Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = ![K_p=0.345](https://tex.z-dn.net/?f=K_p%3D0.345)
The given balanced equilibrium reaction is,
![2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)](https://tex.z-dn.net/?f=2SO_2%28g%29%2BO_2%28g%29%5Crightleftharpoons%202SO_3%28g%29)
The expression of
will be,
![K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28p_%7BSO_3%7D%29%5E2%7D%7B%28p_%7BSO_2%7D%29%5E2%5Ctimes%20%28p_%7BO_2%7D%29%7D)
Now put all the values of partial pressure, we get
![0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}](https://tex.z-dn.net/?f=0.345%3D%5Cfrac%7B%28p_%7BSO_3%7D%29%5E2%7D%7B%2830.6%29%5E2%5Ctimes%20%2813.9%29%7D)
![p_{SO_3}=67.009atm](https://tex.z-dn.net/?f=p_%7BSO_3%7D%3D67.009atm)
Therefore, the partial pressure of
is, 67.009 atm
Answer:
the rotation of earth is determining what part of the sun faces what part of earth making time the part thats away from the sun would be night and the one facing the sun itself would be day
The metal ball lost energy while the putty ball gained energy.
<h3>What is momentum?</h3>
Momentum is the product of mass and velocity of the body. We must note that momentum before collision is equal to momentum after collision.
1) Kinetic energy before collision = 1/2mv^2 = 0.5 * 6 * 4 = 12 J
2) kinetic energy after collision = 0.5 * 6 * 2= 6 J
3) Kinetic energy of putty ball = 0.5 * 6 * 2= 6 J
4) Energy lost by the metal ball = 12 J - 6 J = 6 J
5) Energy gained by the putty ball = 6 J - 0J = 6 J
6) The rest of the energy was converted to heat after the collision.
Learn more about kinetic energy: brainly.com/question/999862
To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.
The angular velocity can be described as
![\omega_f = \omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_0%20%2B%20%5Calpha%20t)
Where,
Final Angular Velocity
Initial Angular velocity
Angular acceleration
t = time
The relation between the tangential acceleration is given as,
![a = \alpha r](https://tex.z-dn.net/?f=a%20%3D%20%5Calpha%20r)
where,
r = radius.
PART A ) Using our values and replacing at the previous equation we have that
![\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%2894rpm%29%28%5Cfrac%7B2%5Cpi%20rad%7D%7B60s%7D%29%3D%209.8436rad%2Fs)
![\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s](https://tex.z-dn.net/?f=%5Comega_0%20%3D%2063rpm%28%5Cfrac%7B2%5Cpi%20rad%7D%7B60s%7D%29%3D%206.5973rad%2Fs)
![t = 11s](https://tex.z-dn.net/?f=t%20%3D%2011s)
Replacing the previous equation with our values we have,
![\omega_f = \omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_0%20%2B%20%5Calpha%20t)
![9.8436 = 6.5973 + \alpha (11)](https://tex.z-dn.net/?f=9.8436%20%3D%206.5973%20%2B%20%5Calpha%20%2811%29)
![\alpha = \frac{9.8436- 6.5973}{11}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B9.8436-%206.5973%7D%7B11%7D)
![\alpha = 0.295rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.295rad%2Fs%5E2)
The tangential velocity then would be,
![a = \alpha r](https://tex.z-dn.net/?f=a%20%3D%20%5Calpha%20r)
![a = (0.295)(0.2)](https://tex.z-dn.net/?f=a%20%3D%20%280.295%29%280.2%29)
![a = 0.059m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.059m%2Fs%5E2)
Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation
![\omega_f^2=\omega_0^2+2\alpha\theta](https://tex.z-dn.net/?f=%5Comega_f%5E2%3D%5Comega_0%5E2%2B2%5Calpha%5Ctheta)
Replacing with our values and re-arrange to find ![\theta,](https://tex.z-dn.net/?f=%5Ctheta%2C)
![\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B%5Comega_f%5E2-%5Comega_0%5E2%7D%7B2%5Calpha%7D)
![\theta = \frac{9.8436^2-6.5973^2}{2*0.295}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B9.8436%5E2-6.5973%5E2%7D%7B2%2A0.295%7D)
![\theta = 90.461rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090.461rad)
That is equal in revolution to
![\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090.461rad%28%5Cfrac%7B1rev%7D%7B2%5Cpi%20rad%7D%29%20%3D%2014.397rev)
The linear displacement of the system is,
![x = \theta*(2\pi*r)](https://tex.z-dn.net/?f=x%20%3D%20%5Ctheta%2A%282%5Cpi%2Ar%29)
![x = 14.397*(2\pi*\frac{0.25}{2})](https://tex.z-dn.net/?f=x%20%3D%2014.397%2A%282%5Cpi%2A%5Cfrac%7B0.25%7D%7B2%7D%29)
![x = 11.3m](https://tex.z-dn.net/?f=x%20%3D%2011.3m)