(1 point) How many moles are contained in 23.5 g of Sb2S3? 2. (1 point) How many formula units are contained in 23.5 g of Sb2S3? 3. (1 point) How many sulfide ions are present in 23.5 g of Sb2S3? 4.(1 point) Given 5.0 x 1024 K2SO4 formula units, how many moles are present? 5.(1 point) Given 5.0 x 1024 K2SO4 formula units, how many grams are present?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the moles of Sb\u2082S\u2083, we first need to determine the molar mass of Sb\u2082S\u2083. The molar mass can be calculated using the atomic masses of antimony (Sb) and sulfur (S).<\/p>\n\n\n\n Molar mass of Sb\u2082S\u2083 = (2 \u00d7 121.76 g\/mol) + (3 \u00d7 32.06 g\/mol) Now, we can calculate the number of moles in 23.5 g of Sb\u2082S\u2083:<\/p>\n\n\n\n [ Thus, there are 0.0692 moles<\/strong> of Sb\u2082S\u2083 in 23.5 g.<\/p>\n\n\n\n The number of formula units is calculated using Avogadro’s number ((6.022 \\times 10^{23}) units\/mol). To find the number of formula units, multiply the number of moles by Avogadro’s number:<\/p>\n\n\n\n [ Thus, there are 4.17 \u00d7 10\u00b2\u00b2 formula units<\/strong> of Sb\u2082S\u2083 in 23.5 g.<\/p>\n\n\n\n Each formula unit of Sb\u2082S\u2083 contains 3 sulfide ions (S\u00b2\u207b). To find the number of sulfide ions, we multiply the number of formula units by 3:<\/p>\n\n\n\n [ Thus, there are 1.25 \u00d7 10\u00b2\u00b3 sulfide ions<\/strong> in 23.5 g of Sb\u2082S\u2083.<\/p>\n\n\n\n To calculate moles, use the formula:<\/p>\n\n\n\n [ Thus, there are 8.3 moles<\/strong> of K\u2082SO\u2084.<\/p>\n\n\n\n First, calculate the molar mass of K\u2082SO\u2084:<\/p>\n\n\n\n Molar mass of K\u2082SO\u2084 = (2 \u00d7 39.10) + 32.06 + (4 \u00d7 16.00) = 78.20 + 32.06 + 64.00 = 174.26 g\/mol<\/p>\n\n\n\n Now, convert moles to grams:<\/p>\n\n\n\n [ Thus, there are 1443.35 grams<\/strong> of K\u2082SO\u2084 in 5.0 \u00d7 10\u00b2\u2074 formula units.<\/p>\n\n\n\n (1 point) How many moles are contained in 23.5 g of Sb2S3? 2. (1 point) How many formula units are contained in 23.5 g of Sb2S3? 3. (1 point) How many sulfide ions are present in 23.5 g of Sb2S3? 4.(1 point) Given 5.0 x 1024 K2SO4 formula units, how many moles are present? 5.(1 […]<\/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-181344","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181344","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=181344"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181344\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181344"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181344"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. How many moles are contained in 23.5 g of Sb\u2082S\u2083?<\/h3>\n\n\n\n
\n
= 243.52 g\/mol + 96.18 g\/mol
= 339.70 g\/mol<\/p>\n\n\n\n
\\text{moles of Sb\u2082S\u2083} = \\frac{\\text{mass of Sb\u2082S\u2083}}{\\text{molar mass of Sb\u2082S\u2083}} = \\frac{23.5 \\, \\text{g}}{339.70 \\, \\text{g\/mol}} = 0.0692 \\, \\text{mol}
]<\/p>\n\n\n\n2. How many formula units are contained in 23.5 g of Sb\u2082S\u2083?<\/h3>\n\n\n\n
\\text{formula units} = 0.0692 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{units\/mol} = 4.17 \\times 10^{22} \\, \\text{formula units}
]<\/p>\n\n\n\n3. How many sulfide ions are present in 23.5 g of Sb\u2082S\u2083?<\/h3>\n\n\n\n
\\text{sulfide ions} = 3 \\times 4.17 \\times 10^{22} = 1.25 \\times 10^{23} \\, \\text{ions}
]<\/p>\n\n\n\n4. Given 5.0 \u00d7 10\u00b2\u2074 K\u2082SO\u2084 formula units, how many moles are present?<\/h3>\n\n\n\n
\\text{moles of K\u2082SO\u2084} = \\frac{\\text{formula units of K\u2082SO\u2084}}{\\text{Avogadro’s number}} = \\frac{5.0 \\times 10^{24}}{6.022 \\times 10^{23}} = 8.3 \\, \\text{mol}
]<\/p>\n\n\n\n5. Given 5.0 \u00d7 10\u00b2\u2074 K\u2082SO\u2084 formula units, how many grams are present?<\/h3>\n\n\n\n
\n
\\text{mass of K\u2082SO\u2084} = 8.3 \\, \\text{mol} \\times 174.26 \\, \\text{g\/mol} = 1443.35 \\, \\text{g}
]<\/p>\n\n\n\nSummary of answers:<\/h3>\n\n\n\n
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