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N76 [4]
3 years ago
15

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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Binary compounds of alkali metals and hydrogen react with water to liberate hydrogen gas. The hydrogen gas from the reaction of
DanielleElmas [232]

Answer: The mass of H_{2} liberated is 0.0383 g and the mass of NaH that reacted 0.455 g.

Explanation:

The given data is as follows.

       Volume = 0.501 L,

 Temperature of gas = 35^{o}C = (35 + 273) K = 308 K,

 Total pressure = 760 mm Hg

  Vapor pressure of water at 35^{o}C = 422 mm Hg

  Partial pressure of H_{2} = (760 - 422) mm Hg

                                     = 717.8 mm Hg

The chemical equation is as follows.

       NaH + H_{2}O \rightarrow NaOH + H_{2}

Using ideal gas equation, we will find the moles of hydrogen gas as follows.

             PV = nRT

         n = \frac{PV}{RT}

As we know,

               1 atm = 760 mm Hg

      717.8 mm Hg = 717.8 \times \frac{1 atm}{760 mm Hg}    

                             = 0.944 atm

Putting the given values into the above formula we will calculate the number of moles as follows.

                    n = \frac{PV}{RT}

                       = \frac{0.944 atm \times 0.510 L}{0.08206 Latm/mol K \times 308 K}

                       = 0.0190 moles

Now, we will find the mass of H_{2} as follows.

       Mass = Moles × Molar mass

                 = 0.0190 moles × 2.01588 g/mol

                 = 0.0383 g

Therefore, mass of hydrogen is 0.0383 g.

As the ratio between H_{2} and NaH is 1:1. So, we will calculate the moles of NaH as follows.

             Mass = Moles × Molar mass

                       = 0.0190 moles × 23.99771 g/mol

                       = 0.455 g

Therefore, mass of NaH is 0.455 g.

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Answer:

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A is the answer because it makes sense.
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