{"id":2728,"date":"2016-08-24T17:27:38","date_gmt":"2016-08-24T17:27:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2728"},"modified":"2016-08-26T18:44:50","modified_gmt":"2016-08-26T18:44:50","slug":"dissociation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/dissociation\/","title":{"raw":"Dissociation","rendered":"Dissociation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NGRlYTMyNzI4ZjE0YzZmMDdjOTliNDliOWRmZGQ0MjM.-hoq\">\r\n \t<li>Define dissociation.<\/li>\r\n \t<li>Be able to write equations for dissociation of ionic compounds.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Have you ever seen trucks pour salt on icy roads?<\/h3>\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212343\/20140811155540779952.jpeg\" alt=\"In order to function, deicing salts must first dissociate into their component ions\" width=\"400\" height=\"375\" \/> Runner on a Frozen Road. Courtesy of <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_40255_-_Runner_on_a_frozen_road_in_Fargo,_North_Dakota.jpg\">FEMA<\/a>.[\/caption]\r\n<p id=\"x-ck12-YjgyY2JkNzliYzViNmExNWQ3NzRmNDRjY2RmOGFkMjA.-jlu\">In many areas, ice on the streets and sidewalks represent a serious walking and driving hazard.\u00a0 One common approach to melting the ice is to put some form of deicing salt on the surface.\u00a0 Materials such as sodium chloride or calcium chloride are frequently employed for this purpose.\u00a0 In order to be effective, the solid material must first dissolve and break up into the ions that make up the compound.<\/p>\r\n\r\n<\/div>\r\n<h2>Dissociation<\/h2>\r\n<p id=\"x-ck12-YTY5YWJmNWJmMWFhMGE5NTQyYThmM2M5NmUwYWFkODA.-e2a\">An ionic crystal lattice breaks apart when it is dissolved in water. \u00a0<strong>Dissociation <\/strong> is the separation of ions that occurs when a solid ionic compound dissolves.\u00a0 It is important to be able to write dissociation equations.\u00a0 Simply undo the crisscross method that you learned when writing chemical formulas of ionic compounds.\u00a0 The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equation.\u00a0 Shown below are dissociation equations for NaCl, Ca(NO<sub>3<\/sub>)<sub>2 <\/sub> , and (NH<sub>4<\/sub>)<sub>3<\/sub>PO<sub>4<\/sub>.<\/p>\r\n<p id=\"x-ck12-ZDlhMzhhZDQ4YmNjYzk3MzQ3ODNiNTg1M2YxZjg2Yzc.-n2h\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">NaCl(<em>s<\/em>) \u2192\u00a0Na<sup>+ <\/sup> (<em>aq<\/em>) + Cl<sup>- <\/sup> (<em>aq<\/em>)<\/p>\r\n<p id=\"x-ck12-ZmJkZTMxYjM3ZGM2NThiNDdjMmJjOWZhY2M3MDllYTc.-4xe\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">Ca(NO<sub>3<\/sub>) <sub> 2 <\/sub> (<em>aq<\/em>)\u00a0\u2192 Ca<sup>2+ <\/sup> (<em>aq<\/em>) + 2NO<sub>3 <\/sub><sup> - <\/sup> (<em>aq<\/em>)<\/p>\r\n<p id=\"x-ck12-OGMzZjBlNGE3OWJkZjVmYjQ1NzJjM2FlODMwZmZlNTk.-bcl\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">(NH<sub>4<\/sub>)<sub>3 <\/sub> PO<sub>4 <\/sub> (<em>s<\/em>) \u2192\u00a03NH<sub>4 <\/sub><sup> + <\/sup> (<em>aq<\/em>) + PO<sub>4 <\/sub><sup> 3- <\/sup> (<em>aq<\/em>)<\/p>\r\n<p id=\"x-ck12-MmYyM2E2ODJhMmQzOGU2OGY0ZmRkYTU2M2QwZDgwOGI.-sfg\">The formula unit of sodium chloride dissociates into one sodium ion and one chloride ion.\u00a0 The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions.\u00a0 This is because of the 2+ charge of the calcium ion.\u00a0 Two nitrate ions, each with a 1\u2212 charge are required to make the equation balance electrically.\u00a0 The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion.\u00a0 Note that the polyatomic ions themselves do not dissociate further, but remain intact.<\/p>\r\n<p id=\"x-ck12-ZjAxMmE0MGVhYTk2ZGY4OWFmNjM2N2Y3YzMxOTZkNzA.-tkq\">Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make the original compound neutral.\u00a0 The 3 subscript of the nitrate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and simply remain as part of its formula after the compound dissociates.\u00a0 Notice that the compounds are solids ( <em> s <\/em> )\u00a0which then become ions in aqueous solution ( <em> aq <\/em> ).<\/p>\r\n\r\n<div id=\"x-ck12-Yzg4MTYyNmNjNmUxZmJkMGU1NzAwM2UzYzg5OGFjMWI.-92c\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2MTk0ODY2Mi03Ni05NS1DLUludENoLTA0LTAzLTA4LUNhbGNpdW0tTml0cmF0ZQ..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212345\/20140811155540873625.jpeg\" alt=\"Calcium nitrate dissociates into calcium ions and nitrate ions in water\" width=\"500\" height=\"283\" longdesc=\"Calcium%20nitrate%20is%20a%20typical%20ionic%20compound.%20In%20an%20aqueous%20solution%20it%20dissociates%20into%20calcium%20ions%20and%20nitrate%20ions.\" \/> Figure 1. Calcium nitrate is a typical ionic compound. In an aqueous solution it dissociates into calcium ions and nitrate ions. By <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Dusi%C4%8Dnan_v%C3%A1penat%C3%BD.JPG\">Ond\u0159ej Mangl<\/a>.[\/caption]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"x-ck12-N2FkYWMyZmU4ZmQ3ODExMWE0ZWM5YjZhZGZmZDNlM2I.-ssx\">Nonelectrolytes do not dissociate when forming an aqueous solution.\u00a0 An equation can still be written that simply shows the solid going into solution.\u00a0 For the dissolving of sucrose:<\/p>\r\n<p id=\"x-ck12-ZjRjY2JkYTcwZGY4NGNiMGM2OWJjNjc5YmVjOGZhNTI.-ro6\" class=\"x-ck12-indent\" style=\"text-align: center;\">C <sub> 12 <\/sub> H <sub> 22 <\/sub> O <sub> 11 <\/sub> (<em>s<\/em>)\u00a0\u2192\u00a0C <sub> 12 <\/sub> H <sub> 22 <\/sub> O <sub> 11 <\/sub> (<em>aq<\/em>)<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ODliNDhlNzA0NmE5NzAwMWY0M2RjNTNiMzNiYmQyZmY.-ser\">\r\n \t<li>Dissociation is the separation of ions that occurs when a solid ionic compound dissolves.<\/li>\r\n \t<li>Nonionic compounds do not dissociate in water.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-knh\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-ZTY2YzQ2NmZlMTQ4OGQ3MWY5MDI2N2M0NDdjM2ZkODQ.-uq9\"><a href=\"http:\/\/www.chemistrylecturenotes.com\/html\/dissociation_of_ionic_compound.html\"> http:\/\/www.chemistrylecturenotes.com\/html\/dissociation_of_ionic_compound.html <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-M2Q5NTJlZjUyNzUyMWFmZWU1NmUzMDc0MDQ1NjhlMGM.-nvq\">\r\n \t<li>What does silver nitrate form upon dissociation in water?<\/li>\r\n \t<li>How would HCl\u00a0dissociate in water?<\/li>\r\n \t<li>Is HF\u00a0primarily dissociated or undissociated in water?<\/li>\r\n \t<li>Do alcohols dissociate in water?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-OTlkZTkxMzQ2YTllYjRjNmQzNDUwZDZiMDliMDQ1ZTY.-9w3\">\r\n \t<li>Define dissociation.<\/li>\r\n \t<li>Do polyatomic ions dissociate when dissolved in water?<\/li>\r\n \t<li>Write the equation for the dissociation of KBr\u00a0in water.<\/li>\r\n \t<li>Write the equation for the dissociation of NH <sub> 4 <\/sub> Cl\u00a0in water.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2>\u00a0Glossary<\/h2>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-NmM3Yzk2ZWNhY2M0Njc3Y2Q3MzRiYzY4ZmY0MTdhOTE.-nrq\">\r\n \t<li><strong> dissociation:\u00a0 <\/strong> The separation of ions that occurs when a solid ionic compound dissolves.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NGRlYTMyNzI4ZjE0YzZmMDdjOTliNDliOWRmZGQ0MjM.-hoq\">\n<li>Define dissociation.<\/li>\n<li>Be able to write equations for dissociation of ionic compounds.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Have you ever seen trucks pour salt on icy roads?<\/h3>\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212343\/20140811155540779952.jpeg\" alt=\"In order to function, deicing salts must first dissociate into their component ions\" width=\"400\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\">Runner on a Frozen Road. Courtesy of <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_40255_-_Runner_on_a_frozen_road_in_Fargo,_North_Dakota.jpg\">FEMA<\/a>.<\/p>\n<\/div>\n<p id=\"x-ck12-YjgyY2JkNzliYzViNmExNWQ3NzRmNDRjY2RmOGFkMjA.-jlu\">In many areas, ice on the streets and sidewalks represent a serious walking and driving hazard.\u00a0 One common approach to melting the ice is to put some form of deicing salt on the surface.\u00a0 Materials such as sodium chloride or calcium chloride are frequently employed for this purpose.\u00a0 In order to be effective, the solid material must first dissolve and break up into the ions that make up the compound.<\/p>\n<\/div>\n<h2>Dissociation<\/h2>\n<p id=\"x-ck12-YTY5YWJmNWJmMWFhMGE5NTQyYThmM2M5NmUwYWFkODA.-e2a\">An ionic crystal lattice breaks apart when it is dissolved in water. \u00a0<strong>Dissociation <\/strong> is the separation of ions that occurs when a solid ionic compound dissolves.\u00a0 It is important to be able to write dissociation equations.\u00a0 Simply undo the crisscross method that you learned when writing chemical formulas of ionic compounds.\u00a0 The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equation.\u00a0 Shown below are dissociation equations for NaCl, Ca(NO<sub>3<\/sub>)<sub>2 <\/sub> , and (NH<sub>4<\/sub>)<sub>3<\/sub>PO<sub>4<\/sub>.<\/p>\n<p id=\"x-ck12-ZDlhMzhhZDQ4YmNjYzk3MzQ3ODNiNTg1M2YxZjg2Yzc.-n2h\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">NaCl(<em>s<\/em>) \u2192\u00a0Na<sup>+ <\/sup> (<em>aq<\/em>) + Cl<sup>&#8211; <\/sup> (<em>aq<\/em>)<\/p>\n<p id=\"x-ck12-ZmJkZTMxYjM3ZGM2NThiNDdjMmJjOWZhY2M3MDllYTc.-4xe\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">Ca(NO<sub>3<\/sub>) <sub> 2 <\/sub> (<em>aq<\/em>)\u00a0\u2192 Ca<sup>2+ <\/sup> (<em>aq<\/em>) + 2NO<sub>3 <\/sub><sup> &#8211; <\/sup> (<em>aq<\/em>)<\/p>\n<p id=\"x-ck12-OGMzZjBlNGE3OWJkZjVmYjQ1NzJjM2FlODMwZmZlNTk.-bcl\" class=\"x-ck12-indent\" style=\"padding-left: 30px;\">(NH<sub>4<\/sub>)<sub>3 <\/sub> PO<sub>4 <\/sub> (<em>s<\/em>) \u2192\u00a03NH<sub>4 <\/sub><sup> + <\/sup> (<em>aq<\/em>) + PO<sub>4 <\/sub><sup> 3- <\/sup> (<em>aq<\/em>)<\/p>\n<p id=\"x-ck12-MmYyM2E2ODJhMmQzOGU2OGY0ZmRkYTU2M2QwZDgwOGI.-sfg\">The formula unit of sodium chloride dissociates into one sodium ion and one chloride ion.\u00a0 The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions.\u00a0 This is because of the 2+ charge of the calcium ion.\u00a0 Two nitrate ions, each with a 1\u2212 charge are required to make the equation balance electrically.\u00a0 The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion.\u00a0 Note that the polyatomic ions themselves do not dissociate further, but remain intact.<\/p>\n<p id=\"x-ck12-ZjAxMmE0MGVhYTk2ZGY4OWFmNjM2N2Y3YzMxOTZkNzA.-tkq\">Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make the original compound neutral.\u00a0 The 3 subscript of the nitrate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and simply remain as part of its formula after the compound dissociates.\u00a0 Notice that the compounds are solids ( <em> s <\/em> )\u00a0which then become ions in aqueous solution ( <em> aq <\/em> ).<\/p>\n<div id=\"x-ck12-Yzg4MTYyNmNjNmUxZmJkMGU1NzAwM2UzYzg5OGFjMWI.-92c\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MTk0ODY2Mi03Ni05NS1DLUludENoLTA0LTAzLTA4LUNhbGNpdW0tTml0cmF0ZQ..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212345\/20140811155540873625.jpeg\" alt=\"Calcium nitrate dissociates into calcium ions and nitrate ions in water\" width=\"500\" height=\"283\" longdesc=\"Calcium%20nitrate%20is%20a%20typical%20ionic%20compound.%20In%20an%20aqueous%20solution%20it%20dissociates%20into%20calcium%20ions%20and%20nitrate%20ions.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Calcium nitrate is a typical ionic compound. In an aqueous solution it dissociates into calcium ions and nitrate ions. By <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Dusi%C4%8Dnan_v%C3%A1penat%C3%BD.JPG\">Ond\u0159ej Mangl<\/a>.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"x-ck12-N2FkYWMyZmU4ZmQ3ODExMWE0ZWM5YjZhZGZmZDNlM2I.-ssx\">Nonelectrolytes do not dissociate when forming an aqueous solution.\u00a0 An equation can still be written that simply shows the solid going into solution.\u00a0 For the dissolving of sucrose:<\/p>\n<p id=\"x-ck12-ZjRjY2JkYTcwZGY4NGNiMGM2OWJjNjc5YmVjOGZhNTI.-ro6\" class=\"x-ck12-indent\" style=\"text-align: center;\">C <sub> 12 <\/sub> H <sub> 22 <\/sub> O <sub> 11 <\/sub> (<em>s<\/em>)\u00a0\u2192\u00a0C <sub> 12 <\/sub> H <sub> 22 <\/sub> O <sub> 11 <\/sub> (<em>aq<\/em>)<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ODliNDhlNzA0NmE5NzAwMWY0M2RjNTNiMzNiYmQyZmY.-ser\">\n<li>Dissociation is the separation of ions that occurs when a solid ionic compound dissolves.<\/li>\n<li>Nonionic compounds do not dissociate in water.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-knh\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-ZTY2YzQ2NmZlMTQ4OGQ3MWY5MDI2N2M0NDdjM2ZkODQ.-uq9\"><a href=\"http:\/\/www.chemistrylecturenotes.com\/html\/dissociation_of_ionic_compound.html\"> http:\/\/www.chemistrylecturenotes.com\/html\/dissociation_of_ionic_compound.html <\/a><\/p>\n<ol id=\"x-ck12-M2Q5NTJlZjUyNzUyMWFmZWU1NmUzMDc0MDQ1NjhlMGM.-nvq\">\n<li>What does silver nitrate form upon dissociation in water?<\/li>\n<li>How would HCl\u00a0dissociate in water?<\/li>\n<li>Is HF\u00a0primarily dissociated or undissociated in water?<\/li>\n<li>Do alcohols dissociate in water?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-OTlkZTkxMzQ2YTllYjRjNmQzNDUwZDZiMDliMDQ1ZTY.-9w3\">\n<li>Define dissociation.<\/li>\n<li>Do polyatomic ions dissociate when dissolved in water?<\/li>\n<li>Write the equation for the dissociation of KBr\u00a0in water.<\/li>\n<li>Write the equation for the dissociation of NH <sub> 4 <\/sub> Cl\u00a0in water.<\/li>\n<\/ol>\n<\/div>\n<h2>\u00a0Glossary<\/h2>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-NmM3Yzk2ZWNhY2M0Njc3Y2Q3MzRiYzY4ZmY0MTdhOTE.-nrq\">\n<li><strong> dissociation:\u00a0 <\/strong> The separation of ions that occurs when a solid ionic compound dissolves.<\/li>\n<\/ul>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2728\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":29,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2728","chapter","type-chapter","status-publish","hentry"],"part":2336,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2728","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/29"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2728\/revisions"}],"predecessor-version":[{"id":3405,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2728\/revisions\/3405"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2336"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2728\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2728"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2728"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2728"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2728"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}