Can you please include the statement or the model?
Answer:
3.10 mole of C3H8O change in entropy is 89.54 J/K
Explanation:
Given data
mole = 3.10 moles
temperature = -89.5∘C = -89 + 273 = 183.5 K
ΔH∘fus = 5.37 kJ/mol = 5.3 ×10^3 J/mol
to find out
change in entropy
solution
we know change in entropy is ΔH∘fus / melting point
put these value so we get change in entropy that is
change in entropy 5.3 ×10^3 / 183.5
change in entropy is 28.88 J/mol-K
so we say 1 mole of C3H8O change in entropy is 28.88 J/mol-K
and for the 3.10 mole of C3H8O change in entropy is 3.10 ×28.88 J/K
3.10 mole of C3H8O change in entropy is 89.54 J/K
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
One does not simply should you do you should