As a rule of thumb, if the [ H3O ] is less than 1% of the value of the [Acid], it does not have to be considered when figuring the concentration of the acid. Use your results in

Explanation:

In strong acids, we see how the [H3O+] comes close to be equal to the concentration of the acid. Whereas in the weak acids, the [H3O+] is appears to be far less than the concentration of the acid.

One of the ways to define and measure the relative strengths of a series of acids is to measure the pH of solutions of acids of equal concentration which is ideal: remembering that the lower the pH, the greater the concentration of hydronium ion, the stronger the acid. Let's check few more variations;

• in a strong monoprotic acid, the [H30+] in solution is equal to the original acid concentration.

• in a weak acid, the [H3O+] is lesser than the original acid concentration. Simply put that the [H3O+] is smaller than if the acid were to be a strong acid of the same concentration.

• In series of weak monoprotic acids (of the category HA) of the same concentration, here we see how the [H30+] increases (while the pH declines) as the acids become stronger.

Other concepts relevant to this discuss is thus;

• the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb).

• for any conjugate acid–base pair, KaKb=Kw.

• smaller values of pKa relates that it gives a larger acid ionization constants and hence stronger acids.

• some acids or bases are listed as having Ka, or Kb, values that are large or very small.

• when aqueous acids are stronger than H3O+ they are completely ionized (HNO3, for example), invariably meaning that their K, values are "large."

• meaning that their conjugate bases (e.g NO3 ) do not produce meaningful concentrations of OH ions, so its Kb values is "very small."

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

can you attach the