The combustion of gasoline produces carbon dioxide and water. assume gasoline to be pure octane (c8h18) and calculate how many kilograms of carbon dioxide are added to the atmosphere per 3.8 kg of octane burned. ( hint: begin by writing a balanced equation for the combustion reaction.) express your answer using two significant figures.

11.73 Kg of CO₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

Next, we shall determine the mass of C₈H₁₈ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C₈H₁₈ = (12×8) + (18×1)

= 96 + 18

= 114 g/mol

Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g

Convert 228 g to kg.

1000 g = 1 Kg

Therefore,

228 g = 228 g × 1 Kg / 1000 g

228 g = 0.228 Kg

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 16 × 44 = 704 g

Convert 704 g to Kg

1000 g = 1 Kg

Therefore,

704 g = 704 g × 1 Kg / 1000 g

704 g = 0.704 Kg

SUMMARY:

From the balanced equation above,

0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂.

Finally, we shall determine the mass of carbon dioxide, CO₂, that will be produced by the reaction of 3.8 kg of octane, C₈H₁₈. This can be obtained as follow:

From the balanced equation above,

0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂.

Therefore, 3.8 kg of C₈H₁₈ will react to produce = (3.8 × 0.704) / 0.228 = 11.73 Kg of CO₂

Thus, 11.73 Kg of CO₂ is added to the atmosphere per 3.8 kg of C₈H₁₈.

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