Answer:
The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.
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 is 3Mg + N2 —--> Mg3 N2
Answer:
Explanation:
Filtration followed by evaporation:
To separate the mixture of sand and sugar, it is best to use the separation technique of filtration then evaporation.
Pour the water into the mixture. The sugar will dissolve with time in the water. Sand is made up of quartz and does not dissolve in water.
After the dissolution, filter the solution to separate the sand using a filter paper.
Dry the sand thereafter then proceed to evaporate the sugar with water solution. Evaporation will turn water into vapor and the sugar crystals will be left behind.
