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Lubov Fominskaja [6]
3 years ago
6

The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per uni

t time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = R max [ E ] t = k 2 ( R max is often written as V max ) If the enzyme molecule has more than one active site, then [ E ] t is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that R max for carbonic anhydrase equals 249 μmol ⋅ L − 1 ⋅ s − 1 and [ E ] t = 2.23 nmol ⋅ L − 1 . Carbonic anhydrase has a single active site.
Chemistry
1 answer:
Nataliya [291]3 years ago
5 0

Answer:

Explanation:

turn over number = R max / [E]t   = K2

From given , R max = 249 * 10 ^ -6 mol. L^-1

T [E]t = 2.23 n mol. L^-1

           =   2.23 * 10^-9 mol. L^-1

Putting values in above equation,

=          111.65 * 10^3           S^-1

Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.

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For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
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Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

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The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

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6 0
3 years ago
100.0 mL of a 0.500 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution
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Answer:

0.1 M

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 100 mL

Initial concentration (C1) = 0.5 M

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Using the dilution formula C1V1 = C2V2, the new concentration of the solution can be obtained as follow:

C1V1 = C2V2

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50 = C2 × 500

Divide both side by 500

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Therefore, the new concentration of the solution is 0.1 M

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