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liberstina [14]
2 years ago
10

A. Draw a graph that shows the relative concentration of BTB from the center of the

Chemistry
1 answer:
rosijanka [135]2 years ago
6 0

Answer:

did you get the answer???

Explanation:

I really need help

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Why is the periodic table shaped the way it is
jasenka [17]
Properties and electron arrangement. Going down each column elements have similar properties and as they move left to right have the same number of electrons in their outer shell
8 0
3 years ago
Diatomic gases such as H2(g), O2(g), and N2(g) contain ________ bonds.
Studentka2010 [4]
Diatomic gases contain covalent bond
6 0
2 years ago
An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
fomenos

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

5 0
2 years ago
The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
worty [1.4K]
<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

4 0
3 years ago
How many nanometers are in 0.0006245101 km?
maw [93]

Answer:

624510100

Explanation:

Doing a conversion factor:

0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]

5 0
3 years ago
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