A 72.4 mL solution of Cu(OH) is neutralized by 47.8 mL of a 0.56 M H2(C204) solution. What is the concentration of the Cu(OH)?​

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A 72.4 mL solution of Cu(OH) is neutralized by 47.8 mL of a 0.56 M H2(C204) solution. What is the concentration of the Cu(OH)?​

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Gia Bảo 3 years 2021-08-08T13:25:25+00:00 1 Answers 21 views 0

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  1. yenoanh
    0
    2021-08-08T13:27:24+00:00
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    Answer: The concentration of Cu(OH)_{2} is  0.369 M.

    Explanation:

    Given: V_{1} = 72.4 mL,     M_{1} = ?

    V_{2} = 47.8 mL,       M_{2} = 0.56 M

    Formula used to calculate the concentration of Cu(OH)_{2} is as follows.

    M_{1}V_{1} = M_{2}V_{2}

    Substitute the values into above formula as follows.

    M_{1}V_{1} = M_{2}V_{2}\\M_{1} \times 72.4 mL = 0.56 M \times 47.8 mL\\M_{1} = \frac{0.56 M \times 47.8 mL}{72.4 mL}\\= 0.369 M

    Thus, we can conclude that the concentration of Cu(OH)_{2} is  0.369 M.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )