Calculate the concentration (M) of sodium ions (Na ) in a solution made by diluting 47.8 mL of a 0.801 M solution of sodium sulfide to a total volume of 336.7 mL. Do not include units in your answer and significant figures matter.

The concentration (M) of sodium ions is 0.2274 M

Explanation:

Step 1: Data given

Initial volume of the solution = 47.8 mL = 0.0478 L

Initial concentration of the solution = 0.801 M

The new volume = 336.7 mL = 0.3367 L

Step 2: The balanced equation

Na2S → 2Na + + S^2-

Step 3: Calculate the new concentration of Na2S

C1*V1 = C2*V2

⇒with C1 = the initial concentration of the solution = 0.801 M

⇒with v1 = the initial volume of the solution = 0.0478 L

⇒with V2 = the new volume = 0.3367 L

⇒with C2 = the new concentration of the solution = TO BE DETERMINED

0.801 M * 0.0478 L = 0.3367 L * C2

C2 = (0.801 M * 0.0478 L) / 0.3367 L

C2 = 0.1137 M

Step 4: Calculate concentration of sodium ions

For 1 mol Na2S we have 2 mol Na+

For 0.1137 M Na2S, the concentration of Na+ ions is 2*0.1137 = 0.2274 M

The concentration (M) of sodium ions is 0.2274 M

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter