4.14 - ions in calcium acetate Both of these conversions...

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Key for Micro Exam 4 Question 14 Question 14: How many total ions are there in 40.0 g of calcium acetate? Answers: (a) 4.52 x 10 23 (b) 5.82 x 10 23 (c) 1.52 x 10 24 (d) 7.22 x 10 23 (e) 2.41 x 10 23 (f) 6.02 x 10 23 (g) 1.51 x 10 23 (h) 1.80 x 10 25 Any time you need to find the number of ions, electrons, bonds, atoms etc, in a mass or volume of sample, you need to use dimensional analysis. Here is a scheme you could follow: g calcium acetate →moles calcium acetate→moles of ions→# of ions But first, in order to do the first conversion, we will need to know the molecular weight of calcium acetate, and to do the second conversion, we will need to know the number of moles of
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Unformatted text preview: ions in calcium acetate. Both of these conversions can be determined if we know the chemical formula of the ionic compound. Ionic compounds are neutral, so we need to add enough of each ion to make a neutral compound. This would occur with one calcium cation and two acetate anions like this: Ca +2 + 2CH 3 COO-1 → Ca(CH 3 COO) 2 Oftentimes the charge on one ion will become the subscript on the other ion when written as an ionic compound. Here is the dimensional analysis: 40g Ca(CH 3 COO) 2 │1 mol Ca(CH 3 COO) 2 │ 3 mols ions │6 x 10 23 ions │160g Ca(CH 3 COO) 2 │1 mol Ca(CH 3 COO) 2 │1 mol ions = 4.5x10 23 ions...
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