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ANTONII [103]
3 years ago
13

The combustion of gasoline produces carbon dioxide and water. assume gasoline to be pure octane (c8h18) and calculate how many k

ilograms of carbon dioxide are added to the atmosphere per 7.0 kg of octane burned. ( hint: begin by writing a balanced equation for the combustion reaction.)
Chemistry
1 answer:
IgorC [24]3 years ago
7 0
<span>Answer: 2 C8H18 + 25 O2 => 16 CO2 + 18 H2O 1.0 kg = 1000 g C8H18 = 1000 g / 114.2293 g/mole = 8.75 moles C8H18 8.75 moles C8H18 produce (16/2) (8.75) = 70 moles CO2 70 moles CO2 = (70 moles) (44.0096 g/mole) = 3081 g CO2 = 3.1 kg CO2</span>
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A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.
irga5000 [103]

Answer:

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT =  (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) =  (sinθ)√(2gH

v₀ = √(gR/(sin2θ))

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2 years ago
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Answer:

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Explanation:

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I can help with that!
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