A FH Escape Chemistry 106! с X + O8 https://docs.google.com/forms/d/e/1FAIpQLSfdH4Cb3Jih0g_TXnrExG4Oh2E11 NuVCU4MS9N 120% Type here to search Create a solution by dissolving 14.77 g of napthalene (nonelectrolyte, MW = 128.17 g/mol) in 155.0 g of diethyl ether, (MW = 74.12 g/mol). What is the mole fraction of napththalene in the solution? You must type the 0 before the decimal. .05223 The vapor pressure of pure ether at 22 degrees Celsius (the temperature of the lab) is 414.0 Torr. What is the vapor pressure of your solution, in Torr? 392.4 The normal boiling point of diethyl ether is 34.7 degrees Celsius. What is the boiling point of your solution in degrees Celsius? The molal boiling constant for diethyl ether is 2 16 degrees Celsius/molal W og || a (28) * * DOW ↓ ED ABP 8:35 PM 5/11/2023 x Ę

A FH Escape Chemistry 106! с X + O8 https://docs.google.com/forms/d/e/1FAIpQLSfdH4Cb3Jih0g_TXnrExG4Oh2E11 NuVCU4MS9N 120% Type here to search Create a solution by dissolving 14.77 g of napthalene (nonelectrolyte, MW = 128.17 g/mol) in 155.0 g of diethyl ether, (MW = 74.12 g/mol). What is the mole fraction of napththalene in the solution? You must type the 0 before the decimal. .05223 The vapor pressure of pure ether at 22 degrees Celsius (the temperature of the lab) is 414.0 Torr. What is the vapor pressure of your solution, in Torr? 392.4 The normal boiling point of diethyl ether is 34.7 degrees Celsius. What is the boiling point of your solution in degrees Celsius? The molal boiling constant for diethyl ether is 2 16 degrees Celsius/molal W og || a (28) * * DOW ↓ ED ABP 8:35 PM 5/11/2023 x Ę

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

Section13.7: Colligative Properties Of Solutions

Problem 13.16E: Suppose that you are closing a cabin in the north woods for the winter and you do not want the water...

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You were tasked to prepare 590 ml of 3.00 M HNO3. If you were provided a stock solution of 70.5% (w/w) HNO3 (MW = 63.0 g) with a specific gravity of 1.42, what volume of this solution (in ml) will you be using?
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using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution2. Calculate the mass of solid CuSO4 dissolved in the solution.3. Calculate the number of moles of solid CuSO4 dissolved in the solution.4. Calculate the mass of water evaporated from the solution.5. Calculate the density of solution, (g solution/mL solution).6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution).7. Calculate the molality of solution (moles CuSO4/kg solvent).8. Calculate the molarity of solution (moles CuSO4/L solution).9. Given that the true molarity is 0.0400 M, calculate the percent error of your result.
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i have hard time geting it stright even after I do the experiment can u help me with the one pls Material ChemicalsSealed NO2 tubes, KSCN solution (0.002 M), acidified Fe(NO3)3 solution (0.2 M), KSCN crystals, Fe(NO3)3 crystals, KOH (concentrated (8 M)), distilled waterApparatus:Bunsen burner, ring stand, wire gauze, matches, thermometer, dropper, test tubes, glass stirring rod, beaker tongs Procedure Part A: Fill a 600 mL beaker to the 3/4 mark with water. Bring to a boil using a Bunsen burner. In another 600 mL beaker, fill to the 1/2 mark with water; add ice in order to lower the temperature to about 0°C. Note the room temperature. Observe the NO2 thoroughly at room temperature. Place the sealed NO2 tube in the cold bath. Note your observations. Remove the test tube from the cold bath. Let the test tube stand until it reaches room temperature. Note your observations. Place the test tube in the hot bath. Note your observations. Remove the test tube from the hot bath. Let the test…
I have an equation and the answer, however im not sure how to do the calculation. Please show how to. Answer is supposed to be 0.20 M m1v1=m2v2 (20mL)V=(13mL)(0.30M) V=0.20 M
using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution? 5.027 g 2. Calculate the mass of solid CuSO4 dissolved in the solution? 0.033 g 3. Calculate the number of moles of solid CuSO4 dissolved in the solution? 2.07 x 10-4 mol4. Calculate the mass of water evaporated from the solution? 0.0414 M5. Calculate the density of solution, (g solution/mL solution)?6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution)?7. Calculate the molality of solution (moles CuSO4/kg solvent)?8. Calculate the molarity of solution (moles CuSO4/L solution)?9. Given that the true molarity is 0.0400 M, calculate the percent error…