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11 February, 06:13

How many moles of solute particles are present in 7.94 mL of 0.887 M NaNO3? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.

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  1. 11 February, 08:53
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    The answer is 7.04278 x 10⁻³, and the log (10) is - 2.15

    Explanation:

    In a 0.887 M NaNO₃ solution there are 0.887 mol in 1 liter of solution, We know that 1 L = 1000 ml, so we can calculate the number of mol which are in 7.94 ml as follows:

    solute mol = 7.94 ml x 0.887 mol / 1000 ml = 7.04278 x 10⁻³ mol

    We take log (10) of 7.04278 x 10⁻³ and we obtain: - 2.15
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