{"id":263845,"date":"2025-07-21T09:24:35","date_gmt":"2025-07-21T09:24:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=263845"},"modified":"2025-07-21T09:24:37","modified_gmt":"2025-07-21T09:24:37","slug":"determine-the-quantity-of-nitrogen-atoms-in-15-0-grams-of-pbno22","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/21\/determine-the-quantity-of-nitrogen-atoms-in-15-0-grams-of-pbno22\/","title":{"rendered":"Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2"},"content":{"rendered":"\n<p>Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2<\/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>To determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2, we need to follow a step-by-step process:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the molar mass of Pb(NO2)2.<\/h3>\n\n\n\n<p>We begin by finding the molar mass of the compound, Pb(NO2)2, by adding the atomic masses of each element in the formula.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Lead (Pb): 207.2 g\/mol<\/li>\n\n\n\n<li>Nitrogen (N): 14.0 g\/mol<\/li>\n\n\n\n<li>Oxygen (O): 16.0 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>The formula, Pb(NO2)2, indicates that there is one Pb atom, two nitrogen atoms, and four oxygen atoms.<\/p>\n\n\n\n<p>So, the molar mass of Pb(NO2)2 is calculated as:Molar&nbsp;mass&nbsp;of&nbsp;Pb(NO2)2=207.2\u2009g\/mol+2\u00d7(14.0\u2009g\/mol)+4\u00d7(16.0\u2009g\/mol)\\text{Molar mass of Pb(NO2)2} = 207.2 \\, \\text{g\/mol} + 2 \\times (14.0 \\, \\text{g\/mol}) + 4 \\times (16.0 \\, \\text{g\/mol})Molar&nbsp;mass&nbsp;of&nbsp;Pb(NO2)2=207.2g\/mol+2\u00d7(14.0g\/mol)+4\u00d7(16.0g\/mol)=207.2+28.0+64.0=299.2\u2009g\/mol= 207.2 + 28.0 + 64.0 = 299.2 \\, \\text{g\/mol}=207.2+28.0+64.0=299.2g\/mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the number of moles of Pb(NO2)2.<\/h3>\n\n\n\n<p>Next, we calculate the number of moles of Pb(NO2)2 in 15.0 grams using the formula:moles&nbsp;of&nbsp;Pb(NO2)2=mass&nbsp;of&nbsp;Pb(NO2)2molar&nbsp;mass&nbsp;of&nbsp;Pb(NO2)2=15.0\u2009g299.2\u2009g\/mol=0.0501\u2009mol\\text{moles of Pb(NO2)2} = \\frac{\\text{mass of Pb(NO2)2}}{\\text{molar mass of Pb(NO2)2}} = \\frac{15.0 \\, \\text{g}}{299.2 \\, \\text{g\/mol}} = 0.0501 \\, \\text{mol}moles&nbsp;of&nbsp;Pb(NO2)2=molar&nbsp;mass&nbsp;of&nbsp;Pb(NO2)2mass&nbsp;of&nbsp;Pb(NO2)2\u200b=299.2g\/mol15.0g\u200b=0.0501mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the number of nitrogen atoms.<\/h3>\n\n\n\n<p>Since each molecule of Pb(NO2)2 contains 2 nitrogen atoms, the number of nitrogen atoms is twice the number of moles of Pb(NO2)2. We multiply the moles of Pb(NO2)2 by 2 to find the number of moles of nitrogen atoms:moles&nbsp;of&nbsp;nitrogen&nbsp;atoms=0.0501\u2009mol\u00d72=0.1002\u2009mol\\text{moles of nitrogen atoms} = 0.0501 \\, \\text{mol} \\times 2 = 0.1002 \\, \\text{mol}moles&nbsp;of&nbsp;nitrogen&nbsp;atoms=0.0501mol\u00d72=0.1002mol<\/p>\n\n\n\n<p>Finally, to determine the number of nitrogen atoms, we multiply the moles of nitrogen atoms by Avogadro&#8217;s number (6.022 \u00d7 10\u00b2\u00b3 atoms\/mol):number&nbsp;of&nbsp;nitrogen&nbsp;atoms=0.1002\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=6.03\u00d71022\u2009atoms\\text{number of nitrogen atoms} = 0.1002 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 6.03 \\times 10^{22} \\, \\text{atoms}number&nbsp;of&nbsp;nitrogen&nbsp;atoms=0.1002mol\u00d76.022\u00d71023atoms\/mol=6.03\u00d71022atoms<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2 is <strong>6.03 \u00d7 10\u00b2\u00b2 atoms<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1334.jpeg\" alt=\"\" class=\"wp-image-263846\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2 The Correct Answer and Explanation is: To determine the quantity of nitrogen atoms in 15.0 grams of Pb(NO2)2, we need to follow a step-by-step process: Step 1: Calculate the molar mass of Pb(NO2)2. We begin by finding the molar mass of the compound, Pb(NO2)2, [&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-263845","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263845","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=263845"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263845\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=263845"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=263845"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=263845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}