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Nana76 [90]
1 year ago
12

The electrons lost from chlorophyll photooxidation are replaced by the oxidation of water. how many electrons are generated from

the oxidation of one water molecule?
Physics
1 answer:
White raven [17]1 year ago
3 0

There are 2 electrons generated from the oxidation of one water molecule.

<h3>Describe photooxidation.</h3>

The process of a substance interacting with oxygen or losing electrons from chemical species under the influence of light is known as photooxidation. Photooxidation happens in plants when there is environmental stress. It is called photooxidative stress as a result. Reactive oxygen species are produced by the absorption of excess excitation energy in plant tissues. Chloroplasts are harmed by the accumulation of these reactive oxygen species, which is a damaging process in plants. High-intensity light and little CO_{2} are the two conditions that cause this photooxidative stress to occur most frequently. It is a procedure that requires light. Photorespiration in C_{3} plants guards against photooxidation.

To know more about Photooxidation visit:

brainly.com/question/14788790

#SPJ4

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Answer:

175.96 g

Explanation:

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= 64 x 9.8 x 7070

= 4.434 x 10⁶ J

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= 1.0557 x 10⁶ cals

= 1.0557 x 10⁶ / 6000 g of butter

= 175.96 g of butter.

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During Cardiovascular Excercise, what do the heart and lungs do?
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A 2-kg toy car accelerates from 0 to 5 m/s2. It
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10 joules of work is done by the object
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3 0
2 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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