One beaker contains 26 mL of 1.60 M HCl, and a second beaker contains 53 mL of 0.40 M NaOH. Both solutions were initially at 20 °C. We pour both beakers into a large insulated container which contains 9 g of ice, initially at -60 °C. Calculate the final temperature of the resulting solution. Note that the heat of neutralization is -57.3 kJ/mol.

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform , CHCI2, is 173.11 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 9.963 grams of the compound were dissolved in 278.4 grams of chloroform, the vapor pressure of the solution was 170.58 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound? chloroform = CHCI3 = 119.40 g/mol. MW = g/mol

The vapor pressure of diethyl ether (ether) is 463.6 mm Hg at 25 °C. How many grams of DDT, C14H9Cls, a nonvolatile, nonelectrolyte (MW = 354.5 g/mol), must be added to 230.0 grams of diethyl ether to reduce the vapor pressure to 458.5 mm Hg ? diethyl ether= CH3CH₂OCH₂CH3 = 74.12 g/mol. g DDT

c) To save money, some state Department of Transportation decided to replace relatively expensive calcium chloride (CaCl2) by more affordable beet juice (mostly made of sugar molecules) in the solutions used to treat the roads before a snow/freezing rain storm. However, the beet juice was not as effective at de-icing the roads as was the calcium chloride solution. When road temperatures were very low, the beet juice treatment did not prevent the water from freezing again. You are called to one of those Departments of Transportation to: 1) Explain that observation and 2) Justify why they should continue to use calcium chloride solutions even if it is more expensive. You may even suggest a cheaper alternative to calcium chloride that will be as effective in road de-icing. Provide your answers below.

A solution was prepared by dissovling 0.800 g of sulfur, S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and boiling point of the solution. Boiling point 118.5 degrees C. Freezing point 16.60 degrees C. Kb = 3.08 degrees C/m. Kf = 3.59 degrees C/m.

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform , CHCI3, is 173.11 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 7.339 grams of the compound were dissolved in 176.5 grams of chloroform, the vapor pressure of the solution was 170.01 mm Hg. The compound was also found to be nonvolatile and a non- electrolyte. What is the molecular weight of this compound? chloroform CHCI3 = 119.40 g/mol. %3D MW g/mol

The freezing point of ethanol, CH3CH2OH, is -117.30°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in ethanol is saccharin . A student dissolves 10.10 grams of saccharin, C7H5NO3S (183.2 g/mol), in 266.8 grams of ethanol. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb (°C/m) Kf (°C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 X Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 X The molality of the solution is ______m.The freezing point of the solution is _____°C. FILL IN THE BLANKS

1.55 g potassium nitrate is added to a calorimeter containing 100.0 mL. water at 22.50 oC.  The temperature is changed to 18.00 oC.   Calculate the heat of dissolution in kJ / mole of potassium nitrate. K = 39.10 g/mole,  N = 14.00 g/mol,  O = 16.00 g/mole. Hint: This is a diluted solution:   D= 1.00 g/mL.,   CP = 4.184 J/ g oC.  heat capacity of calorimeter  is 51.7 J / oC.   N = 14.00 g/mole,  K = 39.10 g/mole,    O = 16.00 g/mole   a. 140. kJ/moleb. -140. kJ/molec. 24.3 kJ/ moled.-176 kJ/mole

The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4,  (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67   Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02   The molality of the solution is  m. The freezing point of the solution is  °C.

The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4,  (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67   Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02   The molality of the solution is  m. The freezing point of the solution is  °C.