Which metal cations form a white precipitate with both H2SO4(aq) and Na2S2O4(aq)? 1. Ba2+ 2. Sr2+ 3. Fe2+ 4. Cu2+
1 answer:
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Answer : The mass percent of ethanol is, 10.38 % and the molality of ethanol is, 2.52 mole/kg
Explanation :
In wine, the solute and solvent are ethanol and water respectively.
Given :
12.8 % ethanol by volume means 12.8 mL ethanol present in 100 mL solution.
Volume of ethanol = 12.8 mL
Volume of solution = 100 mL
Volume of water = 100 - 12.8 = 87.2 mL
Density of ethanol =
Density of water = 1 g/mL
Now we have to calculate the mass of ethanol and water.
and,
Now we have to calculate the total mass of 100 mL of wine.
Total mass of 100 mL of wine = 10.1 + 87.2 = 97.3 g
Now we have to calculate the mass percent of ethanol.
The mass percent of ethanol is, 10.38 %
Now we have to calculate the molality.
Molar mass of ethanol = 46 g/mole
The molality of ethanol is, 2.52 mole/kg
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))
