What is the molecular mass of \( \mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2} * 5 \mathrm{H}_{2} \mathrm{O} \) (answer to 1 decimal place) Your Answer: Answer Question 30 (2 points) Saved
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To find the molecular mass of \( \mathrm{Mg}(\mathrm{HCO}_{3})_{2} \cdot 5 \mathrm{H}_{2} \mathrm{O} \), we first calculate the mass of each element: - Magnesium (Mg): 24.3 g/mol - Carbon (C): 12.0 g/mol (2 carbons in 2 bicarbonates) - Hydrogen (H): 1.0 g/mol (bicarbonate has 1 hydrogen each and there are 2, plus 5 waters) - Oxygen (O): 16.0 g/mol (each bicarbonate has 3 oxygen atoms, and there are 5 waters) Calculating: - Bicarbonate contribution: \( 2 \times (1 \times 12 + 1 \times 1 + 3 \times 16) = 2 \times (12 + 1 + 48) = 2 \times 61 = 122 \, g/mol \) - Water contribution: \( 5 \times (2 \times 1 + 1 \times 16) = 5 \times (2 + 16) = 5 \times 18 = 90 \, g/mol \) Now, add them all together: \[ 24.3 + 122 + 90 = 236.3 \, g/mol \] Thus, the molecular mass of \( \mathrm{Mg}(\mathrm{HCO}_{3})_{2} \cdot 5 \mathrm{H}_{2} \mathrm{O} \) is **236.3 g/mol**.
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