{"id":221153,"date":"2025-05-29T08:15:44","date_gmt":"2025-05-29T08:15:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221153"},"modified":"2025-05-29T08:24:59","modified_gmt":"2025-05-29T08:24:59","slug":"give-your-final-answer-in-whole-number-with-units-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/give-your-final-answer-in-whole-number-with-units-g-mol\/","title":{"rendered":"Calculate The Molecular Weight Of Ca(HCO3)2."},"content":{"rendered":"\n<p>Calculate The Molecular Weight Of Ca(HCO3)2. Give Your Final Answer In Whole Number With Units G\/Mol.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is<\/mark><\/strong>:<\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong><br><strong>162 g\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>To calculate the <strong>molecular weight (molar mass)<\/strong> of <strong>calcium bicarbonate, Ca(HCO\u2083)\u2082<\/strong>, we add together the atomic masses of all the atoms in the compound. These atomic masses are taken from the periodic table.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-Step Breakdown:<\/strong><\/h3>\n\n\n\n<p><strong>1. Identify the elements and their atomic masses:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Element<\/th><th>Symbol<\/th><th>Atomic Mass (g\/mol)<\/th><\/tr><\/thead><tbody><tr><td>Calcium<\/td><td>Ca<\/td><td>40.08<\/td><\/tr><tr><td>Hydrogen<\/td><td>H<\/td><td>1.008<\/td><\/tr><tr><td>Carbon<\/td><td>C<\/td><td>12.01<\/td><\/tr><tr><td>Oxygen<\/td><td>O<\/td><td>16.00<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>2. Analyze the chemical formula:<\/strong><\/p>\n\n\n\n<p><strong>Ca(HCO\u2083)\u2082<\/strong> means:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 calcium (Ca)<\/li>\n\n\n\n<li>2 hydrogen (H)<\/li>\n\n\n\n<li>2 carbon (C)<\/li>\n\n\n\n<li>6 oxygen (O)<\/li>\n<\/ul>\n\n\n\n<p>This is because there are <strong>two<\/strong> bicarbonate (HCO\u2083\u207b) ions, and each contains 1 H, 1 C, and 3 O. So we multiply the atoms in HCO\u2083 by 2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H: 1 \u00d7 2 = 2<\/li>\n\n\n\n<li>C: 1 \u00d7 2 = 2<\/li>\n\n\n\n<li>O: 3 \u00d7 2 = 6<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>3. Multiply and sum:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Calcium (Ca)<\/strong>: 1 \u00d7 40.08 = 40.08<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong>: 2 \u00d7 1.008 = 2.016<\/li>\n\n\n\n<li><strong>Carbon (C)<\/strong>: 2 \u00d7 12.01 = 24.02<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong>: 6 \u00d7 16.00 = 96.00<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Molecular Weight:<\/strong> 40.08+2.016+24.02+96.00=162.116\u2009g\/mol40.08 + 2.016 + 24.02 + 96.00 = 162.116 \\, \\text{g\/mol}40.08+2.016+24.02+96.00=162.116g\/mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>4. Round to the nearest whole number:<\/strong> 162\u2009g\/mol\\boxed{162 \\, \\text{g\/mol}}162g\/mol\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion:<\/strong><\/h3>\n\n\n\n<p>The molecular weight of <strong>calcium bicarbonate (Ca(HCO\u2083)\u2082)<\/strong> is <strong>162 g\/mol<\/strong>. This value is essential in chemistry for converting between grams and moles of a substance in stoichiometric calculations, solution preparation, and chemical reaction analysis. Understanding molecular weight also helps determine the proportions of elements within a compound and is foundational in fields like pharmacology, environmental science, and materials engineering.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-117.jpeg\" alt=\"\" class=\"wp-image-221155\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate The Molecular Weight Of Ca(HCO3)2. Give Your Final Answer In Whole Number With Units G\/Mol. The Correct Answer and Explanation is: Correct Answer:162 g\/mol Explanation: To calculate the molecular weight (molar mass) of calcium bicarbonate, Ca(HCO\u2083)\u2082, we add together the atomic masses of all the atoms in the compound. These atomic masses are taken [&hellip;]<\/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-221153","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221153","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=221153"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221153\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}