Answer:
See below, please
Explanation:
1 mol CH4 ——＞2 mol H2O=2x18 g/mol= 36 g
The question is not clear. You should tell us the concentration of CH4, methane.
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:
Calculate the atomic radii of two touching or overlapping atoms.
Explanation:
No doubt, we can't find the atomic boundary of a single atom, but when atoms are in the form of pairs it becomes very easy to measure the atomic radii of two and then dividing it by 2 to get an estimate of atomic radius of a single atom.
It is also called as covalent radius which is half of the total inter-nuclear distance between two same bonded atoms (Homo-nuclear).
If two adjacent mettalic ions are joined by such pairing then the same half of the distance between the nucleus is termed as metallic radii.
Answer:
<u><em>METALS</em></u>
Lose their valence electrons easily/ ionic by electron loss.
<u><em>NOMETAL</em></u>
Gain or share valence electrons easily/ ionic by electron grain.
