A worker drops a 3pound wrench while working on a scaffold 50 feet above the ground. How long does it take for the wrench t reach the ground? What is the speed of this wrench just when it reached the ground? Neglect air resistance? 26. The vapor pressure of pure ethanol at 20Deg C is 44 mmHG. Alcohol contains 45% ethanol and 55% water by weight. What would be partial vapor pressure of ethanol above a shot glass containing 200ml of alcohol in closed room at temp 20 deg C at 765mm Hg? If this shot glass with alcohol is not covered and let it to evaporate, what would be concentration of ethanol in ppm? MW Ethanol = 46.07 amu & MW water = 18.02 amu?
To determine how long it takes for the wrench to reach the ground, we can use the formula:
h = (1/2)gt^2
where h is the height of the scaffold (50 feet), g is the acceleration due to gravity (32.2 ft/s^2), and t is the time it takes for the wrench to fall.
Plugging in the values, we get:
50 = (1/2)(32.2)t^2
t^2 = 3.093
t ≈ 1.76 seconds
Next, to find the speed of the wrench just as it reaches the ground, we can use the formula:
v = gt
Plugging in the values, we get:
v = 32.2 * 1.76
v = 56.72 ft/s
Therefore, the wrench takes about 1.76 seconds to reach the ground and has a speed of 56.72 ft/s just as it reaches the ground.
Moving on to the second part of the question, we need to determine the partial vapor pressure of ethanol in the closed room.
First, we need to calculate the total pressure exerted by the alcohol in the closed room. The total pressure in the room is 765 mmHg.
The partial pressure of ethanol can be calculated using the ideal gas law:
P1 = (w1/MW1) * Ptotal
where P1 is the partial pressure of ethanol, w1 is the weight percent of ethanol (45%), MW1 is the molecular weight of ethanol (46.07 g/mol), and Ptotal is the total pressure in the room (765 mmHg).
Plugging in the values, we get:
P1 = (0.45 * 46.07 / 100) * 765
P1 ≈ 125.93 mmHg
Therefore, the partial pressure of ethanol in the closed room at 20 deg C is approximately 125.93 mmHg.
To determine the concentration of ethanol in parts per million (ppm) if the shot glass is not covered and allowed to evaporate, we need to calculate the moles of ethanol and water evaporating from the shot glass.
First, we need to calculate the moles of ethanol in the shot glass:
n_ethanol = (0.45 * 200 ml) / 46.07 g/mol
n_ethanol ≈ 1.96 moles
Next, we need to calculate the concentration of ethanol in the room in ppm:
conc_ethanol = (n_ethanol / (n_ethanol + n_water)) * 10^6
conc_ethanol = (1.96 / (1.96 + 11.08)) * 10^6
conc_ethanol ≈ 152,952 ppm
Therefore, the concentration of ethanol in the room would be approximately 152,952 ppm if the shot glass is not covered and allowed to evaporate.