interpretation:
The oxidation number for nonmetal X in
Concept introduction:
The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.
Answer to Problem 1CE
The oxidation number for nonmetal X in
Explanation of Solution
In
Assume the oxidation number of
Calculate the oxidation number of
Therefore, the oxidation number of X in
The oxidation number for nonmetal X in
(d)
Interpretation:
The oxidation number for nonmetal X in
Concept introduction:
The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.
Answer to Problem 1CE
The oxidation number for non metal X in
Explanation of Solution
In
Assume the oxidation number of
Calculate the oxidation number of
Therefore, the oxidation number of X in
The oxidation number for non metal X in
Want to see more full solutions like this?
Chapter 17 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
- Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus (P4) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate (H2PO4-) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.arrow_forwardGiven HFH+(aq)+F(aq)Ka=6.9104HF(aq)+F(aq)H2(aq)K=2.7 calculate K for the reaction 2HF(aq)H+(aq)+HF2(aq)arrow_forwardUsing circles to represent cations and squares to represent anions, show pictorially the reactions that occur between aqueous solutions of (a) Fe3+ and OH-. (b) Na+ and PO43-.arrow_forward
- Write a balanced net ionic equation for the disproportionation reaction of (a) hypochlorous acid to chlorine gas and chlorous acid in acidic solution. (b) chlorate ion to perchlorate and chlorite ions.arrow_forwardPhosphate buffers are important in regulating the pH of intracellular fluids. If the concentration ratio of H2PO4/HPO42 in a sample of intracellular fluid is 1.1: 1, what is the pH of this sample of intracellular fluid? H2PO4(aq)HPO42(aq)+H+(aq)Ka=6.2108arrow_forward. What is the oxidation state of chlorine in each of the following substances? a. CIF c. HCI b. Cl2 d. HClOarrow_forward
- 4.22 Generally, an excess of O2 is needed for the reaction Sn+O2SnO2 . What is the minimum number of moles of oxygen required to oxidize 7.3 moles of tin?arrow_forwardWhat volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl3 is an excess of water? Note that when H3PO3 i5 titrated under these conditions, only one proton of the acid molecule reacts.arrow_forwardCalculate the molarity of AgNO3 in a solution prepared by dissolving 1.44 g AgNO3 in enough water to form 1.00 L solution.arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning