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xxMikexx [17]
3 years ago
11

How does the negative control of the lac operon by glucose function through inducer exclusion? See Section 18.3 (Page 373) . Vie

w Available Hint(s) How does the negative control of the lac operon by glucose function through inducer exclusion? See Section 18.3 (Page 373) . High glucose concentrations prevent the transport of lactose into the cell. High glucose concentrations promote the transport of lactose into the cell. High lactose concentrations prevent the transport of glucose into the cell. High lactose concentrations promote the transport of glucose into the cell.
Chemistry
1 answer:
just olya [345]3 years ago
7 0

Answer:

High glucose concentrations prevent the transport of lactose into the cell.

Explanation:

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Which electromagnetic radiation on the electromagnetic spectrum can be detected by the human eye?
qaws [65]

Answer:

D. Visible Light

Explanation:

D is the right answer because visible light is perceived by humans as color and the rest aren't visible at all. For example, if you broke a bone and have to have an X-ray done, you can't see the x-rays. When you are heating something in the microwave, you can't see the waves. And when you are listening to the radio in the car, you also can't see the waves.

4 0
2 years ago
Read 2 more answers
Match each of Earth's spheres with its definition.
insens350 [35]

Answer:

Thin layer of gases suspended in the air surrounding.

Explanation:

Atmosphere, mark me brainlist.

7 0
2 years ago
Read 2 more answers
An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu
chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

The density of a proton is 6.278\times 10^{14} g/cm^3.

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\pi r^3..[1]

Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA

Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA

Volume of nucleus = V'

V=\frac{4}{3}\pi r'^3..[2]

Dividing [2] by [1]

\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}

=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}

\frac{V'}{V}=4.6656\times 10^{-14}

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm

1 m = 100 cm

Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3

Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g

1 amu = 1.66054\times 10^{-24} g

Density of the proton : d

d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3

The density of a proton is 6.278\times 10^{14} g/cm^3.

5 0
3 years ago
Which of the following statements is typically true for a catalyst? I. The concentration of the catalyst will go down as a react
gladu [14]

Answer: II. The catalyst provides a new pathway in the reaction mechanism.

III. The catalyst speeds up the reaction.

Explanation:

Activation energy is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more molecules can cross the energy barrier and more molecules convert to products.

The catalyst itself does not take part in the chemical reaction and gets regenerated as such at the end of the reaction without getting consumed.

3 0
2 years ago
Using molecular orbital theory, explain why the removal of one electron in O2 strengthens bonding, while the removal of one elec
Nuetrik [128]
First you have a knowledge of bond order which is
 B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2  
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)

6 0
3 years ago
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