Answer:
12 átomos de oxígeno hay presentes
Explanation:
Basados en la reacción:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
<em>6 moles de agua producen 1 mol de glucosa</em>
<em />
Si reaccionan 12 moleculas de agua, se producirán:
12 moleculas H₂O * (1 mol C₆H₁₂O₆ / 6 mol H₂O) =
2 moléculas de glucosa se producen.
Como cada molécula de glucosa tiene 6 átomos de oxígeno:
2 moléculas C₆H₁₂O₆ * (6 átomos Oxígeno / 1 molécula C₆H₁₂O₆) =
<h3>12 átomos de oxígeno hay presentes</h3>
No chemical because, its changeing the stuff
Answer: Please see answer below
Explanation:
The steps of glycogen degradation is as follows from this order.
--->Hormonal signals trigger glycogen breakdown.
1. Glycogen is (de)branched by hydrolysis of α‑1,6‑glycosidic linkages.
2. Blocks consisting of three glucosyl residues are moved by remodeling of α‑1,4‑glycosidic linkages.
3.[Glucose 1‑phosphate is cleaved from the non reducing ends of glycogen and converted to glucose 6‑phosphate.
--->Glucose 6‑phosphate undergoes further metabolic processing
The degradation of Glycogen follows three steps:
(1) the release of glucose 1-phosphate from glycogen,
(2) the remodeling of the glycogen substrate to permit further degradation, and
(3) the conversion of glucose 1-phosphate into glucose 6-phosphate for further metabolism.
(https://www.ncbi.nlm.nih.gov/books/NBK21190)
Answer:
The sun shines with equal intensity on a farm field, an asphalt road and the ocean. So basically asphalt road are heated the most during the day
Explanation:
The heat of water is more than the specific heat of sand.
Therefore sand is hot.
.Water is reflecting solar radiation.
The land retains more heat since the land absorb solar radiation.
Therefore the land surfaces warm faster.
Since water is a slow conductor of heat, it need more energy than the sand. so its temperature is increasing. soil loses heat faster.
The ocean heats slower than land , the land air is more warmer than lean air. After the sun set the land loses heat quickly .The air above it cools.
Answer:
SN2
Explanation:
The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.
The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.
Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.
