Answer:
Explanation:
<em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>i</u></em><em><u>t</u></em><em><u> </u></em><em><u>h</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u>s</u></em><em><u> </u></em><em><u>y</u></em><em><u>o</u></em><em><u>u</u></em><em><u> </u></em><em><u>:</u></em><em><u>)</u></em><em><u>)</u></em><em><u>)</u></em>
I'm almost positive the answer is a.
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:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
<span>v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s </span>, hope this helps:)
