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
He was a British philosopher, and an important experimental and theoretical chemist. He is known for his discovery of hydrogen. He at that time called it "inflammable air".
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
