How many milliliters of an aqueous solution of 0.142 M nickel(II) nitrate is needed to obtain 12.8 grams of the salt?<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n To find the volume of a 0.142 M nickel(II) nitrate solution needed to obtain 12.8 grams of the salt, follow these steps:<\/p>\n\n\n\n Answer:<\/strong> Approximately 493 mL of 0.142 M nickel(II) nitrate solution is needed.<\/p>\n\n\n\n Explanation:<\/strong> How many milliliters of an aqueous solution of 0.142 M nickel(II) nitrate is needed to obtain 12.8 grams of the salt? The correct answer and explanation is: To find the volume of a 0.142 M nickel(II) nitrate solution needed to obtain 12.8 grams of the salt, follow these steps: Answer: Approximately 493 mL of 0.142 […]<\/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-41653","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41653","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=41653"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41653\/revisions"}],"predecessor-version":[{"id":41654,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41653\/revisions\/41654"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41653"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41653"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41653"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
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Formula: Ni + 2 \u00d7 (N + 3 \u00d7 O) = 58.69 + 2 \u00d7 (14.01 + 3 \u00d7 16.00)
= 58.69 + 2 \u00d7 (14.01 + 48.00)
= 58.69 + 2 \u00d7 62.01
= 58.69 + 124.02
= 182.71 g\/mol<\/li>\n<\/ul>\n<\/li>\n\n\n\n
moles = mass \/ molar mass = 12.8 g \/ 182.71 g\/mol \u2248 0.0700 mol<\/li>\n\n\n\n
volume (L) = moles \/ molarity = 0.0700 mol \/ 0.142 mol\/L \u2248 0.4929 L<\/li>\n\n\n\n
0.4929 L \u00d7 1000 mL\/L = 492.9 mL<\/li>\n<\/ol>\n\n\n\n
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Molarity (M) represents the number of moles of solute dissolved per liter of solution. To find how much solution contains a certain mass of nickel(II) nitrate, it is necessary first to convert that mass into moles using the compound’s molar mass. The molar mass is the sum of the atomic masses of all atoms in the formula unit of nickel(II) nitrate, calculated based on periodic table values. After finding the moles of nickel(II) nitrate, the volume of solution required can be found by rearranging the molarity formula. Finally, converting liters to milliliters provides a practical unit for measuring solution volumes in a laboratory setting. This process applies broadly for preparing solutions from solid solutes with known molarity values.<\/p>\n","protected":false},"excerpt":{"rendered":"