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antiseptic1488 [7]
2 years ago
10

A mixture of gases contains 1.50 mol of o2, 0.75 mol of he, and 2.25 mol of n2. the total pressure exerted by the mixture is equ

al to 2.50 atm. what is the partial pressure of the he?
Physics
1 answer:
bezimeni [28]2 years ago
4 0

A mixture of gases contains 1.50 mol of o2, 0.75 mol of he, and 2.25 mol of n2. the total pressure exerted by the mixture is equal to 2.50 atm .The partial pressure of the he will be 4.167 atm .

Each constituent gas in a mixture of gases has a partial pressure which is the notional pressure of that constituent gas if it alone occupied the whole volume of the original mixture at the same temperature.

given

moles of O2 = 1.50 moles

moles of He = 0.75 moles

moles of N2 = 2.25 moles

total moles = 1.50 + 0.75 + 2.25 = 4.50 moles

total pressure exerted by the mixture is =  2.50 atm

partial pressure of He = (moles of He / Total moles) * total pressure

                                     = (0.75 / 4.50) * 25 = 4.167 atm

To learn more about partial pressure here

brainly.com/question/13199169

#SPJ4

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Complete Question

A thin film in air is made by putting a thin layer of acetone (n = 1.25) on a layer of water

(n = 1.33). When visible light (400 nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of “purplish” with red light of wavelength 650 nm mixed some blue giving it that appearance and “greenish” light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer:

The thickness of acetone is  d = 104 nm

Explanation:

A diagram showing this process is shown on the first uploaded image

From the question we are told that

     The refractive index of acetone is n_a =1.25

     The refractive index of water is  n_w = 1.33

      The wavelength of the reflected light is  \lambda_r = 650nm =  650  *10^{-9}m

      The wavelength of the destroyed light is  \lambda_g = 520nm =  520 *10^{-9}m

       

Looking at the given data we can see that the

             n_a < n_w

This means that the light which the acetone-water layer would reflect will have a phase shift of \pi

  Again this make us to understand that the light reflected at the acetone layer will also have a phase shift of \pi

Since they would be having the same phase shift the two light would interfere

  For interference the condition for minima is mathematically represented as

             2 n_a d = (m + \frac{1}{2} ) \lambda_g

Where d is the thickness of acetone

                d = \frac{\lambda_g}{4 n_a}

Substituting values

              d = \frac{520 *10^{-9}}{4 * 1.25}

               d = 104 *10^{-9}m

               d = 104 nm

       

     

     

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