How many grams of carbon dioxide are created from the complete combustion of 21.3 L of butane at STP?

165.5 g of CO2

Explanation:

We must first put down the balanced reaction equation:

C4H10(g) + 13/2 O2(g) > 4CO2(g) + 5H2O(g)

From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:

22.4 L of butane created 174 g of CO2

Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell

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