At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?

Explanation:

Hello,

In this case, for the given chemical reaction:

We know the equilibrium constant and equilibrium expression:

That in terms of the reaction extent (ICE procedure) we can write:

In such a way, solving for by using a quadratic equation or solver, we obtain:

Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:

Regards.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j