In a molecule of calcium sulfide, calcium has two valence electron bonds, and a sulfur atom has six valence electrons. How many
lone pairs of electrons are present in the Lewis structure of calcium sulfide? one two three four none
2 answers:
Answer: Option (b) is the correct answer.
Explanation:
Chemical formula of calcium sulfide is CaS. Since there are two valence electrons in calcium and six valence electrons in sulfur.
So, in order to attain stability, calcium will donate its two extra electrons to the sulfur atom. Therefore, calcium sulfide is an ionic compound.
Thus, we can conclude that there will be only two lone pair of electrons in a calcium sulfide molecule.
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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))
