Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
My answer -
I believe that the answer is (A).
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this process is called parellelogram method of resolving vectors.
Answer:
48.4 km, 34.3° north of east
Explanation:
Let's say east is the +x direction and north is the +y direction.
Adding up the x components of the vectors:
x = 20 cos 60 + 30 + 0
x = 40 km
Adding up the y components of the vectors:
y = 20 sin 60 + 0 + 10
y = 27.3 km
The magnitude of the displacement is:
d = √(x² + y²)
d = 48.4 km
The direction is:
θ = atan(y/x)
θ = 34.3° north of east
