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18 June, 21:25

What volume in ML of 12.0M HCL is needed to contain 3.00 moles of HCL?

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Answers (1)
  1. 18 June, 21:37
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    V = 250ml

    Explanation:

    From n = CV

    3 = 12*V

    V = 0.25L = 250ml
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