One of the detrimental effects of the "thermal pollution" of water supplies is that a rise in temperature reduces the amount of dissolved oxygen available to fish. Using the information in the table of Henry's Law constants (K) below, calculate how many litres of water a fish requires at 30.0°C to obtain the same amount of oxygen that it could obtain from 3.05 L of water at 25.0°C. TABLE HENRY'S LAW CONSTANTS (KH) IN AQUEOUS SOLUTION (M/Bar) Gas 0.0 °C 25 °C 30 °C N2 1.1 x 10-3 6.6 x 10-4 4.0 x 10-4 O2 2.5 x 103 1.3 x 10-3 8.8 x 10-4 CO 1.6 x 103 9.5 x 104 4.4 x 10-4 Ar 2.5 x 10-3 1.5 x 10-3 1.0 x 10-3 Не 4.1 x 10-4 3.8 x 10-4 3.8 x 10-4 CO2 7.7 × 10-2 3.4 x 10-2 1.6 x 10-2 i 1.7

The freezing point of water H₂O is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is antifreeze (ethylene glycol). How many grams of antifreeze, CH₂ OHCH₂OH (62.10 g/mol), must be dissolved in 268.0 grams of water to reduce the freezing point by 0.450 °C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kb(°C/m) Kf(°C/m) Water H₂O 0.512 Ethanol CH3 CH₂ OH 1.22 Chloroform CHCl3 3.67 Benzene C6H6 2.53 Diethyl ether CH3 CH₂ OCH2 CH3 2.02 Mass= g 1.86 1.99 5.12

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is glucose.Calculate the vapor pressure of the solution at 25 °C when 6.800 grams of glucose, C6H12O6 (180.2 g/mol), are dissolved in 267.7 grams of water.water = H2O = 18.02 g/mol.VP(solution) =         mm Hg

The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20°C is 4.3 atm. What is the solubility of CO2 in champagne? Assume the Henry's law constant is the same for champagne as it is for water: at 20°C, kH = 3.7 × 10−2 mol/(L·atm).