A chemistry student needs 15.0 g of heptane for an experiment. He has available 0.50 kg of a 44.3% w/w solution of heptane in diethyl ether.Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.Round your answer to 3 significant digits.gx10Nosolution

Step 1: SolutionGiven, mass of solution available = 0.50 kg = 0.50 kg × ( 1000 g / 1 kg ) = 500…

Answered: A chemistry student needs 15.0 g of…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter1: The Nature Of Chemistry

Section: Chapter Questions

Problem 127QRT

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Similar questions

Beakers (a), (b), and (c) are representations of tiny sections (not to scale) of mixtures made from pure benzene and pure water. Select which beaker gives proper representation of the result when the two pure substances are mixed.
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A soft drink contains an unknown mass of citric acid, C3H5O(COOH)3. It requires 6.42 mL of 9.580 × 10−2-M NaOH to neutralize the citric acid in 10.0 mL of the soft drink. C3H5O(COOH)3(aq) + 3 NaOH(aq) → Na3C3H5O(COO)3(aq) + 3 H2O(ℓ) Determine which step in these calculations for the mass of citric acid in 1 mL soft drink is incorrect? Why? n (NaOH) = (6.42 mL)(1L/1000 mL)(9.580 × 10−2 mol/L) n (citric acid) = (6.15 × 10−4 mol NaOH) × (3 mol citric acid/1 mol NaOH) m (citric acid in sample) = (1.85 × 10−3 mol citric acid) × (192.12 g/mol citric acid) m (citric acid in 1 mL soft drink) = (0.354 g citric acid)/(10 mL soft drink) Determine the correct result.
What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
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