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30 March, 11:25

A chemist adds 435.0mL of a 2.28 M zinc nitrate ZnNO32 solution to a reaction flask. Calculate the millimoles of zinc nitrate the chemist has added to the flask. Round your answer to 3 significant digits.

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  1. 30 March, 11:45
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    He added 992 milimoles of zinc nitrate (Zn (NO3) 2)

    Explanation:

    Step 1: Data given

    Volume of zinc nitrate = 435.0 mL = 0.435 L

    Molarity of zinc nitrate = 2.28 M

    Step 2: Calculate moles zinc nitrate

    Moles = molarity * volume

    Moles Zn (NO3) 2 = 2.28 M * 0.435 L

    Moles Zn (NO3) 2 = 0.9918 moles

    Step 3: Convert moles to milimoles

    Moles Zn (NO3) 2 = 0.9918 moles

    Moles Zn (NO3) 2 = 0.9918 * 10^3 milimoles

    Moles Zn (NO3) 2 = 991.8 milimoles ≈ 992 milimoles

    He added 992 milimoles of zinc nitrate (Zn (NO3) 2)
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