D is the answer so now we can understand
The oxidizing agent receives electrons from the reducing agent.
2) The oxidation agent takes electrons from the reducing agent.
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:
2 HCl + Ba(OH)₂ ⇒ BaCl₂ + 2 H₂O
Explanation:
In a complete neutralization reaction, an acid reacts with a base to form neutral salt and water. To form barium chloride, hydrochloric acid (acid) reacts with barium hydroxide (base). The balanced chemical equation is:
2 HCl + Ba(OH)₂ ⇒ BaCl₂ + 2 H₂O
Answer:The binomial nomenclature system combines two names into one to give all species unique scientific names. The first part of a scientific name is called the genus. The second part of a species name is the specific epithet. Species are also organized into higher levels of classification.