Here we have to write a simple equation which describes the action of the enzyme catalase.
The equation is: The concentration of the complex [ES] = ![\frac{[E]0}{1+\frac{Km}{[S]} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BE%5D0%7D%7B1%2B%5Cfrac%7BKm%7D%7B%5BS%5D%7D%20%7D)
Let us consider an enzyme catalyses reaction E + S ⇄ ES → E + P
Where E, S, ES and P are enzyme, substrate, complex and product respectively.
The concentration of the complex [ES] = , where 
is the Michaelis constant.
[E]₀ and [S] is the initial concentration of enzyme and concentration of substrate respectively.
Given in the problem is the mass of the liquid (500 grams) and the volume of the liquid (1000 ml = 1000 cm^3).
We can use these two givens to calculate the density of the liquid using the following rule:
density = mass / volume
density = 500 / 1000 = 0.5 grams / cm^3
Comparing the calculated density with the choices we have, we can deduce that the liquid is most likely to be propane with density 0.494 g / cm^3
Answer:
4.36 g of Carbon
Solution:
Step 1: Calculate the %age of Carbon in given Solid as;
Mass of Carbon = 35.8 g
Mass of Hydrogen = 3.72
Total Mass = 35.8 g + 3.72 = 39.52 g
%age of Carbon = (35.8 g ÷ 39.52 g) × 100
%age of carbon = 90.58 %
Step 2: Calculate grams of Carbon in 4.82 g of given solid as;
Mass of Carbon = 4.82 g × (90.58 ÷ 100)
Mass of Carbon = 4.36 g
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.
Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
Answer:
Hello
The answer is 18,7(b)
Explanation:we don't know the value of o² so at first we should put (x)instead of gr of o²and then write 1molO²/32grO²×2mol SO²/3mol O²×64gr SO²/1molSO²=25 gr SO².and then just find the value of (x).
