Answer:
C is the correct answer and i hope ur having a good day UwU
Explanation:
We use the Raoult's Law to answer this question.
Partial Pressure = Mole Fraction*Total Pressure
When concentration is expressed in ppm, it means 'parts per million'. So, there can be 0.67 mol of ozone per 10⁶ mol of mixture.
Mole fraction = 0.67/10⁶ = 0.67×10⁻⁶
Partial pressure = 0.67×10⁻⁶(735 torr) = <em>4.92×10⁻⁴ torr</em>
The highest frequency sound to which the machine can be adjusted is :
<u>Given data :</u>
Pressure = 10 Pa
Speed of sound = 344 m/s
Displacement altitude = 10⁻⁶ m
<h3>Determine the highest frequency sound ( f ) </h3>
applying the formula below
Pmax =
--- ( 1 )
Therefore :
f = ( Pmax * V ) / 
= ( 10 * 344 ) / 2
* 1.31 * 10⁵ * 10⁻⁶
= 4179.33 Hz
Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .
Learn more about Frequency : brainly.com/question/25650657
<u><em>Attached below is the missing part of the question </em></u>
<em>A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?</em>
The first runner because it is very clear that accelarition depends on the time and we know that the time in this case is pretty simple