What is the chemical equation for the burning of propane if all the products and reactants are in the gas phase? 4h2o(g) + 3co2(g) + 5o2(g) → c3h8(g)c3h8(g) + 5o2(g) → 3co2(g) + 4h2o(g)c3h8(g) + o10(g) → 3co2(g) + 4h2o(l)4h2o(l) + 3co2(g) → c3h8(g) + 5o2(g)answer - b - c3h8(g) + 5o2(g) → 3co2(g) + 4h2o(g)

B

Explanation:

its right

Propane is a hydrocarbon and it will burn in presence of oxygen (it will undergo combustion) to give carbon dioxide and water (all in gas phase).

In presence of oxygen the carbon gets oxidized to carbon dioxide and hydrogen to water.

The general equation for a hydrocarbon undergoing combustion is

Here we are studying the combustion of propane so

x = 3

y = 8  

Propane has molecular formula of C3H8

Therefore the balanced equation will be

Or

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

the answer for sodium is 0.186

the answer for rubidium is 0.251