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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The primary reason for this was that Mendeleev
didn't know that atomic numbers actually existed. Atomic numbers
were only discovered a period after Mendeleev's time. The use of X-rays made it
possible to find the atomic number, and those had not been discovered yet. <span>
<span>The periodic table was then arranged in 1913 by Henry Moseley
in an arrangement according to atomic number.</span></span>
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".
V </span>α T
Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂
V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K
By applying the formula,
2.00 L / 313 K = V₂ / 303 K
V₂ = (2.00 L / 313 K) x 303 K
V₂ = 1.94 L
Hence, the volume of the balloon at 30.0 ⁰C is 1.94 L
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O
let's get the moles of LiOH first
moles= Molarity x Liters
moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH
using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:
0.00423 mol LiOH = 0.00423 mol HBr
now let's find the concentration
molarity= mol/ Liters
0.00423 mol/ 0.01000 Liters= 0.423 M
Molar mass of oxygen gas:
O₂ = 16 * 2 = 32.0 g/mol
1 mole O₂ -------------- 32.0
9.05 mole O₂ ---------- ?
Mass = 9.05 * 32.0
Mass = 289.6 g of O₂
hope this helps!
