Consider an equilibrium (K1) that is established after 10 mL of compound A and 10 mL of compound B are mixed. Now, imagine the equilibrium (K2) where 1 mL of compound A is added to 100 mL of compound B. How are K1 and K2 related algebraically (read this question VERY carefully, at least one more time)

The equilibrium constant K₁ = Equilbrium constant K₂.

The equilibrium constant, K, of a reaction, is defined as:

"The ratio between concentration of products powered to their reaction quotient and concentration of reactants powered to thier reaction quotient".

For the reaction:

aA + bB ⇄ cC + dD

The equilibrium constant, K, is:

Now, assuming the reaction of the problem is 1:1:

A + B ⇄ C + D

The concentrations of the reactants are directly proportional to the volume added. Thus, we can assume that concentration = Volume. Replacing for K₁ and K₂:

In the same way:

Thus, we can say:

Learn more about chemical equilibrium in:

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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.