A jet can travel the 6000 km distance between washington,

1 answer:

6 0

<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>

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The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

denis-greek [22]

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$