Mole fraction of oxygen gas: 0.381
Additional clarification
Given:
2.31 atm Oxygen
3.75 atm Hydrogen
Required:
Mole fraction of Oxygen
Calculation:
According to Dalton’s Law of partial pressures
P tot = P₁ + P₂ +.. + Pₙ
Substituting values:
P tot = P O₂ + P H₂
P tot = 2.31 atm + 3.75 atm
P tot = 6.06 atm
Mole fraction of O₂ (X O₂):
P O₂ = X O₂ x P tot
X O₂ = P O₂ / P tot
X O₂ = 2.31 / 6.06
X O₂ = 0.381
The concentration of acid HX is established at 6.0 M. Calculating the moles of KOH neutralizing the acid yields 0.12 moles KOH. Subsequently, converting this to the moles of acid results in 0.12 moles HX. Thus, the molarity of HX remains 6.0 M.
Answer: Please see answer below
Explanation:
The sequence for glycogen degradation is as follows:[
---> Hormonal signals initiate the breakdown of glycogen.
1. Glycogen undergoes debranching through the hydrolysis of α‑1,6 linkages.
2. Blocks of three glucosyl units are relocated by remodeling α‑1,4 linkages.
3. Glucose 1‑phosphate is derived from the non-reducing ends of glycogen and is transformed into glucose 6‑phosphate.
---> Glucose 6‑phosphate enters further metabolic pathways
Glycogen degradation consists of three stages:
(1) the release of glucose 1-phosphate from glycogen,
(2) transforming the glycogen structure for continued breakdown, and
(3) converting glucose 1-phosphate into glucose 6-phosphate for subsequent metabolism.
(https://www.ncbi.nlm.nih.gov/books/NBK21190)[[TAG_34]][[TAG_35]][[TAG_36]]
Answer:
1984
Explanation:
Utilizing the equation;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2 = half-life of the radioactive isotope
t= age of the wine
Ao= initial activity of the wine
A= activity at time = t
Substituting values, we have 0.693/12.3 = 2.303/t log (5.5/0.688)
0.693/12.3 = 2.079/t
0.056 = 2.079/t
t= 2.079/0.056
t= 37 years
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