{"id":259477,"date":"2025-07-19T02:17:22","date_gmt":"2025-07-19T02:17:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259477"},"modified":"2025-07-19T02:17:24","modified_gmt":"2025-07-19T02:17:24","slug":"how-many-grams-are-there-in-4-89-moles-of-calcium-bromide","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/how-many-grams-are-there-in-4-89-moles-of-calcium-bromide\/","title":{"rendered":"How many grams are there in 4.89 moles of Calcium Bromide"},"content":{"rendered":"\n
How many grams are there in 4.89 moles of Calcium Bromide (CaBr2)? The molar mass of Ca is 40.08 g\/mol, while Br has a molar mass of 79.90 g\/mol.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
To calculate the mass of 4.89 moles of calcium bromide (CaBr2), we need to use the molar mass of CaBr2, which is the sum of the molar masses of calcium (Ca) and bromine (Br).<\/p>\n\n\n\n
Step 1: Calculate the molar mass of CaBr2<\/h3>\n\n\n\n
\n
Molar mass of calcium (Ca) = 40.08 g\/mol<\/li>\n\n\n\n
Molar mass of bromine (Br) = 79.90 g\/mol<\/li>\n<\/ul>\n\n\n\n
Since there are two bromine atoms in CaBr2, we multiply the molar mass of bromine by 2:<\/p>\n\n\n\n
\n
Molar mass of CaBr2 = 40.08 g\/mol + 2 \u00d7 79.90 g\/mol<\/li>\n\n\n\n
Molar mass of CaBr2 = 40.08 g\/mol + 159.80 g\/mol<\/li>\n\n\n\n
Molar mass of CaBr2 = 199.88 g\/mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Use the molar mass to calculate the mass of 4.89 moles<\/h3>\n\n\n\n
Now that we know the molar mass of CaBr2 is 199.88 g\/mol, we can calculate the mass of 4.89 moles of CaBr2 using the formula:mass=moles\u00d7molar mass\\text{mass} = \\text{moles} \\times \\text{molar mass}mass=moles\u00d7molar massmass=4.89\u2009moles\u00d7199.88\u2009g\/mol\\text{mass} = 4.89 \\, \\text{moles} \\times 199.88 \\, \\text{g\/mol}mass=4.89moles\u00d7199.88g\/molmass=977.72\u2009grams\\text{mass} = 977.72 \\, \\text{grams}mass=977.72grams<\/p>\n\n\n\n
Conclusion<\/h3>\n\n\n\n
The mass of 4.89 moles of calcium bromide (CaBr2) is 977.72 grams. This is the total mass of the compound for the given number of moles, based on the molar masses of calcium and bromine.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
How many grams are there in 4.89 moles of Calcium Bromide (CaBr2)? The molar mass of Ca is 40.08 g\/mol, while Br has a molar mass of 79.90 g\/mol. The Correct Answer and Explanation is: To calculate the mass of 4.89 moles of calcium bromide (CaBr2), we need to use the molar mass of CaBr2, […]<\/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-259477","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259477","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=259477"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259477\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259477"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259477"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259477"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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