Example 13:0.29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide at S. T.P. From the given information, calculate the (i) mass of carbon dioxide formed.​

Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10

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