{"id":4401,"date":"2015-06-19T16:00:18","date_gmt":"2015-06-19T16:00:18","guid":{"rendered":"https:\/\/courses.candelalearning.com\/chemistryformajorsx1xmaster\/?post_type=chapter&#038;p=4401"},"modified":"2020-12-16T02:19:19","modified_gmt":"2020-12-16T02:19:19","slug":"videos-stoichiometry","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/videos-stoichiometry\/","title":{"raw":"Videos: Stoichiometry","rendered":"Videos: Stoichiometry"},"content":{"raw":"<h2>Stoichiometry: Chemistry for Massive Creatures\u2014Crash Course Chemistry #6<\/h2>\r\nChemists need stoichiometry to make the scale of chemistry more understandable\u2014Hank is here to explain why, and to teach us how to use it.\r\n\r\nhttps:\/\/youtu.be\/UL1jmJaUkaQ\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/StoichiometryChemistryForMassiveCreatures_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6\" here (opens in new window)<\/a>.\r\n<h2>Stoichiometry Problems: Moles to Moles<\/h2>\r\nShows how to use stoichiometry to determine the number of moles of reactants and products if you are given the number of moles of one of the substances in the reaction.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5518719&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qRVUNCw9fOY&amp;video_target=tpm-plugin-8b7u1xs8-qRVUNCw9fOY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions10of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles\" here (opens in new window)<\/a>.\r\n<h2>Stoichiometry Problems: Grams to Grams<\/h2>\r\nShows how to use stoichiometry to determine the grams of the other substances in the chemical equation if you are given the grams of one of the substances.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5518720&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=bltnuzbs2JA&amp;video_target=tpm-plugin-p1yahu2k-bltnuzbs2JA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions9of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Chemical Reactions (9 of 11) Stoichiometry: Grams to Grams\" here (opens in new window)<\/a>.\r\n<h2>Stoichiometry Problems: Grams to Moles<\/h2>\r\nShows how to use stoichiometry to determine the number of moles of reactants and products if you are given the number of grams of one of the substances in the chemical equation.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5518721&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=09g6nN2PpVE&amp;video_target=tpm-plugin-8hz0sek4-09g6nN2PpVE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions7of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Chemical Reactions (7 of 11) Stoichiometry: Grams to Moles\" here (opens in new window)<\/a>.","rendered":"<h2>Stoichiometry: Chemistry for Massive Creatures\u2014Crash Course Chemistry #6<\/h2>\n<p>Chemists need stoichiometry to make the scale of chemistry more understandable\u2014Hank is here to explain why, and to teach us how to use it.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UL1jmJaUkaQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/StoichiometryChemistryForMassiveCreatures_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Stoichiometry &#8211; Chemistry for Massive Creatures: Crash Course Chemistry #6&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Stoichiometry Problems: Moles to Moles<\/h2>\n<p>Shows how to use stoichiometry to determine the number of moles of reactants and products if you are given the number of moles of one of the substances in the reaction.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5518719&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qRVUNCw9fOY&amp;video_target=tpm-plugin-8b7u1xs8-qRVUNCw9fOY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions10of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Stoichiometry Problems: Grams to Grams<\/h2>\n<p>Shows how to use stoichiometry to determine the grams of the other substances in the chemical equation if you are given the grams of one of the substances.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5518720&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=bltnuzbs2JA&amp;video_target=tpm-plugin-p1yahu2k-bltnuzbs2JA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions9of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Chemical Reactions (9 of 11) Stoichiometry: Grams to Grams&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Stoichiometry Problems: Grams to Moles<\/h2>\n<p>Shows how to use stoichiometry to determine the number of moles of reactants and products if you are given the number of grams of one of the substances in the chemical equation.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5518721&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=09g6nN2PpVE&amp;video_target=tpm-plugin-8hz0sek4-09g6nN2PpVE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/ChemicalReactions7of11Stoichiometry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Chemical Reactions (7 of 11) Stoichiometry: Grams to Moles&#8221; here (opens in new window)<\/a>.<\/p>\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-4401\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6. <strong>Authored by<\/strong>: CrashCourse. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/UL1jmJaUkaQ\">https:\/\/youtu.be\/UL1jmJaUkaQ<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><li>Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles. <strong>Authored by<\/strong>: Step by Step Science. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qRVUNCw9fOY\">https:\/\/youtu.be\/qRVUNCw9fOY<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><li>Chemical Reactions (9 of 11) Stoichiometry: Grams to Grams. <strong>Authored by<\/strong>: Step by Step Science. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/bltnuzbs2JA\">https:\/\/youtu.be\/bltnuzbs2JA<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><li>Chemical Reactions (7 of 11) Stoichiometry: Grams to Moles. <strong>Authored by<\/strong>: Step by Step Science. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/09g6nN2PpVE\">https:\/\/youtu.be\/09g6nN2PpVE<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":78,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6\",\"author\":\"CrashCourse\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/UL1jmJaUkaQ\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Chemical Reactions (10 of 11) Stoichiometry: Moles to Moles\",\"author\":\"Step by Step Science\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/qRVUNCw9fOY\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Chemical Reactions (9 of 11) Stoichiometry: Grams to Grams\",\"author\":\"Step by Step Science\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/bltnuzbs2JA\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Chemical Reactions (7 of 11) Stoichiometry: Grams to Moles\",\"author\":\"Step by Step Science\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/09g6nN2PpVE\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4401","chapter","type-chapter","status-publish","hentry"],"part":3026,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/4401","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/78"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/4401\/revisions"}],"predecessor-version":[{"id":6901,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/4401\/revisions\/6901"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/3026"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/4401\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=4401"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=4401"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=4401"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=4401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}