The high labeling efficiency can be attributed to the linear chelator backbone of DTPA analogues. As mentioned earlier, there are mainly two labeling strategies: the prelabeling and the postconjugation labeling.īecause of the fast labeling kinetics, DTPA-coupled peptides can be readily and efficiently radiolabeled by incubation for 10–15 min at room temperature, using acetate or a citrate buffer (pH 5–7). The choice of the radiolabeling approach depends on the type of biomolecules to be radiolabeled and the purpose of the study (diagnosis and/or therapy). In view of the very low molar concentration of radiometal solution (often lower than 10 –6 M) used for peptide labeling, there is a need to select optimal labeling conditions that produce a radiolabeled peptide complex in high radiochemical purity in a safe and rapid way hence, the labeling (complexation) protocols are of utmost importance.
Maecke, in Advances in Inorganic Chemistry, 2016 3.1 Radiolabeling Protocols Transformed concentration (theory of reaction invariants) (mol/m 3) z i Spatiotemporal field of arbitrary quantity U Surface source of ith component (mol/m 2/s) t iĬharge transfer number of ith component TĮlements of transformation matrices (theory of reaction invariants) u Volumetric source of ith component (mol/m 3/s) S bnd, i Stoichiometric coefficient of ith component in jth reaction S Rate of jth chemical reaction (mol/m 3/s) R Number of components ( χ = c), reactions ( χ = r), resistances ( χ = R), reaction invariants ( χ = Y), subdomains ( χ = Ω) … q Mobility of ith component (ion) (mol/m 2/J/s) = (mol s/kg) n ˙ iĭimensionless unit vector normal to the boundary N χ Kinetic rate constant of jth chemical reaction (units depends on definition) K j Intensity of molar flux of ith component (ion) (mol/m 2/s) k j Intensity of molar flux of kth invariant (mol/m 2/s) I Volumetric and area fraction, respectively F Molar concentration of ith component (mol/m 3) D iĭiffusion coefficient of ith component (m 2/s) D In this lesson, we will solve additional problems with greater complexity to give the student additional practice with these methods an concepts.Vapor–liquid equilibrium (VLE) constant or Kozeny constant This is a very simple problem to illustrate how the concepts of calculating molar concentration can be used to solve problems in chemistry. In other words, if we use 0.15 moles of NaCl and add water to make the final solution volume of 1.5 liters, we know that the molar concentration of the final solution will be 0.1 M NaCl as specificied in the problem statement. Moles of NaCl = 0.15 moles of NaCl required to make this solution. Multipy both sides of this relation by 1.5 to clear this number from the right hand side and we arrive at the answer: Putting this info into the equation above we arrive at the following:Ġ.1 M NaCl = (moles of NaCl) / (1.5 L of solution) Final molar concentration of 0.1 M NaCl
In this problem, we wish to calculate how many moles of solute (in this case NaCl) are required to make the specified solution that has these characteristics: Molar Concentration = (Moles of Solute) / (Volume of Solution) We learned in the last lesson that the molar concentration of any solution is: The key here is that you want to know how many moles of NaCl are needed. Specifically, the molar concentration is used to solve more complex chemistry problems that involve more than one step to solve.įor example, lets say that you would like to calculate how many moles of NaOH are required to prepare 1.5 L of solution with molar concentration of 0.1 M NaCl? How do you solve this problem? How does calculating molar concentration fit into the process?
In this lesson, we continue to work problems that involve calculating molar concentration of a solution.
0 kommentar(er)
