Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calculations and the glassware used to perform the dilution.)

= 0.2 mL.

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= × 0.010

= 0.2 mL.

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

yes it is balanced

a is the answer to this question

explanation: