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Cloud [144]
2 years ago
6

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting sol

ution 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.
Chemistry
1 answer:
Anarel [89]2 years ago
3 0

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 = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

The given data is as follows.

     A_{1} = 0.412 (first point)

     A_{2} = 0.642 (second point)

     C_{1} = 0.240 ppm (first point)

     C_{2} = 0.475 ppm (second point)

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

           Slope = \frac{\text{difference in absrobance at two points}}{\text{difference in concentrations at those points}}

                     = \frac{0.642 - 0.412}{0.475 - 0.240}

                      = \frac{0.23}{0.235}

                     = 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 = 0.24 \times 0.978 + 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 = x \times 0.978 + 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.

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Write a balanced half-reaction describing the reduction of gaseous dibromine to aqueous bromide anions.
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<u>Answer:</u> The balanced half reaction is written below.

<u>Explanation:</u>

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

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Br_2+2e^-\rightarrow 2Br^{-}

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2 years ago
Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2
svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S),         Spring Constant (N/m)

      W,                   24

      X,                    35

      Y,                    22

      Z,                    15

The equation for elastic potential energy, E_e, is E_e = 0.5 \times k \times x^2

The above equation can also be written as E_e =\dfrac{1}{2}  \times k \times x^2

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, E_e, of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, E_e, therefore the correct order is as follows;

X > W > Y > Z

9 0
2 years ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

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If it an alkyne and gives 3-ethylhexane  on hydrogenation then the reaction will be

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