{"id":248536,"date":"2025-07-08T13:02:44","date_gmt":"2025-07-08T13:02:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248536"},"modified":"2025-07-08T13:02:46","modified_gmt":"2025-07-08T13:02:46","slug":"the-formula-for-nickelii-bromide-is-nibr2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/the-formula-for-nickelii-bromide-is-nibr2\/","title":{"rendered":"The formula for nickel(II) bromide is NiBr2."},"content":{"rendered":"\n

The formula for nickel(II) bromide is NiBr2. 1. How many GRAMS of Ni2+ are present in 2.06 moles of nickel(II) bromide? 2. How many MOLES of Br- ion are present in 2.22 grams of nickel(II) bromide?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

    \n
  1. How many grams of Ni\u00b2\u207a are present in 2.06 moles of nickel(II) bromide?<\/strong><\/li>\n<\/ol>\n\n\n\n

    To find the number of grams of Ni\u00b2\u207a in 2.06 moles of NiBr\u2082, we need to follow these steps:<\/p>\n\n\n\n

      \n
    • First, recognize that in nickel(II) bromide, NiBr\u2082, each formula unit contains one Ni\u00b2\u207a ion.<\/li>\n\n\n\n
    • Therefore, the number of moles of Ni\u00b2\u207a will be the same as the number of moles of NiBr\u2082.<\/li>\n\n\n\n
    • We know that the molar mass of Ni (Nickel) is approximately 58.69 g\/mol.<\/li>\n<\/ul>\n\n\n\n

      So, to calculate the grams of Ni\u00b2\u207a: Grams of Ni\u00b2\u207a=moles of Ni\u00b2\u207a\u00d7molar mass of Ni\\text{Grams of Ni\u00b2\u207a} = \\text{moles of Ni\u00b2\u207a} \\times \\text{molar mass of Ni}Grams of Ni\u00b2\u207a=moles of Ni\u00b2\u207a\u00d7molar mass of Ni Grams of Ni\u00b2\u207a=2.06\u2009moles\u00d758.69\u2009g\/mol\\text{Grams of Ni\u00b2\u207a} = 2.06 \\, \\text{moles} \\times 58.69 \\, \\text{g\/mol}Grams of Ni\u00b2\u207a=2.06moles\u00d758.69g\/mol Grams of Ni\u00b2\u207a=120.92\u2009g\\text{Grams of Ni\u00b2\u207a} = 120.92 \\, \\text{g}Grams of Ni\u00b2\u207a=120.92g<\/p>\n\n\n\n

      Thus, 2.06 moles of nickel(II) bromide contain 120.92 grams<\/strong> of Ni\u00b2\u207a.<\/p>\n\n\n\n

      \n\n\n\n
        \n
      1. How many moles of Br\u207b ion are present in 2.22 grams of nickel(II) bromide?<\/strong><\/li>\n<\/ol>\n\n\n\n

        To calculate the number of moles of Br\u207b ions in 2.22 grams of NiBr\u2082, we need to:<\/p>\n\n\n\n

          \n
        • Find the molar mass of NiBr\u2082. The molar mass of Ni is 58.69 g\/mol, and the molar mass of Br (Bromine) is approximately 79.90 g\/mol.<\/li>\n<\/ul>\n\n\n\n

          The molar mass of NiBr\u2082 is: Molar mass of NiBr\u2082=58.69\u2009g\/mol+(2\u00d779.90\u2009g\/mol)\\text{Molar mass of NiBr\u2082} = 58.69 \\, \\text{g\/mol} + (2 \\times 79.90 \\, \\text{g\/mol})Molar mass of NiBr\u2082=58.69g\/mol+(2\u00d779.90g\/mol) Molar mass of NiBr\u2082=58.69\u2009g\/mol+159.80\u2009g\/mol=218.49\u2009g\/mol\\text{Molar mass of NiBr\u2082} = 58.69 \\, \\text{g\/mol} + 159.80 \\, \\text{g\/mol} = 218.49 \\, \\text{g\/mol}Molar mass of NiBr\u2082=58.69g\/mol+159.80g\/mol=218.49g\/mol<\/p>\n\n\n\n

            \n
          • Now, calculate the moles of NiBr\u2082 in 2.22 grams:<\/li>\n<\/ul>\n\n\n\n

            Moles of NiBr\u2082=mass of NiBr\u2082molar mass of NiBr\u2082\\text{Moles of NiBr\u2082} = \\frac{\\text{mass of NiBr\u2082}}{\\text{molar mass of NiBr\u2082}}Moles of NiBr\u2082=molar mass of NiBr\u2082mass of NiBr\u2082\u200b Moles of NiBr\u2082=2.22\u2009g218.49\u2009g\/mol=0.01016\u2009mol\\text{Moles of NiBr\u2082} = \\frac{2.22 \\, \\text{g}}{218.49 \\, \\text{g\/mol}} = 0.01016 \\, \\text{mol}Moles of NiBr\u2082=218.49g\/mol2.22g\u200b=0.01016mol<\/p>\n\n\n\n

              \n
            • Since there are 2 Br\u207b ions per formula unit of NiBr\u2082, the number of moles of Br\u207b ions will be twice the number of moles of NiBr\u2082:<\/li>\n<\/ul>\n\n\n\n

              Moles of Br\u207b=2\u00d7moles of NiBr\u2082=2\u00d70.01016\u2009mol=0.02032\u2009mol\\text{Moles of Br\u207b} = 2 \\times \\text{moles of NiBr\u2082} = 2 \\times 0.01016 \\, \\text{mol} = 0.02032 \\, \\text{mol}Moles of Br\u207b=2\u00d7moles of NiBr\u2082=2\u00d70.01016mol=0.02032mol<\/p>\n\n\n\n

              Thus, 2.22 grams of nickel(II) bromide contain 0.02032 moles of Br\u207b<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              In summary:<\/p>\n\n\n\n

                \n
              1. There are 120.92 grams<\/strong> of Ni\u00b2\u207a in 2.06 moles of NiBr\u2082.<\/li>\n\n\n\n
              2. There are 0.02032 moles of Br\u207b<\/strong> in 2.22 grams of NiBr\u2082.<\/li>\n<\/ol>\n\n\n\n
                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                The formula for nickel(II) bromide is NiBr2. 1. How many GRAMS of Ni2+ are present in 2.06 moles of nickel(II) bromide? 2. How many MOLES of Br- ion are present in 2.22 grams of nickel(II) bromide? The Correct Answer and Explanation is: To find the number of grams of Ni\u00b2\u207a in 2.06 moles of NiBr\u2082, […]<\/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-248536","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248536","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=248536"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248536\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}