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1.1 kJ/mol Std entropy change of fusion, Δ fus S o: 12.1 J/(mol·K) Std enthalpy change of vaporization, Δ vap H o: 8.17 kJ/mol Std entropy change of vaporization, Δ vap S o? J/(mol·K) Solid properties Std enthalpy change of formation, Δ f H o solid? kJ/mol Standard molar entropy, S o solid? J/(mol K) Heat capacity, c p? J/(mol K) Liquid ...
Since the density of dry air at 101.325 kPa at 20 °C is [10] 0.001205 g/cm 3 and that of water is 0.998203 g/cm 3 we see that the difference between true and apparent relative densities for a substance with relative density (20 °C/20 °C) of about 1.100 would be 0.000120. Where the relative density of the sample is close to that of water (for ...
At IUPAC standard temperature and pressure (0 °C and 100 kPa), dry air has a density of approximately 1.2754 kg/m 3. At 20 °C and 101.325 kPa, dry air has a density of 1.2041 kg/m 3. At 70 °F and 14.696 psi, dry air has a density of 0.074887 lb/ft 3.
Density (g cm-3) Boiling point (°C) K b (°C⋅kg/mol) Freezing point (°C) K f (°C⋅kg/mol) Data source; Aniline: 184.3 3.69 –5.96 –5.87 K b & K f [1] Lauric acid: 298.9 44 –3.9 Acetic acid: 1.04 117.9 3.14 16.6 –3.90 K b [1] K f [2] Acetone: 0.78 56.2 1.67 –94.8 K b [3] Benzene: 0.87 80.1 2.65 5.5 –5.12 K b & K f [2 ...
1 Nm 3 of any gas (measured at 0 °C and 1 atmosphere of absolute pressure) equals 37.326 scf of that gas (measured at 60 °F and 1 atmosphere of absolute pressure). 1 kmol of any ideal gas equals 22.414 Nm 3 of that gas at 0 °C and 1 atmosphere of absolute pressure ... and 1 lbmol of any ideal gas equals 379.482 scf of that gas at 60 °F and ...
The standard unit is the meter cubed per kilogram (m 3 /kg or m 3 ·kg −1). Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely ...
In case of air, using the perfect gas law and the standard sea-level conditions (SSL) (air density ρ 0 = 1.225 kg/m 3, temperature T 0 = 288.15 K and pressure p 0 = 101 325 Pa), we have that R air = P 0 /(ρ 0 T 0) = 287.052 874 247 J·kg −1 ·K −1. Then the molar mass of air is computed by M 0 = R/R air = 28.964 917 g/mol. [11]
The annotation, d a°C/b°C, indicates density of solution at temperature a divided by density of pure water at temperature b known as specific gravity. When temperature b is 4 °C, density of water is 0.999972 g/mL.