{"id":120822,"date":"2023-10-01T17:40:36","date_gmt":"2023-10-01T17:40:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=120822"},"modified":"2023-10-01T17:40:38","modified_gmt":"2023-10-01T17:40:38","slug":"chem-104-module-1-module-6-exam-newest-questions-and-answers-2023-2024-verified-by-expert","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2023\/10\/01\/chem-104-module-1-module-6-exam-newest-questions-and-answers-2023-2024-verified-by-expert\/","title":{"rendered":"CHEM 104 MODULE 1 \u2013 MODULE 6 Exam Newest Questions and Answers (2023 \/ 2024) (Verified by Expert)"},"content":{"rendered":"\n<p>Chem 104 Module 1 to 6 Exam answers Portage learning 2022<br>^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^<br>Module 1:<br>Question 1<br>In the reaction of gaseous N2<br>O5<br>to yield NO2<br>gas and O2<br>gas as shown below, the following data table<br>is obtained:<br>\u2192 4 NO2 (g) + O2 (g)<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using the [O2<br>] data from the table, show the calculation of the instantaneous rate early in the<br>reaction (0 secs to 300 sec).<\/li>\n\n\n\n<li>Using the [O2<br>] data from the table, show the calculation of the instantaneous rate late in the reaction<br>(2400 secs to 3000 secs).<\/li>\n\n\n\n<li>Explain the relative values of the early instantaneous rate and the late instantaneous rate.<br>Your Answer:<br>2 N2<br>O5 (g)<br>Data Table #2<br>Time (sec) [N2<br>O5<br>] [O2<br>]<br>0 0.300 M 0<br>300 0.272 M 0.014 M<br>600 0.224 M 0.038 M<br>900 0.204 M 0.048 M<br>1200 0.186 M 0.057 M<br>1800 0.156 M 0.072 M<br>2400 0.134 M 0.083 M<br>3000 0.120 M 0.090 M<\/li>\n\n\n\n<li>rate = (0.014 &#8211; 0) \/ (300 &#8211; 0) = 4.67 x 10-5 mol\/Ls<\/li>\n\n\n\n<li>rate = (0.090 &#8211; 0.083) \/ (3000 &#8211; 2400) = 1.167 x 10-5 mol\/Ls<\/li>\n\n\n\n<li>The late instantaneous rate is smaller than the early instantaneous rate.<\/li>\n<\/ol>\n\n\n\n<p>Question 2<br>The following rate data was obtained for the hypothetical reaction: A + B \u2192 X + Y<br>Experiment # [A] [B] rate<br>1 0.50 0.50 2.0<br>2 1.00 0.50 8.0<br>3 1.00 1.00 64.0<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Determine the reaction order with respect to [A].<\/li>\n\n\n\n<li>Determine the reaction order with respect to [B].<\/li>\n\n\n\n<li>Write the rate law in the form rate = k [A]n<br>[B]m (filling in the correct exponents).<\/li>\n\n\n\n<li>Show the calculation of the rate constant, k.<br>Your Answer:<br>rate = k [A]x<br>[B]y<br>rate 1 \/ rate 2 = k [0.50]x<br>[0.50]y<br>\/ k [1.00]x<br>[0.50]y<br>2.0 \/ 8.0 = [0.50]x<br>\/ [1.00]x<br>0.25 = 0.5x<br>x = 2<br>rate 2 \/ rate 3 = k [1.00]x<br>[0.50]y<br>\/ k [1.00]x<br>[1.00]y<br>8.0 \/ 64.0 = [0.50]y<br>\/ [1.00]y<br>0.125 = 0.5y<br>y = 3<br>rate = k [A]2<br>[B]3<br>2.0 = k [0.50]2<br>[0.50]3<br>k = 64<br>Question 3<br>ln [A] &#8211; ln [A]0<br>= &#8211; k t 0.693 = k t<br>1\/2<br>An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day<br>sample. The t1\/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the<br>paper.<\/li>\n<\/ol>\n\n\n\n<p>Your Answer:<br>0.693 = k t1\/2<br>0.693 = k (5720)<br>k = 1.21 x 10-4<br>ln [A] &#8211; ln [A]0<br>= &#8211; k t<br>ln 19.8 &#8211; ln 100 = &#8211; 1.21 x 10-4 t<br>t = 13, 384 years<br>Question 4<br>Using the potential energy diagram below, state whether the reaction described by the diagram is<br>endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer.<br>Your Answer:<br>The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it<br>has relatively large Eact.<br>Question 5<br>Show the calculation of Kc<br>for the following reaction if an initial reaction mixture of 0.800 mole of CO<br>and 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of<br>H2<br>O and corresponding amounts of CO, H2<br>, and CH4<br>.<br>CO (g) + 3 H2 (g) CH4 (g) + H2<br>O (g)<br>Your Answer:<br>0.309 mole of H2<br>O formed = 0.309 mole of CH4<br>formed<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Chem 104 Module 1 to 6 Exam answers Portage learning 2022^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^Module 1:Question 1In the reaction of gaseous N2O5to yield NO2gas and O2gas as shown below, the following data tableis obtained:\u2192 4 NO2 (g) + O2 (g) Question 2The following rate data was obtained for the hypothetical reaction: A + B \u2192 X + YExperiment # [&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-120822","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/120822","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=120822"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/120822\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=120822"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=120822"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=120822"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}