In the laboratory, you are given an unknown solution containing only three ions. You use the following qualitative analysis scheme: First, to a test tube labeled #1 containing all three ions, you add 6 M HCl and a precipitate forms. Next, you decant the remaining solution into a clean test tube labeled #2, and then add H2S and 0.2 M HCl and a precipitate forms. Finally, you decant the remaining solution into a clean test tube labeled #3 and add (NH4)2S at pH 8 and a precipitate forms. Which of the following lists three ions that could have been in your unknown solution? a. Ag+, Mg2+, and K+
b. Ag+, Cu2+, and Ca2+
c. Ag+, Cu2+, and Na+
d. Ag+, Cu2+, and Fe3+

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

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)