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Diano4ka-milaya
19 days ago
11

The Michaelis-Menten equation is an expression of the relationship between the initial velocity,V0, of an enzymatic reaction and

substrate concentration, [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation to an expression from which the effect of [S] on the rate can be more readily determined. Match the condition (e.g. [S] = Km) with the statement(s) that describe it:
1. Doubling [S] will almost double the rate.
2. Half of the active sites are occupied by substrate.
3. About 90% of the active sites are occupied by substrate.
4. Doubling [S] will have little effect on the rate.
5. Less than 10% of the active sites are occupied by substrate.
6. This condition will result in the highest rate.
Chemistry
1 answer:
alisha [2.7K]19 days ago
4 0
Option 2: At [S] = Km, 50% of active sites are engaged by substrates.
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Which sentence correctly describes an aspect of the Antarctic treaty system
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The correct choice is C. It prohibits the exploitation of mineral resources. 
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Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

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1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

\nu = 0.163 \times 10^{17} s^{-1}

or, \nu = 1.63 \times 10^{16} s^{-1}

It is known that \nu = \frac{c}{\lambda}

1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}

\lambda = 1.84 \times 10^{-8} m

According to the De-Broglie equation \lambda = \frac{h}{p}

with p = m \times \nu

So, \lambda = \frac{h}{m \times \nu}

m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m} = 3.6 \times 10^{-26} J/m

Squaring both sides gives us:

(m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}

12.96 \times 10^{-52} = m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where m = mass of the electron

Therefore, m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

=\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

=1.42 \times 10^{-21} J

Since K.E = \frac{1}{2}m \nu^{2}

= \frac{1.42 \times 10^{-21} J}{2}

=0.71 \times 10^{-21} J

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is 7.1 \times 10^{-22} J.

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