What information is necessary to determine the atomic mass of the element chlorine?
(1) the atomic mass of each artificially produced isotope of chlorine, only
(2) the relative abundance of each naturally occurring isotope of chlorine, only
(3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine
(4) the atomic mass and the relative abundance of each naturally occurring and artificially produced isotope of chlorine

Answer is: (3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine.

m(³⁵Cl) = 34.97 amu; the average atomic mass of chlorine-35.  

m(³⁷Cl) = 36.97 amu; the average atomic massof chlorine-37.

ω(³⁵Cl) = 75.76% ÷ 100% = 0.7576; fractional abudance of chlorine-35.

ω(³⁷Cl) = 24.24% ÷ 100% = 0.2424; fractional abudance of chlorine-37.

m(Cl) = m(³⁵Cl) · ω(³⁵Cl) + m(³⁷Cl) · ω(³⁷Cl).  

m(Cl) = 34.97 amu · 0.7576 + 36.97 amu · 0.2424.  

m(Cl) = 35.4548 amu; average atomic mass of chlorine.

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