The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is 
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:

We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:

Hence, the concentration of copper fluoride in the solution is 
Answer:
Explanation:
turn over number = R max / [E]t = K2
From given , R max = 249 * 10 ^ -6 mol. L^-1
T [E]t = 2.23 n mol. L^-1
= 2.23 * 10^-9 mol. L^-1
Putting values in above equation,
= 111.65 * 10^3 S^-1
Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.
<span>rutherfordium element # 104</span>