A chemist prepares a solution of vanadium chloride by measuring out of into a volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution. Be sure your answer is rounded to significant digits.

The question is incomplete, the complete question is:

A chemist prepares a solution of vanadium (III) chloride (VCl3) by measuring out 0.40g of VCl3 into a 50.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits.

0.153M of anions

Explanation:

First we calculate the concentration of the solution. From m/M= CV

m=given mass, M= molar mass, C =concentration of solution, V= volume of solution

Molar mass of compound= 51 + 3(35.5)= 157.5gmol-1

0.4g/157.5gmol-1= C×50/1000

C= 2.54×10-3/0.05= 0.051M

But 1 mole of VCl3 contains 3 moles of anions

Therefore, 0.051M will contain 3×0.051M of anions= 0.153M of anions

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.