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!
If you mean hydrate as in <em>MgSO4 · 7H2O, </em>then simply find the molar mass of each element you see.
For the example above, that means you would add the molar mass (found on the periodic table) of Mg, then S, then 4(O), 14(H), and 7(O).
The results would be your molar mass for the hydrate.
I hope this is what you meant by your question!
Answer: The partial pressure of 
is 1.86 atm
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
The given balanced equilibrium reaction is,
Pressure at eqm. 0.973 atm 0.548atm x atm
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 1.86 atm
Thus, the partial pressure of is 1.86 atm
Answer:
The moon to the direct right of the Earth relative to the Sun
Explanation:
