{"id":23761,"date":"2022-12-15T14:54:56","date_gmt":"2022-12-15T14:54:56","guid":{"rendered":"http:\/\/www.kaptest.com\/blog\/prep\/?p=23761"},"modified":"2025-02-07T17:53:54","modified_gmt":"2025-02-07T17:53:54","slug":"mcat-general-chemistry-quiz","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/mcat\/mcat-general-chemistry-quiz\/","title":{"rendered":"MCAT General Chemistry Quiz"},"content":{"rendered":"<p>Once you&#8217;ve familiarized yourself with the <a title=\"mcat general chemistry\" href=\"https:\/\/www.kaptest.com\/study\/mcat\/whats-tested-on-the-mcat-chemistry-and-physics\/\">general chemistry<\/a> that will be tested on the MCAT, try out this quiz.<br \/>\n<span style=\"font-weight: 400;\"><div  class='avia-builder-widget-area clearfix  avia-builder-el-0  el_before_av_promobox  avia-builder-el-first '><div id=\"custom_html-39\" class=\"widget_text widget clearfix widget_custom_html\"><div class=\"textwidget custom-html-widget\"><div><div class='op-interactive' id='639b934fae66b7643dc8915d' data-title='MCAT General Chemistry Quiz' data-url='https:\/\/kaplannorthamerica.outgrow.us\/639b934fae66b7643dc8915d?vHeight=1' data-width='100%'><\/div><script src='\/\/dyv6f9ner1ir9.cloudfront.net\/assets\/js\/nloader.js'><\/script><script>initIframe('639b934fae66b7643dc8915d');<\/script><\/div><\/div><\/div><\/div><\/span><br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-1  el_after_av_sidebar  el_before_av_toggle_container '>\t\t<div class='avia-promocontent'><p>\n<strong>Question 1<\/strong><br \/>\nWhat is the highest-energy orbital of elements with valence electrons in the n = 3 shell?<br \/>\nA.&nbsp;s-orbital<br \/>\nB.&nbsp;p-orbital<br \/>\nC.&nbsp;d-orbital<br \/>\nD.&nbsp;f-orbital<\/p>\n<\/div><\/div><br \/>\n<div  class=\"togglecontainer   toggle_close_all  avia-builder-el-2  el_after_av_promobox  el_before_av_promobox \" >\n<section class=\"av_toggle_section\"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\"  >    <div role=\"tablist\" class=\"single_toggle\" data-tags=\"{All} \"  >        <p data-fake-id=\"#toggle-id-1\" class=\"toggler \"  itemprop=\"headline\"    role=\"tab\" tabindex=\"0\" aria-controls=\"toggle-id-1\">Answer 1<span class=\"toggle_icon\" >        <span class=\"vert_icon\"><\/span><span class=\"hor_icon\"><\/span><\/span><\/p>        <div id=\"toggle-id-1\" class=\"toggle_wrap \"  >            <div class=\"toggle_content invers-color \"  itemprop=\"text\"   ><p><b>C:&nbsp;<\/b>When n = 3, l = 0, 1, or 2. The highest value for l in this case is 2, which corresponds to the d subshell. Although the&nbsp;<i>3d<\/i>&nbsp;block appears to be part of the fourth period, it still has the principal quantum number n = 3. In general, the subshells within an energy shell increase in energy as follows: s &lt; p &lt; d &lt; f (although there is no 3f subshell).<\/p>\n            <\/div>        <\/div>    <\/div><\/section>\n<\/div><br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-3  el_after_av_toggle_container  el_before_av_toggle_container '>\t\t<div class='avia-promocontent'><p>\n<strong>Question 2<\/strong><br \/>\nWhat is the most specific characterization of the reaction shown?<br \/>\n<a href=\"http:\/\/www.kaptest.com\/blog\/prep\/wp-content\/uploads\/sites\/21\/2019\/03\/Screen-Shot-2019-03-16-at-11.57.07-PM.png\"><img loading=\"lazy\" class=\"alignnone size-full wp-image-23762\" src=\"http:\/\/www.kaptest.com\/blog\/prep\/wp-content\/uploads\/sites\/21\/2019\/03\/Screen-Shot-2019-03-16-at-11.57.07-PM.png\" alt=\"mcat general chemistry\" width=\"438\" height=\"38\"><\/a><br \/>\nA.&nbsp;Single-displacement<br \/>\nB.&nbsp;Neutralization<br \/>\nC.&nbsp;Double-displacement<br \/>\nD.&nbsp;Oxidation\u2013reduction<\/p>\n<\/div><\/div><br \/>\n<div  class=\"togglecontainer   toggle_close_all  avia-builder-el-4  el_after_av_promobox  el_before_av_promobox \" >\n<section class=\"av_toggle_section\"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\"  >    <div role=\"tablist\" class=\"single_toggle\" data-tags=\"{All} \"  >        <p data-fake-id=\"#toggle-id-2\" class=\"toggler \"  itemprop=\"headline\"    role=\"tab\" tabindex=\"0\" aria-controls=\"toggle-id-2\">Answer 2<span class=\"toggle_icon\" >        <span class=\"vert_icon\"><\/span><span class=\"hor_icon\"><\/span><\/span><\/p>        <div id=\"toggle-id-2\" class=\"toggle_wrap \"  >            <div class=\"toggle_content invers-color \"  itemprop=\"text\"   ><p><b>B:&nbsp;<\/b>This reaction is a classic example of a neutralization reaction, in which an acid and a base react to form a salt and, usually, water. Although this reaction also fits the criteria for a double-displacement reaction, choice (C), in which two molecules essentially exchange ions with each other, neutralization is a more specific description of the process.<\/p>\n            <\/div>        <\/div>    <\/div><\/section>\n<\/div><br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-5  el_after_av_toggle_container  el_before_av_toggle_container '>\t\t<div class='avia-promocontent'><p>\n<strong>Question 3<\/strong><br \/>\nWhich of the following actions does NOT affect the equilibrium position of a reaction?<br \/>\nA.&nbsp;Adding or removing heat.<br \/>\nB.&nbsp;Adding or removing a catalyst.<br \/>\nC.&nbsp;Increasing or decreasing concentrations of reactants.<br \/>\nD.&nbsp;Increasing or decreasing volumes of reactants.<\/p>\n<\/div><\/div><br \/>\n<div  class=\"togglecontainer   toggle_close_all  avia-builder-el-6  el_after_av_promobox  el_before_av_promobox \" >\n<section class=\"av_toggle_section\"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\"  >    <div role=\"tablist\" class=\"single_toggle\" data-tags=\"{All} \"  >        <p data-fake-id=\"#toggle-id-3\" class=\"toggler \"  itemprop=\"headline\"    role=\"tab\" tabindex=\"0\" aria-controls=\"toggle-id-3\">Answer 3<span class=\"toggle_icon\" >        <span class=\"vert_icon\"><\/span><span class=\"hor_icon\"><\/span><\/span><\/p>        <div id=\"toggle-id-3\" class=\"toggle_wrap \"  >            <div class=\"toggle_content invers-color \"  itemprop=\"text\"   ><p><b>B:&nbsp;<\/b>The equilibrium of a reaction can be changed by several factors. Adding or subtracting heat, choice (A), would shift the equilibrium based on the enthalpy change of the reaction. Increasing reactant concentrations would shift the equilibrium in the direction of the product, and the opposite would occur if reactant concentrations were decreased, eliminating choice (C). Changing the volume of a reactant would affect any reaction with gaseous reactants or products, eliminating choice (D). While adding or removing a catalyst would change the reaction rates, it would not change where the equilibrium lies.<\/p>\n            <\/div>        <\/div>    <\/div><\/section>\n<\/div><br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-7  el_after_av_toggle_container  el_before_av_toggle_container '>\t\t<div class='avia-promocontent'><p>\n<b>Question 4<\/b><br \/>\nAt standard temperature and pressure, a chemical process is at equilibrium. What is the free energy of reaction (\u0394G) for this process?<br \/>\nA.&nbsp;\u0394G&gt; 0<br \/>\nB.&nbsp;\u0394G&lt; 0<br \/>\nC.&nbsp;\u0394G = 0<br \/>\nD.&nbsp;There is not enough information to determine the free energy of the reaction.<\/p>\n<\/div><\/div><br \/>\n<div  class=\"togglecontainer   toggle_close_all  avia-builder-el-8  el_after_av_promobox  el_before_av_promobox \" >\n<section class=\"av_toggle_section\"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\"  >    <div role=\"tablist\" class=\"single_toggle\" data-tags=\"{All} \"  >        <p data-fake-id=\"#toggle-id-4\" class=\"toggler \"  itemprop=\"headline\"    role=\"tab\" tabindex=\"0\" aria-controls=\"toggle-id-4\">Answer 4<span class=\"toggle_icon\" >        <span class=\"vert_icon\"><\/span><span class=\"hor_icon\"><\/span><\/span><\/p>        <div id=\"toggle-id-4\" class=\"toggle_wrap \"  >            <div class=\"toggle_content invers-color \"  itemprop=\"text\"   ><p><b>C:&nbsp;<\/b>Standard temperature and pressure indicates 0\u00b0C and 1 atm. Gibbs free energy is temperature dependent, but if a reaction is at equilibrium, \u0394G = 0.<\/p>\n            <\/div>        <\/div>    <\/div><\/section>\n<\/div><br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-9  el_after_av_toggle_container  el_before_av_toggle_container '>\t\t<div class='avia-promocontent'><p>\n<strong>Question 5<\/strong><br \/>\nWhich of the following correctly lists the enthalpy changes for these three steps, respectively?<br \/>\n<em>The process of formation of a salt solution can be better understood by breaking the process into three steps:<\/em><br \/>\n<em>1. Breaking the solute into its individual components<\/em><br \/>\n<em>2. Making room for the solute in the solvent by overcoming intermolecular forces in the solvent<\/em><br \/>\n<em>3. Allowing solute\u2013solvent interactions to occur to form the solution<\/em><br \/>\nA.&nbsp;Endothermic, exothermic, endothermic<br \/>\nB.&nbsp;Exothermic, endothermic, endothermic<br \/>\nC.&nbsp;Exothermic, exothermic, endothermic<br \/>\nD.&nbsp;Endothermic, endothermic, exothermic<\/p>\n<\/div><\/div><br \/>\n<div  class=\"togglecontainer   toggle_close_all  avia-builder-el-10  el_after_av_promobox  avia-builder-el-last \" >\n<section class=\"av_toggle_section\"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\"  >    <div role=\"tablist\" class=\"single_toggle\" data-tags=\"{All} \"  >        <p data-fake-id=\"#toggle-id-5\" class=\"toggler \"  itemprop=\"headline\"    role=\"tab\" tabindex=\"0\" aria-controls=\"toggle-id-5\">Answer 5<span class=\"toggle_icon\" >        <span class=\"vert_icon\"><\/span><span class=\"hor_icon\"><\/span><\/span><\/p>        <div id=\"toggle-id-5\" class=\"toggle_wrap \"  >            <div class=\"toggle_content invers-color \"  itemprop=\"text\"   ><p><b>D:&nbsp;<\/b>The first step will most likely be endothermic because energy is required to break molecules apart. The second step is also endothermic because the intermolecular forces in the solvent must be overcome to allow incorporation of solute particles. The third step will most likely be exothermic be-cause polar water molecules will interact with the dissolved ions, creating a stable solution and releasing energy.<\/p>\n            <\/div>        <\/div>    <\/div><\/section>\n<\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Once you&#8217;ve familiarized yourself with the general chemistry that will be tested on the MCAT, try out this quiz.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[2],"tags":[],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/23761"}],"collection":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/comments?post=23761"}],"version-history":[{"count":4,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/23761\/revisions"}],"predecessor-version":[{"id":47840,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/23761\/revisions\/47840"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=23761"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=23761"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=23761"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}