A 3.41 time 10 to the -6g sample is known to contain 4.67 times time 10 to the 16 molecules. what is this compound<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To determine what compound<\/strong> the sample is, we can use the mass of the sample<\/strong> and the number of molecules<\/strong> to calculate its molar mass<\/strong>, then compare that to known molar masses of compounds.<\/p>\n\n\n\n Use Avogadro\u2019s number<\/strong>: 6.022\u00d71023 molecules\/mol6.022 \\times 10^{23} \\text{ molecules\/mol} Moles=4.67\u00d71016 molecules6.022\u00d71023 molecules\/mol=7.76\u00d710\u22128 mol\\text{Moles} = \\frac{4.67 \\times 10^{16} \\text{ molecules}}{6.022 \\times 10^{23} \\text{ molecules\/mol}} = 7.76 \\times 10^{-8} \\text{ mol}<\/p>\n\n\n\n Molar Mass=MassMoles=3.41\u00d710\u22126 g7.76\u00d710\u22128 mol\u224843.95 g\/mol\\text{Molar Mass} = \\frac{\\text{Mass}}{\\text{Moles}} = \\frac{3.41 \\times 10^{-6} \\text{ g}}{7.76 \\times 10^{-8} \\text{ mol}} \\approx 43.95 \\text{ g\/mol}<\/p>\n\n\n\n A molar mass of approximately 44 g\/mol<\/strong> suggests a few possibilities. The most common compound with a molar mass close to this is:<\/p>\n\n\n\n \ud83d\udd39 Carbon dioxide (CO\u2082)<\/strong> 12.01\u2009(C)+2\u00d716.00\u2009(O)=44.01 g\/mol12.01 \\, \\text{(C)} + 2 \\times 16.00 \\, \\text{(O)} = 44.01 \\text{ g\/mol}<\/p>\n\n\n\n This problem involves basic principles of chemistry \u2014 using mass<\/strong>, number of particles<\/strong>, and molar mass<\/strong> to identify a substance. By converting the given number of molecules to moles using Avogadro\u2019s number, and then applying the definition of molar mass (mass per mole), we calculate a molar mass of approximately 44 g\/mol. The only common compound that matches this molar mass is carbon dioxide (CO\u2082), making it the best and most accurate answer.<\/p>\n","protected":false},"excerpt":{"rendered":" A 3.41 time 10 to the -6g sample is known to contain 4.67 times time 10 to the 16 molecules. what is this compound The correct answer and explanation is: To determine what compound the sample is, we can use the mass of the sample and the number of molecules to calculate its molar mass, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-23355","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23355","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=23355"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23355\/revisions"}],"predecessor-version":[{"id":23356,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23355\/revisions\/23356"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23355"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23355"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23355"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Convert molecules to moles<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Calculate Molar Mass<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 3: Identify the compound<\/strong><\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer: The compound is Carbon Dioxide (CO\u2082)<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\n