I think there are 3
1) lack of membrane-bound organelles
2) unicellular
3) small (usually microscopic) size.
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 
3Ba + N2 -----> Ba3N2
Ba is in the second group and metal, its charge = +2 in the compound.
N is non-metal, it is in 15th group, so it has 5 electrons on the last level, so it can take 3 electrons more(8-5=3). N has charge (-3) in compound.
Ba and N form compound Ba3N2.