What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m l of solution?
1 answer:
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻ M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol 15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2 Ca(NO3)2 -------> Ca²⁺ +2NO3⁻ 1 mol 2 mol0.100 mol 0.200 mol We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution, so 0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M Concentration of NO3⁻ is 0.667 M.
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Answer:
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Explanation:
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The answer is either b or d .