A chemistry student adds a quantity of an unknown solid compound X to 5.00 L of distilled water at 15 degrees Celsius. After 10 minutes of stirring, only a little X has dissolved. The student then drains off the solution and evaporates the water under vacuum. A precipitate is left behind. The student washes, dries, and weighs the precipitate. It weighs 0.17 kg. Can you calculate the solubility of X in water at 15 degrees Celsius?

34 g/L

Explanation:

From the question, only 0.17Kg of the solvent dissolved in the water.

Solubility is defined as the amount of solute that dissolves in 1L of solvent.

But 0.17Kg is the same as 170g of solute

Hence 170g of solute dissolved in 5L of solvent

x g of solute will dissolve in 1L of solvent

x= 170/5 = 34 g/L

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