At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M and [ NO ] = 0.500 M . N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re‑established?

The answer to the question is the final concentration of NO after equilibrium is re‑established = 0.7143 moles

Explanation:

The reaction is given by

O₂(g) + N₂(g) → 2NO(g)

Kc =

= (0.5^2)/(0.1×0.1) = 25

However, initially, y moles each of O and N gave off x moles each to form 2·x moles of NO, therefore.

Hence we have 2·x = 0.5 or x = 0.25

and y = 0.1+0.25= 0.35

When the concentration is increased to 0.8, we have each mole of NO giving out 2·a moles while each O₂ and N₂  increases by a moles which gives

(0.8-2·a)^2/(0.1+a)^2 = 25

Taking square roots of both sides and cross multiplying, we have 7×a=0.3

Which gives a = 0.04286 moles.

Therefore the number of moles of NO at equilibrium

= 0.8-2×0.04286 = 0.7143 moles

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)