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3 years ago

In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)

Chemistry

1 answer:

Veronika [31]3 years ago

4 0

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

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Select the keyword or phrase that will best complete each sentence. A tetrahedral carbon is____hybridized while a linear carbon

aivan3 [116]

Answer:

Explanation:

A tetrahedral carbon is__sp³__hybridized while a linear carbon is__sp___hybridized. Two different compounds that have the same molecular formula are known as___isomers____. Pi (π) bonds are generally_weaker (because they overlapped side ways)___than sigma (sigma) bonds. Hybridization is the combination of two or more__atomic ____orbitals to form the same number of__hybrid (combined s and p)__orbitals, each having the same shape and energy. A_pi (π)____bond is formed by side-by-side overlap of two p orbitals. The_electronegativity___is a measure of an atom's attraction for electrons in a bond and indicates how much a particular atom ''wants" electrons. Two Lewis structures that have the same atomic placement and a structure but a different arrangement of pi electrons are called_Resonance structure____. All single bonds are___sigma__bonds.

7 0

4 years ago

Solid to Gas

miskamm [114]

Answer:

c. c. c c c c c cçcc. ccccccccc

Explanation:

ccc cc

7 0

3 years ago

Need help on 17,19,20 please.

pentagon [3]

Answer:

c, maybe d, and I think b.

Explanation:

Im sorry if wrong

5 0

3 years ago

Which process is expected to have an increase in entropy?

olganol [36]

Answer: Option (B) is the correct answer.

Explanation:

Degree of randomness of the molecules of a substance is known as entropy. More is the kinetic energy between the molecules of a substance more will be the degree of randomness.

Therefore, when a substance is present in a gaseous state then it has the maximum entropy. In liquid state, molecules are closer to each other so, there is less randomness between them.

On the other hand, in solid state molecules are much more closer to each other as they arr held by strong intermolecular forces of attraction. Therefore, they have very less entropy.

When liquid water is formed from gaseous hydrogen and oxygen molecules then gas is changing into liquid. So, there is decrease in entropy.

When $N_{2}O_{4}$ decomposes then the reaction will be as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Since, 1 mole is producing 2 moles. This means that degree of randomness is increasing as both the molecules are present in gaseous form.

In formation of a precipitate, aqueous solution is changing into solid state. Hence, degree of randomness is decreasing.

Rusting of iron also leads to the formation of solid as it forms $Fe_{2}O_{3}.xH_{2}O$ .

Thus, we can conclude that decomposition of $N_{2}O_{4}$ gas to $NO_{2}$ gas is the process that is expected to have an increase in entropy.

4 0

3 years ago

For the reaction 3C2H2(g)---> C6H6(l) at 25 C the standard enthalpy change is -631 kj and the standard entropy change is -430

Black_prince [1.1K]

Given :

Temperature , T = 25°C = (25 + 273) K = 298 K .

The standard enthalpy change , $\Delta H^{\circ}=-631\ kJ/mol$ .

The standard entropy change , $\Delta S^{\circ}=-430\ J/K=-0.430\ kJ/K$ .

To Find :

The standard free energy change at 25° C .

Solution :

Standard free energy is given by :

$\Delta G^o=\Delta H^o-T\Delta S^o\\\\\Delta G^o=-631-[298\times (-0.430)]\ kJ\\ \\\Delta G^o=-631-(-128.14)\ kJ\\\\\Delta G^o=-502.86\ kJ$

Therefore , the standard free energy change at 25° C is -502.86 kJ .

Hence , this is the required solution .

3 0

4 years ago