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uranmaximum [27]
2 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]2 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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g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

6 0
2 years ago
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