Match each front with its description. when a cold air mass slides under a warm
air mass
warm front
when a cold air mass and a warm air mass
meet and do not move
stationary front
when two cold fronts force a warm front up
and block it from the surface
cold front
when a warm air mass slides over a cold air
occluded front
mass

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.

a electronegativity decreases beacuse the size of nucleus is bigger.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.