The answer is Physical Change.
Answer:
0.00125 moles H₃X
Solution and Explanation:
In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.
Volume of NaOH used = final burette reading - initial burette reading
= 39.18 ml - 3.19 ml
= 35.99 ml or 0.03599 L
Step 1: Moles of NaOH used
Number of moles = Molarity × Volume
Molarity of NaOH = 0.1041 M
Moles of NaOH = 0.1041 M × 0.03599 L
= 0.00375 mole
Step 2: Balanced equation for the reaction between triprotic acid and NaOH
The balanced equation is;
H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)
Step 3: Moles of the triprotic acid (H₃X used
From the balanced equation;
1 mole of the triprotic acid reacts with 3 moles of NaOH
Therefore; the mole ratio of H₃X to NaOH is 1 : 3.
Therefore;
Moles of Triprotic acid = 0.00375 mole ÷ 3
= 0.00125 moles
Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.
Answer:c
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Answer: Glycolysis is stimulated by a high concentration of fructose-2,6-bisphosphate, and the gluconeogenesis is stimulated by a low concentration of fructose-2,6-bisphosphate.
Explanation: Fructose-2, 6-bisphosphate (F2, 6P) is an allosteric activator of the key enzyme in the glycolysis cycle, phosphofructokinase (PFK). F2, 6P also acts as an inhibitor of fructose bisphosphate phosphatase (FBPase) in gluconeogenesis. The concentration of F2, 6P is governed by the balance between its synthesis and breakdown, catalysed by phosphofructokinase-2 (PFK-2) and fructose-bisphosphatase-2 (FBPase-2), respectively. These enzymes are found in a dimeric protein and are controlled by a phosphorylation/dephosphorylation mechanisms. Phosphorylation of the dimeric protein results in an increased concentration of FBPase-2, leading to a decreased concentration of F2, 6P, thus activating the gluconeogenesis cycle. The concentration of PFK-2 is increased when the dephosphorylation of the dimeric protein takes place, leading to the increased concentration of F2, 6P, thus stimulating glycolysis cycle.
