Assume seawater is a solution of 0.570 M NaCl. Calculate the minimum pressure required at 10.00°C to desalinate seawater so that it contains no more than 177 mg NaCl per liter.

The minimum pressure required is 13.2 atm.

Explanation:

P = ∆CRT

Initial concentration of NaCl in sea water = 0.570 M

Final concentration of NaCl in sea water = concentration in g/L ÷ MW = 177 mg/L × 1 g/1000 mg ÷ (58.5 g/mol) = 0.00303 mol/L = 0.00303 M

∆C (change in concentration of NaCl in sea water) = 0.57 - 0.00303 = 0.56697 M

R is gas constant = 0.082057 L.atm/mol.K

T is temperature = 10 °C = 10+273 = 283 K

P (pressure) = 0.56697×0.082057×283 = 13.2 atm

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