A 54.0 mL aliquot of a 1.60 M solution is diluted to a total volume of 218 mL. A 109 mL portion of that solution is diluted by adding 161 mL of water. What is the final concentration? Assume the volumes are additive g

The final concentration is 0.16 M.

Explanation:

According to the dilution law:

where,

= molarity of stock solution = 1.60 M

= volume of stock  solution = 54.0 ml

= molarity of diluted solution = ?

= volume of diluted solution = 218 ml

Putting these values:

Now 109 ml of this diluted solution is further diluted by adding 161 mL of water.

Again applying dilution law:

Thus the final concentration is 0.16 M

okay yes , aha.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2