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
#SPJ4
Answer:
Explanation:
Hello!
In this case, since the decomposition of sodium hydrogen carbonate is:
Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:
Best regards!
The balanced chemical equation of the reactions given is as follows:
- 2LiHCO3 -----> Li2CO3 + H2O + CO2
- 2 N2 + 5 O2 -----> 2 N2O5
- MgBr2 + KOH -----> KBr + Mg(OH)2
- Mn + 2 CuCl -----> 2 Cu + MnCl2
- 8 Zn + S8 -----> 8 ZnS
- 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
- 2 K + 2 H2O -----> 2 KOH + H2
- C5H12 + 8 O2 -----> 6H2O + 5 CO2
- 2 KOH + H2CO3 -----> 2 H2O + K2CO3
- C4H802 + 6 O2 -----> 4 H20 + 4 CO2
- 16 Al + 3 S8 ---> 8 Al2S3
<h3>How to balance chemical equations</h3>
Balancing of chemical equations is the process of adding numerical coefficients in front of moles of reactants and products to ensure that the moles of atoms of elements of the reactants is equal to the moles of atoms of products formed.
The balanced chemical equation of the reactions given is as follows:
- 2LiHCO3 -----> Li2CO3 + H2O + CO2
- 2 N2 + 5 O2 -----> 2 N2O5
- MgBr2 + KOH -----> KBr + Mg(OH)2
- Mn + 2 CuCl -----> 2 Cu + MnCl2
- 8 Zn + S8 -----> 8 ZnS
- 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
- 2 K + 2 H2O -----> 2 KOH + H2
- C5H12 + 8 O2 -----> 6H2O + 5 CO2
- 2 KOH + H2CO3 -----> 2 H2O + K2CO3
- C4H802 + 6 O2 -----> 4 H20 + 4 CO2
- 16 Al + 3 S8 ---> 8 Al2S3
Learn more about balancing of chemical equations at: brainly.com/question/15428811
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
