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m(NaCl) = 2 mol/L × 0.1 L × 58 g/mol = 11.6 g. To create the solution, 11.6 g NaCl is placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL. The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore ...
Dry water or empty water, a form of "powdered liquid", is an air–water emulsion in which water droplets are surrounded by a silica coating. [1] Dry water consists of 95% liquid water, but the silica coating prevents the water droplets from combining and turning back into a bulk liquid. [2] The result is a white powder.
The term molality is formed in analogy to molarity which is the molar concentration of a solution. The earliest known use of the intensive property molality and of its adjectival unit, the now-deprecated molal, appears to have been published by G. N. Lewis and M. Randall in the 1923 publication of Thermodynamics and the Free Energies of Chemical Substances. [3]
Molecular weight (M.W.) (for molecular compounds) and formula weight (F.W.) (for non-molecular compounds), are older terms for what is now more correctly called the relative molar mass (M r). [8] This is a dimensionless quantity (i.e., a pure number, without units) equal to the molar mass divided by the molar mass constant .
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
Up to 99.63 °C (the boiling point of water at 0.1 MPa), at this pressure water exists as a liquid. Above that, it exists as water vapor. Note that the boiling point of 100.0 °C is at a pressure of 0.101325 MPa (1 atm ), which is the average atmospheric pressure.
1 mole of O 2 → 2 moles of MnO(OH) 2 → 2 mole of I 2 → 4 mole of S 2 O 2− 3. Therefore, after determining the number of moles of iodine produced, we can work out the number of moles of oxygen molecules present in the original water sample. The oxygen content is usually presented in milligrams per liter (mg/L).
The transpiration ratio is the ratio of the mass of water transpired to the mass of dry matter produced; the transpiration ratio of crops tends to fall between 200 and 1000 (i.e., crop plants transpire 200 to 1000 kg of water for every kg of dry matter produced). [10]