Chlorine has a characteristic penetrating and irritating odor. The gas is greenish yellow in color and the liquid is clear amber. The data on physical properties of chlorine as determined by different investigators show some variations. Atomic and Molecular PropertiesAtomic Symbol - Cl Atomic Weight - 35.453 Atomic Number - 17 Molecular Weight of Cl2 - 70.906
FlammabilityChlorine is neither explosive nor flammable. Chlorine will support combustion under certain conditions. Many materials that burn in oxygen (air) atmospheres will also burn in chlorine atmospheres. Many organic chemicals react readily with chlorine, sometimes violently. An important specific compound of concern is hydrogen. Chlorine reacts explosively with hydrogen in a range of 4% to 93% hydrogen. The reaction is initiated very easily much the same way as hydrogen and oxygen. See Pamphlet 121 for more information. ValenceChlorine usually forms compounds with a valence of -1 but it can combine with a valence of +1, +2, +3, +4, +5, or +7. Chemical ReactionsReactions with Water Reactions with Metals Reactions with Organic Compounds Physical PropertiesSee CI Pamphlet 1. Taken from Chlorine Basics (Pamphlet 1). This can be downloaded from our bookstore.
Density is mass per unit volume. Finding the density of a gas is the same as finding the density of a solid or liquid. You have to know the mass and the volume of the gas. The tricky part with gases is that you are often given pressures and temperatures with no mention of volume. You have to figure it out from the other information.
This example problem will show how to calculate density of a gas when given the type of gas, the pressure, and the temperature. Question: What is the density of oxygen gas at 5 atm and 27 °C? First, let's write down what we know: Gas is oxygen gas or O2.Pressure is 5 atm Temperature is 27 °C Let's start with the Ideal Gas Law formula. PV = nRT whereP = pressureV = volumen = number of moles of gas R = gas constant (0.0821 L·atm/mol·K) T = absolute temperature If we solve the equation for volume, we get: V = (nRT)/P We know everything we need to find the volume now except the number of moles of gas. To find this, remember the relationship between number of moles and mass. n = m/MM wheren = number of moles of gasm = mass of gas MM = molecular mass of the gas This is helpful since we needed to find the mass and we know the molecular mass of oxygen gas. If we substitute for n in the first equation, we get: V = (mRT)/(MMP) Divide both sides by m: V/m = (RT)/(MMP) But density is m/V, so flip the equation over to get: m/V = (MMP)/(RT) = density of the gas. Now we need to insert the values we know. MM of oxygen gas or O2 is 16+16 = 32 grams/moleP = 5 atmT = 27 °C, but we need absolute temperature. TK = TC + 273 T = 27 + 273 = 300 K m/V = (32 g/mol · 5 atm)/(0.0821 L·atm/mol·K · 300 K)m/V = 160/24.63 g/L m/V = 6.5 g/L Answer: The density of the oxygen gas is 6.5 g/L. Calculate the density of carbon dioxide gas in the troposphere, knowing the temperature is -60.0 °C and the pressure is 100.0 millibar. First, list what you know:
Right off the bat, you can see some units don't match up and that you need to use the periodic table to find the molar mass of carbon dioxide. Let's start with that.
There is one carbon atom and two oxygen atoms, so the molar mass (M) of CO2 is 12.0 + (2 x 16.0) = 44.0 g/mol Converting mbar to atm, you get 100 mbar = 0.098 atm. Converting °C to K, you get -60.0 °C = 213.15 K. Finally, all of the units agree with those found in the ideal gas constant:
Now, plug the values into the equation for the density of a gas: ρ = PM/RT = (0.098 atm)(44.0 g/mol) / (0.0821 L·atm/mol·K)(213.15 K) = 0.27 g/L
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