A5.00-ml sample of blood was treated with trichloroacetic acid to precipitate proteins. after centrifugation, the resulting solution was brought to a ph of 3 and was extracted with two 5-ml portions of methyl isobutyl ketone containing the organic lead complexing agent apcd. the extract was aspirated directly into an air-acetylene flame yielding an absorbance of 0.454 at 283.3 nm. five-milliliter aliquots of standard solutions containing 0.240 and 0.475 ppm pb were treated in the same way and yielded absorbances of 0.412 and 0.642. calculate the concentration pb (ppm) in the sample assuming that beer’s law is followed.

Explanation:

According to the Beer-lambert' s law,

        Absrobance = absrobptitvity × lenght × concentration

So, if we will plot a graph between absrobance and concentration then we will obtain a straight line

Hence, formula to calculate slope of the graph is as follows.

   The slope of graph =

The given data is as follows.

      = 0.412 (first point)

      = 0.642 (second point)

      = 0.240 ppm (first point)

      = 0.475 ppm (second point)

Hence, putting these values into the above formula we will calculate the value of slope as follows.

           Slope =

                     =

                      =

                     = 0.978

It is known that for the graph, line equation can be written as follows.

                       y = mx + c  ........... (1)

where,      C = intercept

                 m = slope

Hence, calculate the value of intercept as follows.

           0.412 = + c

               c = 0.178

As it is given that the absorbance values 0.454 (the y-axis value). Therefore, putting this value into equation (1) we get the following.

                     y = mx + c

                  0.454 = + 0.178

                   0.276 = 0.978x

                         x = 0.282 ppm

Thus, we can conclude that the concentration of Pb in the given sample is 0.282 ppm.

a:

by the combination of an unlimited number of elements

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.