Which has the highest boiling point .33 m NH3 or .10 m Na2SO4
1 answer:
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
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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))
