{"id":250373,"date":"2025-07-10T09:05:19","date_gmt":"2025-07-10T09:05:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=250373"},"modified":"2025-07-10T09:05:21","modified_gmt":"2025-07-10T09:05:21","slug":"a-certain-liquid-has-a-density-of-1-25-g-cm3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/a-certain-liquid-has-a-density-of-1-25-g-cm3\/","title":{"rendered":"A certain liquid has a density of 1.25 g\/cm3."},"content":{"rendered":"\n<p>A certain liquid has a density of 1.25 g\/cm3. Which drawing below most closely represents the volume of this liquid needed to obtain 7.50 g of the liquid?<\/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 find the volume of the liquid needed to obtain 7.50 grams of it, we can use the formula:Density=MassVolume\\text{Density} = \\frac{\\text{Mass}}{\\text{Volume}}Density=VolumeMass\u200b<\/p>\n\n\n\n<p>Rearranging the formula to solve for volume:Volume=MassDensity\\text{Volume} = \\frac{\\text{Mass}}{\\text{Density}}Volume=DensityMass\u200b<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass = 7.50 g<\/li>\n\n\n\n<li>Density = 1.25 g\/cm\u00b3<\/li>\n<\/ul>\n\n\n\n<p>Substitute the values into the formula:Volume=7.50\u2009g1.25\u2009g\/cm3=6.00\u2009cm3\\text{Volume} = \\frac{7.50 \\, \\text{g}}{1.25 \\, \\text{g\/cm}^3} = 6.00 \\, \\text{cm}^3Volume=1.25g\/cm37.50g\u200b=6.00cm3<\/p>\n\n\n\n<p>So, the volume of liquid needed is 6.00 cm\u00b3.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>Density is the mass per unit volume. The higher the density, the smaller the volume required for a given mass. In this case, for every 1.25 grams of the liquid, the volume is 1 cm\u00b3. Therefore, to obtain 7.50 grams, you need a volume of 6.00 cm\u00b3, which is calculated by dividing the mass by the density.<\/p>\n\n\n\n<p>This volume (6.00 cm\u00b3) would likely correspond to a container that holds approximately 6 cubic centimeters of the liquid. Depending on the available drawings, you would select the one that visually represents this volume. The drawing would typically be a small rectangular or cubic container that visually represents about 6 cm\u00b3 of liquid.<\/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-banner6-111.jpeg\" alt=\"\" class=\"wp-image-250374\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A certain liquid has a density of 1.25 g\/cm3. Which drawing below most closely represents the volume of this liquid needed to obtain 7.50 g of the liquid? The Correct Answer and Explanation is: To find the volume of the liquid needed to obtain 7.50 grams of it, we can use the formula:Density=MassVolume\\text{Density} = \\frac{\\text{Mass}}{\\text{Volume}}Density=VolumeMass\u200b [&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-250373","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250373","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=250373"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250373\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=250373"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=250373"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=250373"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}