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daser333 [38]
3 years ago
5

Examination of the first few successive ionization energies for a given element usually reveals a large jump between two ionizat

ion energies. For example, the successive ionization energies of magnesium show a large jump between IE2 and IE3. The successive ionization energies of aluminum show a large jump between IE3 and IE4. Explain why these jumps occurs and how you might predict them.
Chemistry
1 answer:
gladu [14]3 years ago
7 0

Answer:

See Explanation

Explanation:

Ionization energy refers to the energy required to remove an electron from an atom. Metals have lower ionization energy than non metals since ionization energy increases across a period.

One thing that we must have in mind is that it takes much more energy to remove an electron from an inner filled shell than it takes to remove an electron from an outermost incompletely filled shell.

Now let us consider the case of magnesium which has two outermost electrons. Between IE2 and IE3 we have now moved to an inner filled shell(IE3 refers to removal of electrons from the inner second shell) and a lot of energy is required to remove an electron from this inner filled shell, hence the jump.

For aluminium having three outermost electrons, there is a jump between IE3 and IE4 because IE4 deals with electron removal from a second inner filled shell and a lot of energy is involved in the process hence the jump.

Hence a jump occurs each time electrons are removed from an inner filled shell.

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Explanation:

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 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}&#10;

Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s&#10;}{K_{M} ln (\frac{45 mmol/L&#10;}{9 mmol/L&#10;}) + (45 mmol/L - 9 mmol/L&#10;)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

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