4. you have a very concentrated solution (12 m) of potassium chloride (kci). you need it to be at the least concentration possible for the experiment you are
about to conduct. the problem is you forgot to order lab supplies, so you only have 2 l of distilled water left. what would be the final concentration if you
added the two 2 l of distilled water to the 0.5 l of 12 m kci?
a 3.0 m
b 24 m
c 2.4m
d 48 m

C. 2.4 M

Explanation:

Because you started with 12M solution of KCl, that means that there was a concentration of 12 mols of KCl per Liter of solution. (12mol/1L =12 M). Since there was only 0.5L of solution, there was only 6 mols of KCl because there is only 12 mols per 1 Liter, so half of that volume would have half the amount of solute to keep that true. (6mol/.5L = 12 M). With the new water added, the formula becomes (6mols KCl/ 2.5L of water =2.4 M)

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