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uranmaximum [27]
3 years ago
11

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressur

e is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate
Physics
1 answer:
Natalka [10]3 years ago
5 0

Answer:  a. 17.7 KJ/Mol

b. T=210K

Explanation:

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.

the question isnt completely originally, but we could look at the likely derivation from the questions

(a) the standard enthalpy of vaporization

using the clausius clapeyron equation

In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )

In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )

Therefore, Delta H (vap) = 17.7 KJ/Mol

b. Also the boiling point

What is the normal boiling point of arsine?

At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr

substitution into the equation as stated in question 1

ln(760/253)=17700/8.314(1/189.55-1/T)

T=210K

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consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

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3 years ago
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2 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

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