For an exothermic solution process, AHmix > AHsolute AHsolvent AHmix > AHsolute + AHsolvent O AHmix < AHsolute + AHsolvent AHmix = AHsolute + AHsolvent
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Q: List the Enthalpy changes accompanying the solution process?
A: ANSWERE IS DISCUSSED BELOW :
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please answer clearly and explain.
For an exothermic solution process, ______
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- 1. What is the molarity of glucose (mol wt =180) in a solution containing 6.70g of glucose and 150 g of water? The final solution has a specific gravity of 1.015. Round off to 4 decimal places. 2. What is the osmolality of 12.5% by weight solution of AlCl (MW=133 g/mole)? 3. What is the normality (N) by w/v solution of CaCl containing 5 g of CaCl and 105g of water (CaCl MW=111 g/mole)? The specific gravity of the final solution is 1.35.Osmotic pressures are often reported in units of atmospheres or mm Hg. The latter impliesthat the height of a column of liquid can be used as a measure of pressure. This is, in fact, thebasis of the torricellian barometer from which the units of “torr” come. The pressure, P, isrelated to the height of the column by:P = ρghwhere ρ is the density of the liquid, g is the acceleration due to gravity and h is the height. Whatwill be the height (in mm) of a column of aqueous solution with a density of 0.9987 g/mL, if theosmotic pressure of the solution at 25 °C is 1.4 mm Hg? The density of mercury at thistemperature is 13.8 g/mLIn the determination of molar mass of an unknown substance by ebullioscopic constant, 30 mL of acetone (C3H6O) was placed in a test tube with thermometer and glass tubing and subjected to water bath. Upon boiling, the temperature reads 56 deg C. For the boiling point of unknown-acetone solution, you prepared the solution by mixing 1.60 g of unknown solute in the 30 mL acetone and subjected it again to water bath. The boiling temperature of the solution is 56.68 deg C. The density of acetone = 0.9849 g/mL and the Kb of acetone = 1.67 degC/kg
- In the determination of molar mass of an unknown substance by ebullioscopic constant, 30 mL of acetone (C3H6O) was placed in a test tube with thermometer and glass tubing and subjected to water bath. Upon boiling, the temperature reads 56 deg C. For the boiling point of unknown-acetone solution, you prepared the solution by mixing 1.60 g of unknown solute in the 30 mL acetone and subjected it again to water bath. The boiling temperature of the solution is 56.68 deg C. The density of acetone = 0.9849 g/mL and the Kb of acetone = 1.67 degC/kg What is the value for delta Tb or the change in boiling temperature (in degrees Celsius)?In the determination of molar mass of an unknown substance by ebullioscopic constant, 30 mL of acetone (C3H6O) was placed in a test tube with thermometer and glass tubing and subjected to water bath. Upon boiling, the temperature reads 56 deg C. For the boiling point of unknown-acetone solution, you prepared the solution by mixing 1.60 g of unknown solute in the 30 mL acetone and subjected it again to water bath. The boiling temperature of the solution is 56.68 deg C. The density of acetone = 0.9849 g/mL and the Kb of acetone = 1.67 degC/kg What is the value for delta Tb or the change in boiling temperature (in degrees Celsius)? Final answer must be rounded off to 2 decimal places, and shall NOT have any unit.In the determination of molar mass of an unknown substance by ebullioscopic constant, 30 mL of acetone (C3H6O) was placed in a test tube with thermometer and glass tubing and subjected to water bath. Upon boiling, the temperature reads 56 deg C. For the boiling point of unknown-acetone solution, you prepared the solution by mixing 1.60 g of unknown solute in the 30 mL acetone and subjected it again to water bath. The boiling temperature of the solution is 56.68 deg C. The density of acetone = 0.9849 g/mL and the Kb of acetone = 1.67 degC/kg - What is the value for delta Tb or the change in boiling temperature (in degrees Celsius)? Final answer must be rounded off to 2 decimal places, and shall NOT have any unit. - How many moles of solute is present in the solution given in the video? Final answer must be rounded off to 2 decimal places, and shall NOT have any units. - What is the molal concentration of the solution? Final answer must be rounded off to 2 decimal places, and shall NOT…
- An experimental attempt was to convert 2000 cm^3 of a solution consists of 0.96 ethanol into another solution consisting of 0.56 ethanol. If the density of water at the experiment's temperature and pressure is 997 kg/ m^3, How much H2O should be added to the 2000 cm^3 solution? What is the solution's volume after conversion? Given data: In the 0.96 ethanol solution :V ethanol = 0.816 cm^3/g and Vwater = 1.273 cm/g In the 0.58 ethanol solution: V ethanol = 0.953 cm^3/g and :V water = 1.243 cm/gGiven the following data for Mass of test tube and stearic acid = 14.17 gMass of test tube = 11.40 gFreezing point of strearic acid = 69.59o CMass of weighing paper + naphthalene =1.230 gMass of weighing paper = 0.920 gFreezing point solution = 64.00o CKf = 4.5o C/m Determine the following1. mass of stearic acid in g (2 decimal places); _____2. mass of naphthalene in g (2 decimal places); _____3. freezing point depression (2 decimal places); _____4. molality of solution (3 significant figures); _____5. moles of naphthalene (3 significant figures); _____6. molar mass of naphthalene, experimentally (3 significant figures); _____7. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); _____You have a container of 17.42 M conc. acetic acid (CH3COOH(aq)) with density 1.05 g/mL, find the molality of a 150ml sample of the solution. molality = moles solute/Kg solvent moles solute = (0.150 L)(17.42 mol/L) = 2.613 mol CH3COOH molar mass of CH3COOH = 60.06 g/mol grams of solution = (150mL)(1.05 g/mL) = 157.5 g solution grams of solute = (2.613 mol)(60.06 g/mol) = 156.9 g CH3COOH so the total grams of water in this solution would be: g H2O = g solution - g CH3COOH = 157.5g - 156.9 g = 0.6g H2O 0.6g H2O = 0.0006 Kg H2O so molality = (2.613 mol CH3COOH) /(0.0006 kg H2O) = 4355 molal this seems rather high to me but if its 'concentrated' acetic acid, I know that you can get it to about 98% as glacial acetic acid so perhaps there is 4355 moles of acetic acid per kg of water. Did I do this correctly? I know that typically the solvent is the larger component of a solution but the fact that it gives me CH3COOH(aq) would imply that H2O is the solvent still in this situation, I have…
- Stock hydrofluoric acid solution is 49.0% HF and has a specific gravity of 1.17. What is the molarity of the solution?You are wanting to apply a soil conditioner to your garden. 10mL of conditioner cover 3m^(2) of your garden. However, before you use it you must dilute the concentrate by placing 10ml of Seasol in a 9L bucket. How many litres of the diluted mixture do you need to cover a garden of 80m^(2).You weigh out 80g of NaOH pellets and dilute to 1 liter. MW = 40 g/mol. What is the normality of the solution?
