13.48 Calculate the number of moles of solute present in each of the following aqueous solutions: 185 mL of 1.50 M HNO3 50.0 mg that is 1.25 m NaCl.

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13.48 Calculate the number of moles of solute present in each of the following aqueous solutions: 185 mL of 1.50 M HNO3 50.0 mg that is 1.25 m NaCl

13.31 Explain why carbonated beverages must be stored in sealed containers. Once the beverage has been opened, why does it maintain more carbonation when refrigerated than at room temperature?

Henry’s Law Example Calculate the concentration (solubility) of CO2 in a 2-liter that is bottled with a partial pressure of 4.0 atm @ 25 ºC. The constant for CO2 in water is 3.1 × 10-2 mol/L-atm

Henry’s Law Example Calculate the concentration (solubility) of CO2 in a 2-liter that is bottled with a partial pressure of 4.0 atm @ 25 ºC. The constant for CO2 in water is 3.1 × 10-2 mol/L-atm Sg = k Pg

Henry’s Law Example Calculate the concentration (solubility) of CO2 in a 2-liter that is bottled with a partial pressure of 4.0 atm @ 25 ºC. The constant for CO2 in water is 3.1 × 10-2 mol/L-atm Sg = k Pg Sg = (3.1 × 10-2 mol/L-atm)(4.0 atm)

Henry’s Law Example Calculate the concentration (solubility) of CO2 in a 2-liter that is bottled with a partial pressure of 4.0 atm @ 25 ºC. The constant for CO2 in water is 3.1 × 10-2 mol/L-atm Sg = k Pg Sg = (3.1 × 10-2 mol/L-atm)(4.0 atm) Sg = 0.124 = 0.12 mol/L

Henry’s Law Example Calculate the concentration (solubility) of CO2 in a 2-liter that is bottled with a partial pressure of 4.0 atm @ 25 ºC. The constant for CO2 in water is 3.1 × 10-2 mol/L-atm Sg = k Pg Sg = (3.1 × 10-2 mol/L-atm)(4.0 atm) Sg = 0.124 = 0.12 mol/L Concentration = 0.12 M

Complete the following thought questions on the example side of your notes: 1) As you increase temperature, solubility of a solid___________. 2) As you lower the temp, solubility of a gas ___________. 3) Solutions that form spontaneously are considered____________. 4) When more bonds form than break, the system (solution) ____________energy.

Determine the molality of glucose in the solution given the following information. 4.35 g glucose (C6H12O6) in 25.0 mL of water. Water has a density of 1.00 g/mL. (pg 544) What is molality? How do you get the units you need from the given data?

Homework 13.39) Calculate the molarity for each of the following: 0.540 g Mg(NO3)2 in 250.0 mL of solution. 22.4 g LiClO4 • 3H2O in 125 mL of solution.