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Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%wewol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many mL of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m¡vj = m2v2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI х 100 %3D %НСІ = Tu mol HCI 1000 mL Then substitute this value of m, into the equation m2V2 and solve for v1 m2V2 (3%) (v2) V = 36.46 g HCI m1 Mol HCI 1 mol HCI m1 |(100) 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!

Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%wewol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many mL of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m¡vj = m2v2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI х 100 %3D %НСІ = Tu mol HCI 1000 mL Then substitute this value of m, into the equation m2V2 and solve for v1 m2V2 (3%) (v2) V = 36.46 g HCI m1 Mol HCI 1 mol HCI m1 |(100) 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!

Chemistry
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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Solve for v1 in a single step using Excel to avoid causing rounding error!

 
Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%(w/vol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many ml of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m1v1 = m2V2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI m1 х 100 %3D %HСІ %3D mol HCI 1000 mL Then substitute this value of m1 into the equation mįv1 = m2V2 and solve for v1 m2V2 (3%) (v2) V1 m1 36.46 g HCI m1 Mol HCI (100) 1 mol HCI 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!
Transcribed Image Text:Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%(w/vol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many ml of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m1v1 = m2V2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI m1 х 100 %3D %HСІ %3D mol HCI 1000 mL Then substitute this value of m1 into the equation mįv1 = m2V2 and solve for v1 m2V2 (3%) (v2) V1 m1 36.46 g HCI m1 Mol HCI (100) 1 mol HCI 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!
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