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Recognizing change

Fiban-induced ihrombocytopenia Drag (ligand) reacts with membrane glycoprotein Ilb/lIIa and induces a conformational change recognized by antibody Epitifibatide, tirofiban... [Pg.67]

On the other hand it is well known that the hnman eye can adapt to changing inspection conditions in recognizing flaws while a camera system is not able to adapt automatically. [Pg.628]

It is generally recognized that the flexibility of a bulk polymer is related to the flexibility of the chains. Chain flexibility is primarily due to torsional motion (changing conformers). Two aspects of chain flexibility are typically examined. One is the barrier involved in determining the lowest-energy conformer from other conformers. The second is the range of conformational motion around the lowest-energy conformation that can be accessed with little or no barrier. There is not yet a clear consensus as to which of these aspects of conformational flexibility is most closely related to bulk flexibility. Researchers are advised to first examine some representative compounds for which the bulk flexibility is known. [Pg.312]

A selected list of redox indicators will be found in Table 8.26. A redox indicator should be selected so that its if" is approximately equal to the electrode potential at the equivalent point, or so that the color change will occur at an appropriate part of the titration curve. If n is the number of electrons involved in the transition from the reduced to the oxidized form of the indicator, the range in which the color change occurs is approximately given by if" 0.06/n volt (V) for a two-color indicator whose forms are equally intensely colored. Since hydrogen ions are involved in the redox equilibria of many indicators, it must be recognized that the color change interval of such an indicator will vary with pH. [Pg.1160]

Figure A6.1 and Table A6.1 show how a solute s distribution changes during the first four steps of a countercurrent extraction. Now we consider how these results can be generalized to give the distribution of a solute in any tube, at any step during the extraction. You may recognize the pattern of entries in Table A6.1 as following the binomial distribution... Figure A6.1 and Table A6.1 show how a solute s distribution changes during the first four steps of a countercurrent extraction. Now we consider how these results can be generalized to give the distribution of a solute in any tube, at any step during the extraction. You may recognize the pattern of entries in Table A6.1 as following the binomial distribution...
We recall some of the ideas of kinetics from the summary given in Sec. 5.2 and recognize that the rates of initiator decomposition can be developed in terms of the reactions listed in the Table 6.1. Using the change in initiator radical concentration d[I-]/dt to monitor the rates, we write the following ... [Pg.353]

Polymer propagation steps do not change the total radical concentration, so we recognize that the two opposing processes, initiation and termination, will eventually reach a point of balance. This condition is called the stationary state and is characterized by a constant concentration of free radicals. Under stationary-state conditions (subscript s) the rate of initiation equals the rate of termination. Using Eq. (6.2) for the rate of initiation (that is, two radicals produced per initiator molecule) and Eq. (6.14) for termination, we write... [Pg.362]

Finally we recognize that a 1°C temperature variation can be approximated as dT and that (dRp/Rp) X 100 gives the approximate percent change in the rate of polymerization. Taking average values of E from the appropriate tables, we obtain E j = 145, E = 16.8, and Ep = 24.9 kJ mol . For thermally initiated polymerization... [Pg.368]

This result should be compared with Eq. (8.28) for the case of the ideal mixture. It is reassuring to note that for n = 1, Eq. (8.36) reduces to Eq. (8.28). Next let us consider whether a change of notation will clarify Eq. (8.36) still more. Recognizing that the solvent, the repeat unit, and the lattice site all have the same volume, we see that Ni/N is the volume fraction occupied by the solvent in the mixture and nN2/N is the volume fraction of the polymer. Letting be the volume fraction of component i, we see that Eq. (8.36) becomes... [Pg.517]


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