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goblinko [34]
4 years ago
9

Suppose that a certain drug company manufactured a compound that had nearly the same structure as a substrate for a certain enzy

me but that could not be acted upon chemically by the enzyme. What type of interaction would the compound have with the enzyme
Chemistry
1 answer:
ioda4 years ago
5 0

Answer: Reversible competitive inhibition

Explanation:

In the case of reversible competitive inhibition, an inhibitor molecule competes with the substrate for binding to the active site of the enzyme. The inhibitor blocks the active site of the enzyme. Thus the enzyme substrate complex do not form. The structure of the inhibitor is similar to the substrate thus also have the binding affinity with the enzyme. The process is reversible because the inhibitor will leave the enzyme it exerts no permanent effect on the enzyme.

The given situation is the example of reversible competitive inhibition as substrate remain unchanged and the enzyme was not able to act on the substrate chemically may be due to inhibition of the function of the enzyme.

You might be interested in
Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.
N76 [4]

Answer:

1 . Al_2O_3

2. TiO_2

Explanation:

The more stable the ionic compound, the more is it lattice energy.

  • The more the charge on the cation and the anion, the greater is the lattice energy.
  • The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide (Sc_2O_3) is an oxide in which Sc^{3+} behaves as cation and O^{2-} behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . Al_2O_3

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation Al^{3+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

2. TiO_2

This is because the charge on the cation Ti^{4+} is greater than that of Sc^{3+} and also the size of the cation Ti^{4+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

3 0
3 years ago
How does changing the temperature affect the chemical reaction?
Rufina [12.5K]

Answer:

C

Explanation:

Temperature is directly related to kinetic energy (KE). As we raise temperature, we are raising KE, as well. Particles with more KE move more quickly and with more force.

This means that these particles are more likely to collide with each other and react to allow the chemical reaction to follow through. In turn, if the chemical reaction is more likely to go to completion, the reaction rate increases, eliminating A and B.

The concentration of the solute is not affected by the temperature; in other words, temperature will not increase or decrease the amount of solute in the solution, so eliminate D.

Thus the answer is C.

Hope this helps!

4 0
4 years ago
She collected 20 cm³ of oxygen. What volume of hydrogen could she also have collected at the same time?
loris [4]

Answer:

Assuming it was collected from the atmosphere it would be virtually nothing

Explanation:

hydrogen makes up 0.000055% of the atmosphere while oxygen makes up 23 percent. 20/400000 cm^3 of hydrogen

7 0
3 years ago
How many oxygen atoms are in 4 Fe(C2H302)3?
statuscvo [17]

Answer:

24

Explanation:

5 0
4 years ago
An acetic acid buffer containing 0.50 M acetic acid (CH3COOH) and 0.50 M sodium acetate (CH3COONa) has a pH of 4.74. What will t
ad-work [718]

Answer:

pH = 4.71

Explanation:

We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:

pH = pKa + log [CH3COONa] / [CH3COOH]

<em>Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.</em>

As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:

pH = pKa + log 1

4.74 = pKa

To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl

Moles CH3COONa: Initial CH3COONa - Moles HCl.

<em>Moles CH3COOH: </em>

0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =

0.052 moles CH3COOH

<em>Moles CH3COONa: </em>

0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =

0.048 moles CH3COONa

Using H-H equation:

pH = 4.74 + log [0.048 moles] / [0.052 moles]

<h3>pH = 4.71</h3>
5 0
3 years ago
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