How much concentrated 15.54 M sulfuric acid is need to prepare 6.1 mL of a 2.28 M solution?

Question

How much concentrated 15.54 M sulfuric acid is need to prepare 6.1 mL of a 2.28 M solution?

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Mộc Miên 4 years 2021-08-09T11:30:36+00:00 1 Answers 19 views 0

Answers ( )

  1. minhkhang
    0
    2021-08-09T11:32:33+00:00
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    Answer:

    0.895 mL

    Explanation:

    Using the formula;

    C1V1 = C2V2

    Where;

    C1 = initial concentration (M)

    C2 = final concentration (M)

    V1 = initial volume (mL)

    V2 = final volume (mL)

    According to the information provided in this question;

    C1 = 15.54M

    V1 = ?

    C2 = 2.28M

    V2 = 6.1 mL

    Using C1V1 = C2V2

    V1 = C2V2 ÷ C1

    V1 = (2.28 × 6.1) ÷ 15.54

    V1 = 13.908 ÷ 15.54

    V1 = 0.895 mL

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