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 
This question comes with four answer choices:
<span>A. H2O + H2O ⇄ 2H2 + O2
B. H2O + H2O⇄ H2O2 + H2
C. H2O + H2O ⇄ 4H+ + 2O2-
D. H2O + H2O ⇄ H3O+ + OH-
Answer: option </span><span>D. H2O + H2O ⇄ H3O+ + OH-
(the +sign next to H3O is a superscript, as well as the - sing next to OH)
Explanation:
The self-ionization of water, or autodissociation, produces the two ions H3O(+) and OH(-). The presence of ions is what explain the electrical conductivity of pure water.
</span><span>In this, one molecule of H2O loses a proton (H+) (deprotonates) to become a hydroxide ion, OH−. Then, he <span>hydrogen ion, H+</span>, immediately protonates another water molecule to form hydronium, H3O+.
</span>
N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol
n = 7,44mol
M = 17g/mol
m = n * M = 7,44mol * 17g/mol = 126,48g
An atomic mass unit is defined as precisely 1/12 the mass of an atom of carbon-12.