40SU|||A chemist prepares a solution of iron (II) bromide (FeBr₂) by measuring out 46.4 mg of FeBr₂ into a 50. mL volumetric flask and filling to the mark withdistilled water.Microsoft OWCalculate the molarity of Br anions in the chemist's solution.Be sure your answer is rounded to the correct number of significant digits.O CHEMICAL REACTIONSCalculating ion molarity using solute massMicrosoftMicrosoft6.52.210...escmolLExplanationBEOO@Check9,208280103 NOV6x10$4X%5atvSMacBook ProA6&72022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility*668Å(9W)0QUAr

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Calculating ion molarity using solute mass A chemist prepares a solution of iron (II) bromide (FeBr₂) by measuring out 46.4 mg of FeBr₂ into a 50. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Br anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits. mol L d E

Calculating ion molarity using solute mass A chemist prepares a solution of iron (II) bromide (FeBr₂) by measuring out 46.4 mg of FeBr₂ into a 50. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Br anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits. mol L d E

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter4: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 20Q: A student wants to prepare 1.00 L of a 1.00-M solution of NaOH (molar mass = 40.00 g/mol). If solid...

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A student wants to prepare 1.00 L of a 1.00-M solution of NaOH (molar mass = 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M NaOH is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and mass be determined?
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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.
The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 106 parts of solution. Mathematically, by mass: ppm=gsolutegsolution=mgsolutekgsolution In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0 g of solute per 1.0 mL, which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 ppb Hg in H2O b. 1.0 ppb CHCl3 in H2O c. 10.0 ppm As in H2O d. 0.10 ppm DDT (C14H9Cl5) in H2O