Calculate the molar mass of HgO. Round your answer to one decimal place.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To calculate the molar mass of HgO (mercury(II) oxide)<\/strong>, we need to sum the atomic masses of mercury (Hg)<\/strong> and oxygen (O)<\/strong> based on their values from the periodic table:<\/p>\n\n\n\n $$ $$ The molar mass<\/strong> of a compound is the mass of one mole of that compound, measured in grams per mole (g\/mol)<\/strong>. To find it, we add together the atomic masses of all atoms present in one molecule (or formula unit) of the substance.<\/p>\n\n\n\n In this case, HgO<\/strong> is a binary compound composed of one atom of mercury (Hg)<\/strong> and one atom of oxygen (O)<\/strong>.<\/p>\n\n\n\n Since the chemical formula of mercury(II) oxide is HgO, the compound contains:<\/p>\n\n\n\n When calculating the molar mass:<\/p>\n\n\n\n $$ Thus, the molar mass of HgO<\/strong> is 216.6 g\/mol<\/strong>, which is the mass of one mole of HgO molecules. This value is useful in stoichiometry, allowing chemists to relate grams of a substance to moles for chemical reactions.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the molar mass of HgO. Round your answer to one decimal place. The Correct Answer and Explanation is: To calculate the molar mass of HgO (mercury(II) oxide), we need to sum the atomic masses of mercury (Hg) and oxygen (O) based on their values from the periodic table: Molar Mass of HgO: $$\\text{Molar mass […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-216829","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216829","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=216829"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216829\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216829"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216829"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216829"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Molar Mass of HgO:<\/h3>\n\n\n\n
\\text{Molar mass of HgO} = \\text{Atomic mass of Hg} + \\text{Atomic mass of O}
$$<\/p>\n\n\n\n
\\text{Molar mass of HgO} = 200.6 + 16.0 = \\boxed{216.6 \\text{ g\/mol}}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: 216.6 g\/mol<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83e\uddea Explanation (Approximately 300 Words):<\/h3>\n\n\n\n
\n
\n
\n
\\text{Hg: } 1 \\times 200.6 = 200.6 \\text{ g\/mol} \\
\\text{O: } 1 \\times 16.0 = 16.0 \\text{ g\/mol} \\
\\text{Total: } 200.6 + 16.0 = 216.6 \\text{ g\/mol}
$$<\/p>\n\n\n\n