Answer:
The Michaelis‑Menten equation is given as
v₀ = Kcat X [E₀] X [S] / (Km + [S])
where,
Kcat is the experimental rate constant of the reaction; [s] is the substrate concentration and
Km is the Michaelis‑Menten constant.
Explanation:
See attached image for a detailed explanation
The ionic character of any compound depend on the lattice energy as well as the electronegativity of element present in that compound.
More would be the lattice energy more would be ionic nature of that compound.
The lattice energy of any compound is inversely proportional to the ionic radii cation and anion.
In given case the ionic radii of oxide in both oxides would be equal therefore the lattice energy only depend on the ionic radii of cation.
As the radii of Magnesium less then radii of lithium therefore lattice energy of Magnesium oxide would be more than lithium oxide.
Hence, MgO would be more ionic in nature than 
Answer:
In 1000 ml there is 0.10 moles of Fe 2+
Therefore, in 10 ml there is (0.1/1000)*10= 0.001 mol of Fe2+
mole ratio for rxn Fe2+ : MnO4- is
1 : 2
therefore if 0.001 moles of Fe2+ react then 0.001*2 =0.002 moles of MnO4- react with Fe2+
hence, molarity of MnO4- = (mol*vol)/1000
= 0.002*10.75/1000= 2.15*10-5M
Explanation:
Hope this helps
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.
