How many milliliters of 0.500 M HBr would be required to react with 40.0 mL of 0.300 M Ca(OH)2?

48dm³

Explanation:

Given reaction:

                  Ca(OH)₂ + 2HBr →  CaBr₂ + 2H₂O

Parameters:

Concentration of HBr = 0.5M

 Volume of Ca(OH)₂ = 40mL

 Concentration of  Ca(OH)₂ = 0.3M

Solution:

To solve this problem, we are going to use the mole concept. We solve from the known specie to the unknown.

 We first find the number of moles of the known specie which is the Ca(OH)₂ ;

                number of moles  = concentration x volume

                number of moles  = 0.3 x 40 x 10⁻³ = 0.012moles

From the reaction equation;

           1 mole of Ca(OH)₂  requires 2 moles HBr

          0.012 moles of Ca(OH)₂ will require 0.012 x 2  = 0.024moles of HBr

Now,

 To find the volume of HBr;

          Volume  =

          Volume  =   = 0.048dm³

In mL;

       Volume 0.048 x 1000 = 48dm³

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.