{"id":182737,"date":"2025-01-14T18:32:58","date_gmt":"2025-01-14T18:32:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182737"},"modified":"2025-01-14T18:33:00","modified_gmt":"2025-01-14T18:33:00","slug":"1-calculate-the-molar-mass-of-barium-sulfate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/14\/1-calculate-the-molar-mass-of-barium-sulfate\/","title":{"rendered":"1.)\u00a0\u00a0 Calculate the molar mass of barium sulfate"},"content":{"rendered":"\n<p>1.)\u00a0\u00a0 Calculate the molar mass of barium sulfate, BaSO4. a.) 233.4 g\/mol b.) 217.4 g\/mol c.) 104.0 g\/mol d.) 96.0 g\/mol 2.) Calculate the molar mass of lead (IV) oxalate, Pb(C2O4)2. a.) 235.21 g\/mol b.) 271.24 g\/mol c.) 347.21 g\/mol d.) 383.24 g\/mol 3.) Ethylene glycol is a coolant widely used in automobiles. If its molecular formula is C2H602, how many moles would be present if you had 2500 g of this in your vehicle? a.) 1.6 x 10$ mol b.) 2.5 x 10-2 mol c.) 40. mol d.) 62 mol 4.) Hydrogen peroxide H2O2 is commonly used to clean bacteria in scrapes and cuts. It can be used as a mouthwash and is an additive in some toothpastes. If a sample contained 5.23 x 1021 molecules of H:0, what would the equivalent be in moles? a.) 8.69 x 103 mol b.) 8.69 x 10-mol c.) 115 mol d.) 8.69 x 1043 mol 5.) Calculate the percent composition of ammonium nitrate, NH NO.. a.) 17.5 %N, 5.0%H, 77.5%O b.) 17.5 % N, 3.2% H, 51.6%O c.) 35.0%N, 1.2 % H, 57.2%O d.) None of the above English (United States) Focus LE<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the answers with detailed explanations:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1) Molar Mass of Barium Sulfate, BaSO\u2084<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The atomic masses of the elements are:<ul><li>Ba = 137.33 g\/mol<\/li><li>S = 32.07 g\/mol<\/li><li>O = 16.00 g\/mol (there are 4 oxygen atoms in BaSO\u2084)<\/li><\/ul>The molar mass of BaSO\u2084 is calculated as: Molar\u00a0Mass\u00a0of\u00a0BaSO\u2084=137.33+32.07+(4\u00d716.00)=137.33+32.07+64.00=233.4\u2009g\/mol\\text{Molar Mass of BaSO\u2084} = 137.33 + 32.07 + (4 \\times 16.00) = 137.33 + 32.07 + 64.00 = 233.4 \\, \\text{g\/mol} Therefore, the correct answer is <strong>a) 233.4 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2) Molar Mass of Lead (IV) Oxalate, Pb(C\u2082O\u2084)\u2082<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The atomic masses of the elements are:<ul><li>Pb = 207.2 g\/mol<\/li><li>C = 12.01 g\/mol (there are 4 carbon atoms in Pb(C\u2082O\u2084)\u2082)<\/li><li>O = 16.00 g\/mol (there are 8 oxygen atoms in Pb(C\u2082O\u2084)\u2082)<\/li><\/ul>The molar mass of Pb(C\u2082O\u2084)\u2082 is calculated as: Molar\u00a0Mass\u00a0of\u00a0Pb(C\u2082O\u2084)\u2082=207.2+(4\u00d712.01)+(8\u00d716.00)\\text{Molar Mass of Pb(C\u2082O\u2084)\u2082} = 207.2 + (4 \\times 12.01) + (8 \\times 16.00) =207.2+48.04+128.00=383.24\u2009g\/mol= 207.2 + 48.04 + 128.00 = 383.24 \\, \\text{g\/mol} Therefore, the correct answer is <strong>d) 383.24 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3) Moles of Ethylene Glycol (C\u2082H\u2086O\u2082) in 2500 g<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The atomic masses of the elements are:<ul><li>C = 12.01 g\/mol (there are 2 carbon atoms in C\u2082H\u2086O\u2082)<\/li><li>H = 1.008 g\/mol (there are 6 hydrogen atoms in C\u2082H\u2086O\u2082)<\/li><li>O = 16.00 g\/mol (there are 2 oxygen atoms in C\u2082H\u2086O\u2082)<\/li><\/ul>The molar mass of C\u2082H\u2086O\u2082 is calculated as: Molar\u00a0Mass\u00a0of\u00a0C\u2082H\u2086O\u2082=(2\u00d712.01)+(6\u00d71.008)+(2\u00d716.00)\\text{Molar Mass of C\u2082H\u2086O\u2082} = (2 \\times 12.01) + (6 \\times 1.008) + (2 \\times 16.00) =24.02+6.048+32.00=62.068\u2009g\/mol= 24.02 + 6.048 + 32.00 = 62.068 \\, \\text{g\/mol} To find the number of moles in 2500 g: Moles=2500\u2009g62.068\u2009g\/mol\u224840.2\u2009mol\\text{Moles} = \\frac{2500 \\, \\text{g}}{62.068 \\, \\text{g\/mol}} \\approx 40.2 \\, \\text{mol} Therefore, the correct answer is <strong>c) 40 mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">4) Moles of Hydrogen Peroxide (H\u2082O\u2082) in 5.23 x 10\u00b2\u00b9 molecules<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One mole of any substance contains 6.022\u00d710236.022 \\times 10^{23} molecules (Avogadro&#8217;s number). To convert molecules to moles: Moles=5.23\u00d710216.022\u00d71023=8.69\u00d710\u22123\u2009mol\\text{Moles} = \\frac{5.23 \\times 10^{21}}{6.022 \\times 10^{23}} = 8.69 \\times 10^{-3} \\, \\text{mol} Therefore, the correct answer is <strong>b) 8.69 x 10\u207b\u00b3 mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">5) Percent Composition of Ammonium Nitrate, NH\u2084NO\u2083<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The atomic masses of the elements are:<ul><li>N = 14.01 g\/mol (there are 2 nitrogen atoms in NH\u2084NO\u2083)<\/li><li>H = 1.008 g\/mol (there are 4 hydrogen atoms in NH\u2084NO\u2083)<\/li><li>O = 16.00 g\/mol (there are 3 oxygen atoms in NH\u2084NO\u2083)<\/li><\/ul>The molar mass of NH\u2084NO\u2083 is calculated as: Molar\u00a0Mass\u00a0of\u00a0NH\u2084NO\u2083=(2\u00d714.01)+(4\u00d71.008)+(3\u00d716.00)\\text{Molar Mass of NH\u2084NO\u2083} = (2 \\times 14.01) + (4 \\times 1.008) + (3 \\times 16.00) =28.02+4.032+48.00=80.052\u2009g\/mol= 28.02 + 4.032 + 48.00 = 80.052 \\, \\text{g\/mol} The percent composition of each element is:<ul><li>Percent N = 28.0280.052\u00d7100\u224835.0%\\frac{28.02}{80.052} \\times 100 \\approx 35.0\\%<\/li><li>Percent H = 4.03280.052\u00d7100\u22485.0%\\frac{4.032}{80.052} \\times 100 \\approx 5.0\\%<\/li><li>Percent O = 48.0080.052\u00d7100\u224860.0%\\frac{48.00}{80.052} \\times 100 \\approx 60.0\\%<\/li><\/ul>Therefore, the correct answer is <strong>d) None of the above<\/strong>, since the composition does not match any given choices exactly.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>a) 233.4 g\/mol<\/li>\n\n\n\n<li>d) 383.24 g\/mol<\/li>\n\n\n\n<li>c) 40 mol<\/li>\n\n\n\n<li>b) 8.69 x 10\u207b\u00b3 mol<\/li>\n\n\n\n<li>d) None of the above<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>1.)\u00a0\u00a0 Calculate the molar mass of barium sulfate, BaSO4. a.) 233.4 g\/mol b.) 217.4 g\/mol c.) 104.0 g\/mol d.) 96.0 g\/mol 2.) Calculate the molar mass of lead (IV) oxalate, Pb(C2O4)2. a.) 235.21 g\/mol b.) 271.24 g\/mol c.) 347.21 g\/mol d.) 383.24 g\/mol 3.) Ethylene glycol is a coolant widely used in automobiles. If its [&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-182737","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182737","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=182737"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182737\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182737"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182737"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}