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Ede4ka [16]
3 years ago
11

A biologist working in a lab adds a compound to a solution that contains an enzyme and substrate. This particular compound binds

reversibly to the enzyme at the active site. Once the compound is bound to the enzyme, the rate of catalysis of substrate to product is greatly reduced. Which of the statements are true of the compound?
Chemistry
1 answer:
ad-work [718]3 years ago
6 0

Answer:

Adding more substrate would overcome  the effect of the compound

Explanation:

  • Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
  • Compounds known as inhibitors slow down the rate of enzyme activity.
  • Inhibitors are classified as competitive and non-competitive inhibitors.
  • Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
  • The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
  • Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.

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Cell notation will list each half-reaction:
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Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

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What best describes the degree to which a material can transmit Heat A:melting point B:boiling point C: thermal conductivity D:
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8 0
2 years ago
Name the following compound:
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Explanation:

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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