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A chemist adds 55.0mL of a ×1.710−5/mmolL mercury(II) iodide solution to a reaction flask. Calculate the micromoles of mercury(II) iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

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Answer:

The answer is 0.017 μmol of mercury iodide

Explanation:

To know the micromoles we must use the following conversion, 1 mmol = 1000μmol. we use a rule of three for the calculation of micromoles:

1mmol------------------1000μmol

1.7x10^-5mmol------- Xμmol

Clearing the X, we have:

X μmol = (1.7x10^-5x1000)/1 = 0.017μmol

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