Answer:
The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.
The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .
These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.
Basically the phosphate pathway reaction is very slow due to enzyme catalysis.
Answer:
No, it is not enthalpy favored since the chemical system gains energy.
Explanation:
The dissolution of ammonium nitrate in water is an endothermic process.
Endothermic process requires the system to gain energy to can dissolve the particles in water.
So, the reaction is not enthalpy favored.
Answer:
0.025M
Explanation:
As you must see in your graph, each concentration of the experiment has an absorbance. Following the Beer-Lambert's law that states "The absorbance of a solution is directely proportional to its concentration".
At 0.35 of absorbance, the plot has a concentration of:
<h3>0.025M</h3>
Answer:
138.96kJ is the maximum electrical work
Explanation:
The maximum electrical work that can be obtained from a cell is obtained from the equation:
W = -nFE
<em>Where W is work in Joules,</em>
<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>
Zn(s) → Zn²⁺(aq) + 2e⁻
F is faraday constant = 96500Coulombs/mol
E is cell potential = 0.72V
Replacing:
W = -2mol*96500Coulombs/mol*0.72V
W = - 138960J =
<h3>138.96kJ is the maximum electrical work</h3>
<em />
The chemical formula for chlorophyll is C55H72O5N4<span>Mg. Only 1 of the 137 atoms comes from magnesium.
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