For alloys of two hypothetical metals a and b, there exist an α, a-rich phase and a β, b-rich phase. from the mass fractions of both phases for two different alloys (given below), which are at the same temperature, determine the composition of the phase boundary (or solubility limit) for (a)α and (b)β phases at this temperature.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.