C should be the answer to the following formula
Explanation:
The given data is as follows.
= 100 mm Hg or
= 0.13157 atm
=
= (1080 + 273) K = 1357 K
=
= (1220 + 273) K = 1493 K
= 600 mm Hg or
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,

![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
=
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
Answer:
solution is clear solution while colloidal is between the solution and suspension. And in suspension particles are suspended.
Explanation:
In solution light can be passed without any scattering of light from solute particles while suspension is cloudy and having larger particle size than colloids, if suspension stands for a while particles will settle down easily.
In colloids light will scattered and dispersed by reflecting with large particles.
Converting mmHg to atm is solved by division.
Example: Convert 745.0 to atm.
Solution- divide the mmHg value by the 760.0 mmHg / atm.
745 mmHg over 760.0 mmHg/atm
atm value is 0.980263
Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.
So your fraction so to speak should read, 0.190789/0.08552632 or 0.190789 over 0.08552632
(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
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