A stock solution of HNO3 is prepared and found to contain 13.1 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is M.

The diluted solution is 0.65 M

Explanation:

Dilution formula to solve the problem is:

M concentrated . V concentrated = M Diluted . V diluted

We can also make a rule of three though.

In 1000 mL of concentrated solution, we have 13.1 moles of  HNO₃

So in 25 mL we have (25 . 13.1 ) / 1000 = 0.327 moles of HNO₃

This moles are also in 0.500 L of diluted solution, the new molarity will be:

0.327 mol / 0.5L = 0.65 M

Molarity of diluted solution must be always lower than concentrated solution, otherwise the excersise is wrong.

If we replace in the formula

13.1 M . 25 mL = 500 mL . M diluted

M diluted = 13.1 m . 25 mL / 500 mL = 0.65 M

asking the goog might u get an answer

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .