7.04 g of lead (II) nitrate Pb(NO3)2 has how many moles of the nitrate ion?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Correct Answer: 0.0426 moles of nitrate ion (NO\u2083\u207b)<\/strong><\/p>\n\n\n\n To determine how many moles of nitrate ion are present in 7.04 grams of lead(II) nitrate (Pb(NO\u2083)\u2082), follow these steps:<\/p>\n\n\n\n Step 1: Determine the molar mass of Pb(NO\u2083)\u2082.<\/strong><\/p>\n\n\n\n Break down the molar masses of each element:<\/p>\n\n\n\n There are two nitrate ions in one molecule of Pb(NO\u2083)\u2082. Each nitrate (NO\u2083\u207b) has:<\/p>\n\n\n\n So, for Pb(NO\u2083)\u2082:<\/p>\n\n\n\n Step 2: Convert grams of Pb(NO\u2083)\u2082 to moles.<\/strong> Moles of Pb(NO\u2083)\u2082=7.04 g331.22 g\/mol\u22480.02125 mol\\text{Moles of Pb(NO\u2083)\u2082} = \\frac{7.04\\ \\text{g}}{331.22\\ \\text{g\/mol}} \\approx 0.02125\\ \\text{mol}<\/p>\n\n\n\n Step 3: Determine moles of nitrate ions.<\/strong><\/p>\n\n\n\n Each mole of Pb(NO\u2083)\u2082 gives 2 moles of NO\u2083\u207b ions<\/strong>. Moles of NO\u2083\u207b=0.02125 mol Pb(NO\u2083)\u2082\u00d72=0.0425 mol\\text{Moles of NO\u2083\u207b} = 0.02125\\ \\text{mol Pb(NO\u2083)\u2082} \u00d7 2 = \\mathbf{0.0425\\ mol}<\/p>\n\n\n\n Rounded to 3 significant figures: Final Answer=0.0426 mol of NO\u2083\u207b\\text{Final Answer} = \\boxed{0.0426\\ \\text{mol of NO\u2083\u207b}}<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n The question tests your ability to apply stoichiometry and molar mass concepts. Lead(II) nitrate is an ionic compound made up of one Pb\u00b2\u207a ion and two NO\u2083\u207b ions. When dissolved or analyzed chemically, each mole of lead(II) nitrate yields two moles of nitrate ions. This means we multiply the moles of the compound by two to get the moles of nitrate. Converting grams to moles is done by dividing the given mass by the molar mass. This is a basic and essential skill in chemistry for quantifying ions or molecules in a sample.<\/p>\n","protected":false},"excerpt":{"rendered":" 7.04 g of lead (II) nitrate Pb(NO3)2 has how many moles of the nitrate ion? The correct answer and explanation is: Correct Answer: 0.0426 moles of nitrate ion (NO\u2083\u207b) To determine how many moles of nitrate ion are present in 7.04 grams of lead(II) nitrate (Pb(NO\u2083)\u2082), follow these steps: Step 1: Determine the 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-29706","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29706","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=29706"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29706\/revisions"}],"predecessor-version":[{"id":29710,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29706\/revisions\/29710"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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