The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
<h3>What is energy?</h3>
Energy is the ability to do work.
The primary source of energy on the earth is the sun.
The energy from the sun is used by producers to produce food on which other organisms depend on.
The energy from the sun gets to the humpback whale through producers such as plankton.
The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
Learn more about energy flow at: brainly.com/question/21786633
Answer:
Follows are the solution to this question:
Explanation:
Please find the image file of the chemical reaction in the attachment:
In a water medium, the CH3- type CH 3Li is a heavy nucleophile that attacks the carbonyl carbon atom to form the alkoxide ion, which will then be protonated to form alcohol.
Answer: The range of wavelengths of light that can be used to cause given phenomenon is .
Explanation:
Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J
Formula used is as follows.
where,
E = energy
h = Planck's constant =
c = speed of light =
Substitute the values into above formula as follows.
Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is .
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
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Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Carbonic anhydrase speeds up<span> the transfer of carbon dioxide from cells to the blood.
Hope this work cuz</span>