Answer:Nothing, the photon just bounces off the surface.
Explanation:
According to Albert Einstein, a photoelectron can only be emitted from a metal surface when the energy of the incident photon is greater than the work function of the metal.
In the scenario described in the question, the work function of the metal is greater than the energy of the photon. Hence, the photon just bounces off the metal surface without emitting any electron.
Energy is stored in chemical bonds during photosynthesis.
During photosynthesis, the radiant energy from the sun is converted to chemical energy in carbohydrates.
Inorganic materials in the form of carbon dioxide and oxygen combine to form carbohydrates in the presence of radiant energy according to the equation below:

The energy is thus, stored in chemical bonds in the carbohydrate and this is what is oxidized during respiration to release the locked energy.
More on photosynthesis can be found here: brainly.com/question/1388366
The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
Answer:
-191.7°C
Explanation:
P . V = n . R . T
That's the Ideal Gases Law. It can be useful to solve the question.
We replace data:
2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°
(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°
T° = 81.3 K
We convert T° from K to C°
81.3K - 273 = -191.7°C
Answer:
The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.
Explanation:
Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:
Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺
Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.
2 Pyruvate ----> 2 AcetylCoA + 2CO₂
Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.