{"id":224525,"date":"2025-06-03T07:29:51","date_gmt":"2025-06-03T07:29:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224525"},"modified":"2025-06-03T07:29:53","modified_gmt":"2025-06-03T07:29:53","slug":"write-a-lewis-structure-for-nitrogen-trifluoride-nf3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/03\/write-a-lewis-structure-for-nitrogen-trifluoride-nf3\/","title":{"rendered":"Write a Lewis structure for nitrogen trifluoride, NF3."},"content":{"rendered":"\n<p>Write a Lewis structure for nitrogen trifluoride, NF3. Name the shape and state whether the molecule is polar or non-polar. How many atoms of carbon are in 0.185 mol of (C3H8)O? Calculate the molar mass of Mg3(PO4)2. What is the mass percent of nitrogen in (NH4)3PO4? Final Exam-7 (Part 1) Final Exam-1 (Part 2) Print Screen Home End Page Up Page Down<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-64.png\" alt=\"\" class=\"wp-image-224526\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s address and explain each question from the image:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>6. Write a Lewis structure for nitrogen trifluoride, NF\u2083. Name the shape and state whether the molecule is polar or non-polar.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure:<\/strong><br>Nitrogen (N) is the central atom, with three fluorine (F) atoms bonded to it. Nitrogen has 5 valence electrons and each fluorine has 7. The structure includes 3 single bonds between N and F, and one lone pair on nitrogen. Each F also has 3 lone pairs.<\/li>\n\n\n\n<li><strong>Shape:<\/strong><br>Trigonal pyramidal (due to one lone pair on N).<\/li>\n\n\n\n<li><strong>Polarity:<\/strong><br><strong>Polar<\/strong> \u2014 Fluorine is more electronegative than nitrogen, and the lone pair creates an asymmetric distribution of electrons.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>7. How many atoms of C are in 0.185 mol of (C\u2083H\u2086)O?<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One molecule of (C\u2083H\u2086)O has 3 carbon atoms.<\/li>\n\n\n\n<li>Atoms of C = 0.185 mol \u00d7 6.022\u00d710236.022 \\times 10^{23} molecules\/mol \u00d7 3 C atoms\/molecule<br>= 3.35\u00d710233.35 \\times 10^{23} atoms of carbon (approx.)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>8. Calculate the molar mass of Mg\u2083(PO\u2084)\u2082.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mg: 24.31 \u00d7 3 = 72.93<\/li>\n\n\n\n<li>P: 30.97 \u00d7 2 = 61.94<\/li>\n\n\n\n<li>O: 16.00 \u00d7 8 = 128.00<\/li>\n\n\n\n<li><strong>Total molar mass<\/strong> = 72.93 + 61.94 + 128.00 = <strong>262.87 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>9. What is the mass percent of N in (NH\u2084)\u2083PO\u2084?<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of (NH\u2084)\u2083PO\u2084:\n<ul class=\"wp-block-list\">\n<li>N: 14.01 \u00d7 3 = 42.03<\/li>\n\n\n\n<li>H: 1.008 \u00d7 12 = 12.096<\/li>\n\n\n\n<li>P: 30.97<\/li>\n\n\n\n<li>O: 16.00 \u00d7 4 = 64.00<\/li>\n\n\n\n<li><strong>Total<\/strong> = 149.10 g\/mol<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Mass percent of N = (42.03 \/ 149.10) \u00d7 100 \u2248 <strong>28.19%<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>300-word Explanation:<\/strong><\/p>\n\n\n\n<p>These chemistry problems explore key concepts in molecular structure, stoichiometry, and composition. In problem 6, the Lewis structure for NF\u2083 illustrates how nitrogen forms three single bonds with fluorine atoms, leaving one lone pair. This lone pair causes electron repulsion, giving NF\u2083 a <strong>trigonal pyramidal<\/strong> shape. The asymmetry due to the lone pair and the highly electronegative fluorine atoms makes the molecule <strong>polar<\/strong>, resulting in a net dipole moment.<\/p>\n\n\n\n<p>In problem 7, the quantity of carbon atoms in a given mole of a compound is calculated by multiplying the number of moles by <strong>Avogadro\u2019s number<\/strong> and the number of carbon atoms per molecule. This showcases the concept of mole-to-atom conversion.<\/p>\n\n\n\n<p>Problem 8 requires calculating the <strong>molar mass<\/strong> of magnesium phosphate. This involves summing the atomic masses of all atoms in the formula unit. Molar mass is essential in converting between grams and moles, a key step in chemical reactions and stoichiometric calculations.<\/p>\n\n\n\n<p>Problem 9 examines the <strong>mass percent composition<\/strong>, a method to express the relative amount of an element in a compound. This involves dividing the mass contribution of the element (nitrogen) by the compound\u2019s total molar mass. This concept is crucial in analyzing mixtures, purity, and nutrient content in chemical and real-world applications.<\/p>\n\n\n\n<p>Together, these problems reinforce an understanding of chemical bonding, molecular shape, polarity, molar calculations, and composition\u2014all foundational topics in general chemistry.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-49.jpeg\" alt=\"\" class=\"wp-image-224527\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Write a Lewis structure for nitrogen trifluoride, NF3. Name the shape and state whether the molecule is polar or non-polar. How many atoms of carbon are in 0.185 mol of (C3H8)O? Calculate the molar mass of Mg3(PO4)2. What is the mass percent of nitrogen in (NH4)3PO4? Final Exam-7 (Part 1) Final Exam-1 (Part 2) Print [&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-224525","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224525","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=224525"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224525\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224525"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224525"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}