A chemistry student needs 125 mL of 0.150 NaOH solution for her experiment but the only solution available in laboratory is 3.02 M describe how student could prepare the solution she needs.

The student needs to measure out 6.21mL of the stock solution and dilute it to 125mL with water.

Explanation:

To obtain the desired solution, the student need to determine how much of the stock solution she will need.

Now, let us calculate the volume of the stock solution needed.

Data obtained from the question include:

Molarity of the stock solution (M1) = 3.02 M

Volume of the stock solution needed (V1) =...?

Molarity of the diluted solution (M2) = 0.150 M

Volume of the diluted solution (V2) = 125mL

Using the dilution formula, the volume of the stock solution needed can be obtained as follow:

M1V1 = M2V2

3.02 x V2 = 0.15 x 125

Divide both side by 3.02

V2 = (0.15 x 125)/3.02

V2 = 6.21 mL.

Therefore, the student needs to measure out 6.21mL of the stock solution and dilute it to 125mL with water.

"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."