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23 April, 23:39

A chemist adds 0.55 L of a 4.69M silver nitrate (AgNO3) olution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

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  1. 24 April, 00:17
    0
    0.438 kg is the mass of solute that the chemist has added to the flask

    Explanation:

    We have the volume and the molarity. Let's calculate the moles, that chemist added.

    Volume. molarity = mol

    0.55 L. 4.69 mol/L = mol → 2.58 moles

    Now we can know the mass of solute (mol. molar mass)

    2.58 m. 169.89 g/m = 438.3 g

    Finally we convert grams to kilograms

    438.3 g / 1000 = 0.438 kg
  2. 24 April, 01:05
    0
    The mass of silver nitrate is 0.438 kg

    Explanation:

    Step 1: Data given

    Volume = 0.55L

    Molarity silver nitrate (AgNO3) = 4.69M

    Molar mass of AgNO3 = 169.87 g/mol

    Step 2: Calculate moles AgNO3

    Number of moles = molarity * volume

    Number of moles = 4.69 M * 0.550 L

    Number of moles AgNO3 = 2.5795 moles

    Step 3: Calculate mass of AgNO3

    Mass AgNO3 = moles AgNO3 * Molar mass AgNO3

    Mass AgNO3 = 2.5795 moles * 169.87 g/mol

    Mass AgNO3 = 438.18 grams ≈ 438 grams = 0.438 kg

    The mass of silver nitrate is 0.438 kg
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