Step 1: Bromination Reaction<\/strong><\/h3>\n\n\n\nThe reaction:
C14H12+Br2\u2192C14H12Br2C_{14}H_{12} + Br_2 \\to C_{14}H_{12}Br_2C14\u200bH12\u200b+Br2\u200b\u2192C14\u200bH12\u200bBr2\u200b<\/p>\n\n\n\n
Given:<\/p>\n\n\n\n
\n- Mass of trans-stilbene: 2.017 g<\/li>\n\n\n\n
- Mass of PHPB: 4.001 g<\/li>\n<\/ul>\n\n\n\n
You\u2019ve calculated the limiting reagent, which is trans-stilbene (C14H12), and it’s given that the theoretical yield is 340 g of C14H12Br2.<\/p>\n\n\n\n
The calculations show a 100% yield in the bromination reaction, which means you’re assuming all the trans-stilbene reacts with bromine, and the full theoretical mass of trans-1,2-dibromo-1,2-diphenylethane is obtained.<\/p>\n\n\n\n
Step 2: Dehydrobromination Reaction<\/strong><\/h3>\n\n\n\nThe reaction:
C14H12Br2+2KOH\u2192C14H10+2KBr+2H2OC_{14}H_{12}Br_2 + 2KOH \\to C_{14}H_{10} + 2KBr + 2H_2OC14\u200bH12\u200bBr2\u200b+2KOH\u2192C14\u200bH10\u200b+2KBr+2H2\u200bO<\/p>\n\n\n\n
Given:<\/p>\n\n\n\n
\n- Mass of KOH: 1.50 g<\/li>\n<\/ul>\n\n\n\n
The limiting reagent calculation leads to diphenylacetylene (C14H10) with a theoretical yield of 178.2 g. Assuming a 100% yield means you get this amount of C14H10 after the reaction.<\/p>\n\n\n\n
Percent Yield<\/strong><\/h3>\n\n\n\nThe percent yield calculation, which uses the actual yield (what you obtain from the experiment) divided by the theoretical yield (calculated from stoichiometry), multiplied by 100, is assumed to be 100% here. This means no losses during the reaction.<\/p>\n\n\n\n
Melting Point of Diphenylacetylene<\/strong><\/h3>\n\n\n\nThe melting point of diphenylacetylene (C14H10) is between 58.8-62.9\u00b0C<\/strong>, which aligns with common literature values for this compound. If your experimental product is pure, it should fall within this range. If the yield is lower or impure, the melting point might be lower than expected.<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
This synthesis pathway involves straightforward stoichiometric calculations with the assumption of perfect yields. The limiting reagents are correctly identified, and theoretical yields align with expectations based on molecular weights and stoichiometry.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-831.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Step 1: C14H12 + Br2 -> C14H12Br2 Mass of trans-stilbene: 2.017 g Mass of PHPB: 4.001 g Limiting reagent calculations: 1 mol C14H12 -> 1 mol C14H12Br2 2.017 g C14H12 -> x mol C14H12Br2 180.25 g C14H12 -> 1 mol C14H12 1 mol C14H12 -> 1 mol C6H6Br3N 3 mol Br -> 1 mol Br2 […]<\/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-257807","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807","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=257807"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257807"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257807"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
C14H12+Br2\u2192C14H12Br2C_{14}H_{12} + Br_2 \\to C_{14}H_{12}Br_2C14\u200bH12\u200b+Br2\u200b\u2192C14\u200bH12\u200bBr2\u200b<\/p>\n\n\n\n
You\u2019ve calculated the limiting reagent, which is trans-stilbene (C14H12), and it’s given that the theoretical yield is 340 g of C14H12Br2.<\/p>\n\n\n\n
The calculations show a 100% yield in the bromination reaction, which means you’re assuming all the trans-stilbene reacts with bromine, and the full theoretical mass of trans-1,2-dibromo-1,2-diphenylethane is obtained.<\/p>\n\n\n\n
Step 2: Dehydrobromination Reaction<\/strong><\/h3>\n\n\n\nThe reaction:
C14H12Br2+2KOH\u2192C14H10+2KBr+2H2OC_{14}H_{12}Br_2 + 2KOH \\to C_{14}H_{10} + 2KBr + 2H_2OC14\u200bH12\u200bBr2\u200b+2KOH\u2192C14\u200bH10\u200b+2KBr+2H2\u200bO<\/p>\n\n\n\n
Given:<\/p>\n\n\n\n
\n- Mass of KOH: 1.50 g<\/li>\n<\/ul>\n\n\n\n
The limiting reagent calculation leads to diphenylacetylene (C14H10) with a theoretical yield of 178.2 g. Assuming a 100% yield means you get this amount of C14H10 after the reaction.<\/p>\n\n\n\n
Percent Yield<\/strong><\/h3>\n\n\n\nThe percent yield calculation, which uses the actual yield (what you obtain from the experiment) divided by the theoretical yield (calculated from stoichiometry), multiplied by 100, is assumed to be 100% here. This means no losses during the reaction.<\/p>\n\n\n\n
Melting Point of Diphenylacetylene<\/strong><\/h3>\n\n\n\nThe melting point of diphenylacetylene (C14H10) is between 58.8-62.9\u00b0C<\/strong>, which aligns with common literature values for this compound. If your experimental product is pure, it should fall within this range. If the yield is lower or impure, the melting point might be lower than expected.<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
This synthesis pathway involves straightforward stoichiometric calculations with the assumption of perfect yields. The limiting reagents are correctly identified, and theoretical yields align with expectations based on molecular weights and stoichiometry.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Step 1: C14H12 + Br2 -> C14H12Br2 Mass of trans-stilbene: 2.017 g Mass of PHPB: 4.001 g Limiting reagent calculations: 1 mol C14H12 -> 1 mol C14H12Br2 2.017 g C14H12 -> x mol C14H12Br2 180.25 g C14H12 -> 1 mol C14H12 1 mol C14H12 -> 1 mol C6H6Br3N 3 mol Br -> 1 mol Br2 […]<\/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-257807","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807","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=257807"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257807"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257807"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
C14H12Br2+2KOH\u2192C14H10+2KBr+2H2OC_{14}H_{12}Br_2 + 2KOH \\to C_{14}H_{10} + 2KBr + 2H_2OC14\u200bH12\u200bBr2\u200b+2KOH\u2192C14\u200bH10\u200b+2KBr+2H2\u200bO<\/p>\n\n\n\n
The limiting reagent calculation leads to diphenylacetylene (C14H10) with a theoretical yield of 178.2 g. Assuming a 100% yield means you get this amount of C14H10 after the reaction.<\/p>\n\n\n\n
Percent Yield<\/strong><\/h3>\n\n\n\nThe percent yield calculation, which uses the actual yield (what you obtain from the experiment) divided by the theoretical yield (calculated from stoichiometry), multiplied by 100, is assumed to be 100% here. This means no losses during the reaction.<\/p>\n\n\n\n
Melting Point of Diphenylacetylene<\/strong><\/h3>\n\n\n\nThe melting point of diphenylacetylene (C14H10) is between 58.8-62.9\u00b0C<\/strong>, which aligns with common literature values for this compound. If your experimental product is pure, it should fall within this range. If the yield is lower or impure, the melting point might be lower than expected.<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
This synthesis pathway involves straightforward stoichiometric calculations with the assumption of perfect yields. The limiting reagents are correctly identified, and theoretical yields align with expectations based on molecular weights and stoichiometry.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Step 1: C14H12 + Br2 -> C14H12Br2 Mass of trans-stilbene: 2.017 g Mass of PHPB: 4.001 g Limiting reagent calculations: 1 mol C14H12 -> 1 mol C14H12Br2 2.017 g C14H12 -> x mol C14H12Br2 180.25 g C14H12 -> 1 mol C14H12 1 mol C14H12 -> 1 mol C6H6Br3N 3 mol Br -> 1 mol Br2 […]<\/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-257807","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807","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=257807"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257807"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257807"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The melting point of diphenylacetylene (C14H10) is between 58.8-62.9\u00b0C<\/strong>, which aligns with common literature values for this compound. If your experimental product is pure, it should fall within this range. If the yield is lower or impure, the melting point might be lower than expected.<\/p>\n\n\n\n This synthesis pathway involves straightforward stoichiometric calculations with the assumption of perfect yields. The limiting reagents are correctly identified, and theoretical yields align with expectations based on molecular weights and stoichiometry.<\/p>\n\n\n\n Step 1: C14H12 + Br2 -> C14H12Br2 Mass of trans-stilbene: 2.017 g Mass of PHPB: 4.001 g Limiting reagent calculations: 1 mol C14H12 -> 1 mol C14H12Br2 2.017 g C14H12 -> x mol C14H12Br2 180.25 g C14H12 -> 1 mol C14H12 1 mol C14H12 -> 1 mol C6H6Br3N 3 mol Br -> 1 mol Br2 […]<\/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-257807","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807","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=257807"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257807\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257807"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257807"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Conclusion<\/h3>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"