Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
The momentum of the second ball was 15 kg.m/s.
<h3>What is inelastic collision?</h3>
In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.
In the given question,
Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.
Initial velocity of 1st ball of mass 3 kg is 1 m/s.
So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.
According to Principle of conservation of linear momentum for this inelastic collision,
Initial momentum of first ball + initial momentum of second ball = final momentum of the system
⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball
= 18 kg.m/s - 3 kg.m/s.
= 15 kg.m/s.
Hence, initial momentum of second ball = 15 kg.m/s.
Learn more about momentum here:
brainly.com/question/24030570
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Answer:
I believe the answer is B.
Answer:
becomes narrower
Explanation:
Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.
As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.
As it is outside the focal point it must be real.Real images must be inverted.
As it is beyond the centre of curvature it is also beyond 2F which means that the image is inside the centre of curvature ( between F and 2F from the mirror ) As the image is closer to the mirror than the object it must be diminished in size.
Hope this helps you :)
