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Alchen [17]
3 years ago
7

During dehydration synthesis, a ________ molecule is produced.

Chemistry
1 answer:
hoa [83]3 years ago
6 0
*** Remember ***<u />  Dehydration synthesis is when two molecules are joined together, by removing a water molecule. Also a key thing to remember is that <em>dehydration</em><u /> can be remembered as "losing water". 

During dehydration synthesis, a <u>   water   </u><u /> molecule is produced. 


I hope this helps! :D

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She should search for DNA

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We would like to estimate how quickly the non-uniformities in gas composition in the alveoli are damped out. Consider an alveolu
Vera_Pavlovna [14]

Answer: It will take 11.775 seconds.

Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is

A = 4.π.r²

r for an alveolus would be: r = 0.00005m or r = 5.10^{-5}m

Finding the area:

A = 4.3.14.(5.10^{-5})²

A = 3.14.10^{-8}m²

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c = 0.9.3.14.10^{-8}

c = 28.26.10^{-9}

The oxygen diffusivity is 2.4.10^{-9}m²/s, that means in 1 second 2.4.10^{-9} of oxygen spread in one alveolus area. So:

1 second = 2.4.10^{-9}m²

t seconds = 28.26.10^{-9}m²

t = \frac{28.26.10^{-9} }{2.4.10^{-9} }

t = 11.775s

For a concentration change at the center to be 90%, it will take 11.775s.

4 0
3 years ago
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What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

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