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:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
N2 = 3*n1
T2 = 2*T1
V1 = V2
(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1
Since P2 is 6P1, it is 6 times greater than original pressure
An atom<span> with a neutral </span>charge<span> is one where the number of electrons is equal to the </span>atomic<span> number
hope i helped</span>
When you ride a vehicle in a fast speed, then your peripheral vision will reduce that is why there is a need for you to follow the direction of the objects when you are travelling in order for you to compensate to the decrease in the field of vision.
