Ten (10.0) grams of hydrogen gas and 80,0 g of nitrogen gas are mixed and to react. how many grams of ammonia will result? n: (g) + 3 h: (g) 2 nhs(g) a 90.0g 56.5g48.6g 284ge7i 4. dilution: to make 2.5 l of 0.10 m hydrochloric acid you would a. dilute 41.7 ml of 6.0 m hci to 2.5 l with water b. deterniine the mass of 0.10 mol hcl and dissotve it in 2.5 i of water e add 50 ml of 6.0 m hcl to 2.45 l of water d dilute 0.25 l of pure hcl to 2.5 l with water e dilute 59.4 ml of 3.0 m hcl to 2.5 l with water

1. When H2 (g) and N2 (g) react, it is produced 12.14 g NH3.

2. 0.1 mol HCl has 3.64 g and 9.84 g 37% HCl (commercial presentation).

3. The result concentration when diluted 50 mL HCl 6 M in 2.45 L water is 0.12 M.

4. The final concentration, when diluted 0.25 L pure HCl in 2.5 L water, is 3.2 M.

5. The final concentration when diluted 59.4 mL of 3 M HCl in 2.5 L water is 0.07 M

Explanation:

1. The ammonia balance equation is 3H2(g) + N2(g) =  2NH3(g). The molar mass are 2, 28 and 17 respectively. The limit reagent is N because it is completely consumed. With 10 g N it has 0.35 mol N and with this amount 0.71 mol NH3 are produced according with equation.

2. The most pure presentation HCl is 37% solution. For to know How many grams to take for to have 0.1 mol, we must to use HCl mass molar (36.4 g/mol). Then, 36.4 / 0.1 = 3.6g HCl pure. From a solution 37% HCl take 9.84 g (100%).

3. Assuming it has the same amount moles HCl in differents volumes, Then in 50 mL there are 0.3 moles HCl. 0.3/2.45L=0.12M.

4. 0.25 L HCl has 295 g HCl. (density HCl 1.18g/mL). In moles 295 g are 8.1 moles HCl. Then 8.1/2.5=3.2M

5. Assuming it has the same amount moles HCl in differents volumes, Then in 59.4 mL there are 1.18 moles HCl. 1.18/2.5L=0.07M.

I hope my answers helps you

answer; i believe the correct answer is (solid gold);

1. they vary with a higher number because boiling is way hotter than melting