How could you show experimentally that the molecular formula of propene is c3h6, not ch2?
2 answers:
Answer:
By performing experiments related with the elevation in boiling point and depression in freezing point.
Explanation:
Hello,
In this case, we use the concept of colligative properties of solutions because they depend on the concentration of solute molecules or ions rather than its identity. They are classified as vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.
Elevation in boiling point and depression in freezing point are suitable for us to determine the molecular formula of propane as shown below:
- Elevation in Boiling Point: by performing an elevation in boiling point experiment, it is possible to quantify ebuilloscopic constant for the solvent, in order words, the molal (referred to molality) boiling point elevation constant of propane, using it as a solute. Then, one applies the formula:
To compute the molality and subsequently propane's molecular mass.
- Depression in freezing point: by performing this one, it is possible to determine the molecular mass of the propane. Thus, we find the molal freezing point depression constant of the solvent to subsequently apply the shown below formula:
Which is suitable to compute the molality as well as in the previous case.
Best regards.
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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))
