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:
i think it is the last option a pure substance that can be separated into different elements by chemical means
Explanation:
Answer:
the changing tempure in rocks causing it to back apart
Hellium argon neon xenon krypton randon oxygen fluorine chlorine bromine
Answer:
Explanation:
17. it goes from solid copper to aqueous copper:
Cu(s) --> Cu₂(aq) + 2e⁻
18. complete ionic:
Cu(s) --> Cu₂(aq) + 2e⁻
19. net ionic, must include only reacting species, so
Cu(s) --> Cu₂(aq) + 2e⁻
20. this type of reaction is dissolution reaction(redox reaction)
copper reduced from Cu²⁺ to Cu.