A chemist prepares a solution of iron(II) bromide (FeBr2) by measuring out 46. g of iron(II) bromide into a 400. mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's iron(II) bromide solution. Round your answer to 2 significant digits.

Solution

Given mass of FeBr2 = 85.5 g

Molar mass of FeBr2 = 215.65

No. of Moles of FeBr2 = Given Mass / Molar mass = 85.5 / 215.65

No. of Moles = 0.3964

Given Volume is 450 mL i.e. 0.450 L

Now Concentration = 0.3964/0.450

Hence Concentration = 0.881 mol/L

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

i believe the answer is true.