0.5927-g sample was dissolved, and the Ca2+ and Ba2+ ions present were precipitated as BaC2O4

.H2O and

CaC2O4

.H2O. The oxalates were then heated in a thermogravimetric apparatus leaving a residue that weighed

0.5127 g in the range of 320°C–400°C and 0.4362 g in the range of 580°C–620°C. Calculate the percentage Ca

and percentage Ba in the sample.​

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

catya that's not even a b or c