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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.At a depth of 100 ft in water, the external pressure is 4.0 atm. If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Solution: At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.At a depth of 100 ft in water, the externa

Problem

At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.

At a depth of 100 ft in water, the external pressure is 4.0 atm. If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Solution

We're being asked to calculate the milliliters of Ngas released into the blood stream.


For this problem, we will do the following steps:


Step 1: Calculate Henry’s Law constant for the gas


Recall that the solubility of a gas is given by Henry’s law:

Sgas = kH·Pgas

where:

Sgas = solubility of the gas (in mol/L or M)

kH = Henry’s law constant for the gas

Pgas = partial pressure of the gas

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