Answer:
ΔH = -110.5kJ
Explanation:
It is possible to obtain enthalpy of combustion of a particular reaction by the algebraic sum of similar reactions (Hess's law). Using:
1. C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5kJ
2. CO(g) + 1/2O₂(g) → CO₂(g) ΔH₂ = -283.0kJ
The sum of 1 -2 gives:
C(s) + <u>O₂(g)</u> + <u>CO₂(g)</u> → <u>CO₂(g)</u> + CO(g) + <u>1/2O₂(g)</u>
C(s) + 1/2O₂(g) → CO(g) ΔH = -393.5kJ - (-283.0kJ) =
<h3>ΔH = -110.5kJ</h3>
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Answer:
a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change
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.
Answer:
12.5 g
Explanation:
If element R's half life is 2 minutes then we need to divide 100 g by 1/2 3 times. 100/2 = 50/2= 25/2= 12.5 g.
Hope this helps!
Answer:
Stronger!
Explanation:
The <u>stronger</u> the forces among the particles in a sample of matter, the more rigid the matter will be.
