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5 October, 06:29

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

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Answers (1)
  1. 5 October, 08:38
    0
    V = 250ml

    Explanation:

    From n = CV

    3 = 12*V

    V = 0.25L = 250ml
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