Tea leaves were boiled in water and around 1.50 gram of caffeine were extracted by 200.0 mL of water. If three portions of 25.0 mL of dichloromethane were used to extract the caffeine from the aqueous layer, how much caffeine will be extracted by the dichloromethane? Solubility of caffeine in water: 2.18 g per 100 mL water Solubility of caffeine in dichloromethane: 18.2 g per 100 mL dichloromethane

Tea leaves were boiled in water and around 1.50 gram of caffeine were extracted by 200.0 mL of water. If three portions of 25.0 mL of dichloromethane were used to extract the caffeine from the aqueous layer, how much caffeine will be extracted by the dichloromethane? Solubility of caffeine in water: 2.18 g per 100 mL water Solubility of caffeine in dichloromethane: 18.2 g per 100 mL dichloromethane

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter13: The Chemistry Of Solutes And Solutions

Section: Chapter Questions

Problem 94QRT

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Tea leaves were boiled in water and around 1.50 grams of caffeine were extracted by 200.0 mL of water. If three portions of 25.0 mL of dichloromethane were used to extract the caffeine from the aqueous layer, how much caffeine will be extracted by the dichloromethane? Solubility of caffeine in water: 2.18 g per 100 mL waterSolubility of caffeine in dichloromethane: 18.2 g per 100 mL dichloromethane
A student carried out the synthesis of aspirin as described using 3.00 g of salicylic acid and 6.0 ml. of acetic anhydride and obtained 1.306 g of aspirin. (a) Calculate the amount of salicylic acid used in moles: (salicylic acid molar mass: 138.12 g/mol) (b) Calculate the mass (g) of acetic anhydride used: (acetic anhydride density: 1.08 g/cm³, Molar mass: 102.09 g/mol) (c) Calculate the amount of acetic anhydride used in moles: (d) Which one is the limiting reactant?
A town requiring 20 m3/s drinking water has two sources, a local well with 60 g/m3 nitrate and a distant reservoir 10 g/m3? What flow rates of well and reservoir water to meet 45 ppm drinking water standards and minimize the use of more expensive reservoir water?
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Step 1: In tube 1, you add 100 ul of plasma to 900 ul of distilled water. Step 2: In tube 2, you then take 100 ul from tube 1 and add 900 ul of distilled water. Step 3: In tube 3, you then take 100 ul from tube 2 and add 900 ul of distilled water. What is the final/total dilution factor in tube 3 at the end of step 3?
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(A) if both sentences are true,(B) if both sentences are false,(C) if the first sentence is true but the second is false, and(D) if the first sentence is false but the second is true. 1. In the Winkler Titration Method, sodium thiosulfate is standardized against a known concentration of magnesium iodate. The normality (N) of a solution is always equal to its molarity (M). 2. In the experiment on determining the dissolved oxygen concentration of a water sample, the compound used to acidify the sample was hydrochloric acid. An acidic solution will have a pH greater than 7. 3. Radioactive wastes, like uranium, must be handled and disposed of according to special regulations designed to protect people and the environment. Nuclear power plants produce too much greenhouse gases due to the burning of uranium.
1. A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein indicator 25 mL of mixture required 19.5 mL of 0.995 M HCl for the endpoint. With methyl orange, 25 mL of solution required 25 mL of the same HCl for the endpoint. Calculate grams per liter of Na2CO3 in the mixture. a.23b.36 c.78 d.112. A 1.2040 g sample containing sodium carbonate and inert material was dissolved in water and titrated to methyl orange end point, requiring 32.50 mL of 0.1020 M HCl. Calculate the %Na2CO4 in the sample. a. 93.36% b. 14.59% c. 12.75% d. 23.41%3. A 0.1510 g KHP (MW = 204.22 g/mol) sample required a volume of 57.04 mL of NaOH solution to reach a phenolphthalein end point. Calculate the molarity of NaOH titrant. a. 0.02593 M b. 0.006481 M c. 0.01296 M d. 0.01012 M