{"id":95,"date":"2022-07-13T17:38:33","date_gmt":"2022-07-13T17:38:33","guid":{"rendered":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/solution-exercise-3\/"},"modified":"2023-03-22T17:52:06","modified_gmt":"2023-03-22T17:52:06","slug":"solution-exercise-3","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/solution-exercise-3\/","title":{"raw":"Solution Exercise 3","rendered":"Solution Exercise 3"},"content":{"raw":"<div class=\"solution-exercise-3\">\r\n<p class=\"import-Normal\">65 percent of the rock is fluorapatite and the mole proportion of F in fluorapatite is the ratio of the atomic weight of F to the molecular weight of fluorapatite, which is 19\/504. So, the percent of F in the rock from fluorapatite would be:<\/p>\r\n<p class=\"import-Normal\" style=\"text-align: center;\">[latex]\\displaystyle 0.65\\left (\\frac{19}{504}\\right )100\\textrm{%}=2.45\\textrm{%}[\/latex]<\/p>\r\n<p class=\"import-Normal\">5 percent of the rock is fluorite and the mole proportion of F in fluorite is 38\/78, the ratio of the atomic weight of F to the molecular weight of fluorite. So, the percent of F in the rock from fluorite would be:<\/p>\r\n<p class=\"import-Normal\" style=\"text-align: center;\">[latex]\\displaystyle 0.05\\left (\\frac{38}{78}\\right )100\\textrm{%}=2.44\\textrm{%}[\/latex]<\/p>\r\n<p class=\"import-Normal\">And the sum from these two minerals would be 4.89 percent. However, these two minerals comprise only 80 percent of the rock so that the total F in the rock would be:<\/p>\r\n<p class=\"import-Normal\" style=\"text-align: center;\">(0.8) 4.89% = 3.9% F<\/p>\r\n<p class=\"import-Normal\" style=\"text-align: right;\"><a href=\"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/exercise-3\/\">Return to Exercise 3<\/a><\/p>\r\n\r\n<\/div>","rendered":"<div class=\"solution-exercise-3\">\n<p class=\"import-Normal\">65 percent of the rock is fluorapatite and the mole proportion of F in fluorapatite is the ratio of the atomic weight of F to the molecular weight of fluorapatite, which is 19\/504. So, the percent of F in the rock from fluorapatite would be:<\/p>\n<p class=\"import-Normal\" style=\"text-align: center;\">[latex]\\displaystyle 0.65\\left (\\frac{19}{504}\\right )100\\textrm{%}=2.45\\textrm{%}[\/latex]<\/p>\n<p class=\"import-Normal\">5 percent of the rock is fluorite and the mole proportion of F in fluorite is 38\/78, the ratio of the atomic weight of F to the molecular weight of fluorite. So, the percent of F in the rock from fluorite would be:<\/p>\n<p class=\"import-Normal\" style=\"text-align: center;\">[latex]\\displaystyle 0.05\\left (\\frac{38}{78}\\right )100\\textrm{%}=2.44\\textrm{%}[\/latex]<\/p>\n<p class=\"import-Normal\">And the sum from these two minerals would be 4.89 percent. However, these two minerals comprise only 80 percent of the rock so that the total F in the rock would be:<\/p>\n<p class=\"import-Normal\" style=\"text-align: center;\">(0.8) 4.89% = 3.9% F<\/p>\n<p class=\"import-Normal\" style=\"text-align: right;\"><a href=\"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/chapter\/exercise-3\/\">Return to Exercise 3<\/a><\/p>\n<\/div>\n","protected":false},"author":1,"menu_order":42,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-95","chapter","type-chapter","status-publish","hentry"],"part":172,"_links":{"self":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapters\/95","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":10,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapters\/95\/revisions"}],"predecessor-version":[{"id":329,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapters\/95\/revisions\/329"}],"part":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/parts\/172"}],"metadata":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapters\/95\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/wp\/v2\/media?parent=95"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/pressbooks\/v2\/chapter-type?post=95"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/wp\/v2\/contributor?post=95"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/books.gw-project.org\/fluoride-in-groundwater\/wp-json\/wp\/v2\/license?post=95"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}