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3 years ago

Will the reaction of addition to an inhibitor be fast or slow?

Chemistry

1 answer:

ad-work [718]3 years ago

7 0

Answer:hope we can be friends

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Although phlorizin inhibition of Na+-glucose cotransport occurs within a few seconds, 3H-phlorizin binding to the sodium-coupled glucose transport protein(s) requires several minutes to reach equilibrium (the fast-acting slow-binding paradigm). Using kinetic models of arbitrary dimension that can be reduced to a two-state diagram according to Cha’s formalism, we show that three basic mechanisms of inhibitor binding can be identified whereby the inhibitor binding step either (A) represents, (B) precedes, or (C) follows the rate-limiting step in a binding reaction. We demonstrate that each of mechanisms A–C is associated with a set of unique kinetic properties, and that the time scale over which one may expect to observe mechanism C is conditioned by the turnover number of the catalytic cycle. In contrast, mechanisms A and B may be relevant to either fast-acting or slow-binding inhibitors.

Explanation:

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Balance the following equation:<br> K+ H2O + H2 + KOH

alex41 [277]

Answer:

2K + 2H2O → H2 + 2KOH

Explanation:

Find how many atoms you have on both sides then add 2 to both sides.

Reactant: Products:

K: 1+1=2 K: 1+1=2

H: 2+2=4 H: 3+1=4

O: 1+1=2 O: 1+1=2

Therefore it is balanced. Hope this helps

6 0

3 years ago

Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

____ [38]

Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

8 0

3 years ago

The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0

3 years ago

The sugar ribose is an important component of RNA. Ribose is 40.0% C, 6.7% H, and 53.3\% by mass. How many grams of each element

Brums [2.3K]

Answer:

C_5H_10O_5

Explanation:

5 0

3 years ago

In a chemical reaction, water decomposes to form hydrogen and oxygen. What is the correct term for the water?

gulaghasi [49]

The answer is: [D]: a reactant.
________________________

7 0

3 years ago

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