Answer is: (Knowing the code allows you to understand the message)
Answer:
E = 3.6×10⁻¹⁹ J
Explanation:
Given data:
Wavelength = 550 nm (550 ×10⁻⁹ nm)
Energy of wave = ?
Solution:
Formula:
E = h c/λ
c = 3×10⁸ m/s
h = 6.63×10⁻³⁴ Js
Now we will put the values in formula.
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s /550 ×10⁻⁹ nm
E = 19.89×10⁻²⁶ J.m /550 ×10⁻⁹ nm
E = 0.036×10⁻¹⁷ J
E = 3.6×10⁻¹⁹ J
Hehe I love Titanium!! ;) Anyways, the chemical formula of titanium (II) Oxide is TiO
Answer:
A,C,D,B
Explanation:
1killometer=1000m
1mm=0.001m
1cm=0.01m
base unit of length is meter
In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.
Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).
The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.
This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.
For more information on phosphorylation click on the link below:
brainly.com/question/7465103
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