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Sloan [31]
4 years ago
6

Consider four small molecules, A–D, which have the following binding affinities for a specific enzyme (these numbers are the equ

ilibrium constants Kd for the dissociation of the enzyme/molecule complex). Which binds most tightly to the enzyme? Which binds least tightly?
A) 4.5 μM

B) 13 nM

C) 8.2 pM

D) 6.9 mM
Chemistry
1 answer:
ehidna [41]4 years ago
8 0

Answer:

Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.

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Explanation:

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2Br^- (aq) + I_2(s)  Br_2(l) + 2I^- (aq)

E^ocell=oxidation potential of anode + reduction potential of cathode\\

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deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

E^ocell= (-1.07+0.53)V\\=-0.54V

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

Answer:

Keq=6.13x10^33

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Answer:

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Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
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Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

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4 0
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Alexeev081 [22]

Answer:

Ionic Bonding: The formation of an Ionic bond is the result of the transfer of one or more electrons from a metal onto a non-metal.

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Explanation:

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