Bromine has two stable isotopes: 79Br and 81Br with masses of 78.9183371 amu and 80.9162906 amu. If the average atomic mass for Bromine on Earth is 79.904 amu, what is the mass of 79Br in a 10.0 g sample?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the mass of the 79Br isotope in a 10.0 g sample of bromine, we need to understand how the average atomic mass of bromine is calculated based on the relative abundances of its isotopes, 79Br and 81Br.<\/p>\n\n\n\n Bromine exists naturally as a mixture of two isotopes:<\/p>\n\n\n\n The average atomic mass of bromine on Earth is 79.904 amu<\/strong>. This average is a weighted average of the masses of the two isotopes, weighted by their relative abundances in nature.<\/p>\n\n\n\n Let\u2019s assume:<\/p>\n\n\n\n Now, we can set up the equation for the weighted average atomic mass: Average Atomic Mass=(x\u22c578.9183371\u2009amu)+((1\u2212x)\u22c580.9162906\u2009amu)\\text{Average Atomic Mass} = (x \\cdot 78.9183371 \\, \\text{amu}) + ((1 – x) \\cdot 80.9162906 \\, \\text{amu})<\/p>\n\n\n\n We are given that the average atomic mass of bromine is 79.904 amu, so we can plug this value into the equation: 79.904=(x\u22c578.9183371)+((1\u2212x)\u22c580.9162906)79.904 = (x \\cdot 78.9183371) + ((1 – x) \\cdot 80.9162906)<\/p>\n\n\n\n Expanding and simplifying the equation: 79.904=78.9183371x+80.9162906\u221280.9162906×79.904 = 78.9183371x + 80.9162906 – 80.9162906x 79.904=80.9162906\u2212(80.9162906\u221278.9183371)x79.904 = 80.9162906 – (80.9162906 – 78.9183371)x 79.904\u221280.9162906=\u2212(80.9162906\u221278.9183371)x79.904 – 80.9162906 = – (80.9162906 – 78.9183371)x \u22121.0122906=\u2212(1.9979535)x-1.0122906 = -(1.9979535)x x=1.01229061.9979535\u22480.5068x = \\frac{1.0122906}{1.9979535} \\approx 0.5068<\/p>\n\n\n\n Thus, the fraction of 79Br in the sample is approximately 0.5068<\/strong> (or 50.68%).<\/p>\n\n\n\n Now that we know the fraction of 79Br, we can calculate the mass of 79Br in a 10.0 g sample of bromine: Mass of 79Br=0.5068\u00d710.0\u2009g=5.068\u2009g\\text{Mass of 79Br} = 0.5068 \\times 10.0 \\, \\text{g} = 5.068 \\, \\text{g}<\/p>\n\n\n\n The mass of 79Br in the 10.0 g sample is approximately 5.068 g<\/strong>.<\/p>\n\n\n\n The problem involves calculating the proportion of 79Br in a sample of bromine based on its average atomic mass, which is a weighted average of the isotopes’ masses. By using the equation for a weighted average, we solve for the fraction of 79Br, then apply it to the total mass of the sample to find the amount of 79Br. This method allows us to determine the specific amount of each isotope in a sample of bromine based on its natural isotopic distribution.<\/p>\n","protected":false},"excerpt":{"rendered":" Bromine has two stable isotopes: 79Br and 81Br with masses of 78.9183371 amu and 80.9162906 amu. If the average atomic mass for Bromine on Earth is 79.904 amu, what is the mass of 79Br in a 10.0 g sample? The correct answer and explanation is : To determine the mass of the 79Br isotope in […]<\/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-210876","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210876","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=210876"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210876\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=210876"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=210876"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=210876"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understanding the Problem<\/h3>\n\n\n\n
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Step 2: Let\u2019s Set Up the Equation<\/h3>\n\n\n\n
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Step 3: Solve for x (the Fraction of 79Br)<\/h3>\n\n\n\n
Step 4: Determine the Mass of 79Br in 10.0 g<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n