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.
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Answer:
1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm
Explanation:
An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
- P= 1 atm
- V= 22.4 L
- n= ?
- R= 0.082
- T=273 K
Reemplacing:
1 atm* 22.4 L= n* 0.082 *273 K
Solving:
n= 1 mol
Another way to get the same result is by taking the STP conditions into account.
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<u><em>1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm</em></u>
Answer:
B
Explanation:
Man with variety backgrounds
<span>My only guess is obtain a metal and heat it in a boiling water bath (of known temperature) this will be your initial temperature. Now obtain a calorimeter cup with water of known temperature as well. Place the metal into the calorimeter cup and record the temperature after 5 minutes. You now have delta T, mass of the metal, and Q. Solve for C.
Hope this helps xox :)</span>
