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oee [108]
3 years ago
7

Bottlenose dolphins use echolocation pulses with a frequency of about 100 kHz, higher than the frequencies used by most bats. Wh

y might you expect these water-dwelling creatures to use higher echolocation frequencies than bats?
a. Word Bank-b. smallerc. inversed. directe. greater
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

1. greater

2. direct

3. smaller

4. inverse

Explanation:

The speed of sound in water is greater than in air; hence for the same frequency the sound wavelength in water is <u>greater </u>than in air (for the given frequency the wavelength is in the <u>direct </u>proportion with the speed of sound).

To "see" an object via the echolocation creature needs to use sound with the wavelength <u>smaller </u>than the size of an object viewed.

That means to "see" objects of the same size dolphin and bat need to use ultrasound of the same wavelength, hence dolphin needs to use higher frequency (for the given speed of sound the wavelength is in <u>inverse </u>proportion with the frequency).

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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
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Answer:

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Explanation:

Applying Clausius Clapeyron Equation

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

Where;

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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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