Average masses

The atomic mass is the average mass of an element in atomic mass units ("amu"). 

The following masses, in grams, were recorded in an experiment to determine the average mass of a U.S. penny. 

To determine Ks we need to know the average mass of the cereal samples and the relative standard deviation for the %(w/w) ash. The average mass of the five cereal samples is 1.0007 g. The average %(w/w) ash and the absolute standard deviation are, respectively, 1.298% and 0.03194. The percent relative standard deviation, therefore, is... 

True Density or Specific Gravity. The average mass per unit volume of the individual particles is called the tme density or specific gravity. This property is most important when volume or mass of the filled composition is a key performance variable. The tme density of fillers composed of relatively large, nonporous, spherical particles is usually determined by a simple Hquid displacement method. Finely divided, porous, or irregular fillers should be measured using a gas pycnometer to assure that all pores, cracks, and crevices are penetrated. 

Density Gradient Separation. Based on specific density, each cell in a test tube finds its own position (see Fig. 1), ie, red cells at the bottom, then granulocytes, monocytes, lymphocytes, platelets, and plasma on top. Table 2 Hsts average mass density of the cellular components of blood. The actual numbers vary slightly from person to person. 

The diametei of average mass and surface area are quantities that involve the size raised to a power, sometimes referred to as the moment, which is descriptive of the fact that the surface area is proportional to the square of the diameter, and the mass or volume of a particle is proportional to the cube of its diameter. These averages represent means as calculated from the different powers of the diameter and mathematically converted back to units of diameter by taking the root of the moment. It is not unusual for a polydispersed particle population to exhibit a diameter of average mass as being one or two orders of magnitude larger than the arithmetic mean of the diameters. In any size distribution, the relation ia equation 4 always holds. 

If the velocity of the flowing stream is uniform over a veiy deep region of hquid (total thickness, 6r V ), the time-averaged mass-transfer coefficient according to penetration theoiy is given oy... 

An approximate solution to this expression can be obtained if we assume m /ms to be constant, i.e., an average mass ratio in the thiekening zone from top to bottom. This results in the following expression ... 

Nitric acid test t 65 wt.% HNOj Five 48 h exposures to boiling solution refreshed after period Average mass loss per unit area of five testing periods -1- 0-99 to -1- 1-20 1. Chromium-depleted areas 2. <7-phase 3. Chromium carbide... 

Atomic mass (Section 1.1) The weighted average mass of an element s naturally occurring isotopes. 

This brief discussion of some of the many effects and interrelations involved in changing only one of the operating variables points up quite clearly the reasons why no exact analysis of the dispersion of gases in a liquid phase has been possible. However, some of the interrelationships can be estimated by using mathematical models for example, the effects of bubble-size distribution, gas holdup, and contact times on the instantaneous and average mass-transfer fluxes have recently been reported elsewhere (G5, G9). 

With the aid of the surface-renewal concept, Huang and Kuo also derived the equation for the average mass-transfer rate ... 

Comparison of Eq. (184) with Eq. (183) shows the effect of size distribution for the case of fast chemical reaction with simultaneous diffusion. This serves to emphasize the error that may arise when one applies uniform-drop-size assumptions to drop populations. Quantitatively the error is small, because 1 — is small in comparison with the second term in the brackets [i.e., kL (kD)112). Consequently, Eq. (184) and Eq. (183) actually give about the same result. In general, the total average mass-transfer rate in the disperser has been evaluated in this model as a function of the following parameters ... 

Characteristic length [Eq. (121)] L Impeller diameter also characteristic distance from the interface where the concentration remains constant at cL Li Impeller blade length N Impeller rotational speed also number of bubbles [Eq, (246)]. N Ratio of absorption rate in presence of chemical reaction to rate of physical absorption when tank contains no dissolved gas Na Instantaneous mass-transfer rate per unit bubble-surface area Na Local rate of mass-transfer per unit bubble-surface area Na..Average mass-transfer rate per unit bubble-surface area Nb Number of bubbles in the vessel at any instant at constant operating conditions N Number of bubbles per unit volume of dispersion [Eq. (24)] Nb Defined in Eq. (134)... 

Nt Total average mass-transfer rate in the disperser... 

Show that under the conditions specified in Problem 10.7 and assuming the Higbie model of surface renewal, the average mass flux at the interface is given bv ... 

Given that, from the penetration theory for mass transfer across an interface, the instantaneous rale ol mass transfer is inversely proportional to the square root of the time of exposure, obtain a relationship between exposure lime in the Higbie mode and surface renewal rate in the Danckwerts model which will give the same average mass transfer rate. The age distribution function and average mass transfer rate from the Danckwerts theory must be deri ved from first principles. 

In the Danckwerts model, it is assumed that elements of the surface have an age distribution ranging from zero to infinity. Obtain the age distribution function for this model and apply it to obtain the average, mass Iransfer coefficient at the surface, given that from the penetration theory the mass transfer coefficient for surface of age t is VlD/(7rt, where D is the diffusivity. 

K, where Tb.onb is the average mass liquid temperature at the cross-section where ONB occurs. 

The tabulated molar mass of an element divided by Avogadro s number is the average mass per atom of that element, but it is not the exact mass of an individual nucleus. There are two reasons for this. First, molar masses refer to neutral atoms. The tabulated molar mass of an element includes the mass of its electrons in addition to the mass of its nucleus. Consequently, the mass of Z electrons must be subtracted from the isotopic molar mass in computing the energy of formation of a nuclide. Second, molar masses of the elements are weighted averages of... 

Two forms of the same element are called isotopes. The isotopes of an element have the same atomic number but have different atomic masses. Iron has several isotopes that, when weighted by their naturally occurring abundance, gives an average mass of 55.845 amu. A simple example would be an element with only two isotopes, one with a mass of 10 amu, the other of 12 amu. If the isotopes were equally common, then the average atomic mass for that element would be 11. If 90% of the element occurred naturally as the isotope with a mass of 10 amu, then the average atomic mass would be 10.2, as calculated below ... 

The atomic mass of an element is the weighted average mass of the isotopes of that element. Based on this definition, which of these does NOT show the correct atomic mass for an element ... 

Measure the average mass of a split pea and calculate its volume. 

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