The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and
4595 L of hydrogen gas at STP, heat the mixture to run the reaction,
then isolate the ammonia from the reaction mixture. What volume of
NH3 in liters, measured at STP, would be produced? Assume the
reaction goes to completion.
N2 (9) + 3 H2 (9) - 2 NH3(g)

At 40 âc the solution has at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.g of kno3 per 100 g of water and it can contain up to at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.g of kno3 per 100 g of water. at 0 âc the solubility is ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.kno3 per 100 g of water, so ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.gkno3 per 100 g of water will precipitate out of solution. a kno3 solution containing 55 g of kno3 per 100.0 g of water is cooled from 40 ∘c to 0 ∘c. what will happen during cooling?

Nonpoint source pollution is difficult to control because it

The data below were determined for the reaction shown below. s2o82– + 3i – (aq) → 2so42– + i3– expt. # [s2o82–] (m) [i –] (m) initial rate 1 0.038 0.060 1.4 × 10 – 5 m/s 2 0.076 0.060 2.8 × 10 – 5 m/s 3 0.076 0.030 1.4 × 10 – 5 m/s the rate law for this reaction must be: