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Ahat [919]
3 years ago
15

A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the secon

d half she drives 60 mi/h. what is her average speed on this trip?
Physics
1 answer:
Gemiola [76]3 years ago
5 0
The average speed of a moving object is the rate of change of a certain distance with respect with time. It is equal to the total distance that was traveled by the object over the total time it takes to travel that distance. For this problem we need to assume that the total distance that was traveled would be equal to 120 miles. So, for the first half of the distance or 60 miles at a speed of 30 miles per hour, the time taken would be two hours. For the remaining 60 miles at a speed of 60 miles per hour, 1 hour is total time traveled. So, we calculate the average speed as follows:

Average speed = total distance / total time
Average speed = 120 miles / 2 hr + 1 hr
Average speed = 40 mi / hr
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A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
ohaa [14]
Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}

V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters
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Categorize these resources as either renewable or Non-renewable
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3 years ago
The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

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3 years ago
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