<u>Answer:</u> The boiling point of water in Tibet is 69.9°C
<u>Explanation:</u>
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
= final pressure = 240. mmHg
= Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature or normal boiling point of water = ![100^oC=[100+273]K=373K](https://tex.z-dn.net/?f=100%5EoC%3D%5B100%2B273%5DK%3D373K)
= final temperature = ?
Putting values in above equation, we get:
![\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B240%7D%7B760%7D%29%3D%5Cfrac%7B40700J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B373%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D%5C%5C%5C%5C-1.153%3D4895.36%5B%5Cfrac%7BT_2-373%7D%7B373T_2%7D%5D%5C%5C%5C%5CT_2%3D342.9K)
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:


Hence, the boiling point of water in Tibet is 69.9°C
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Answer:
Al: 3
S: 6
Explanation:
Al is in Group III. This means that it has 3 valence electrons.
S is in Group VI. This means that it has 6 valence electrons.
I hope this helps :)
So,
To find the number of moles in 90g of NaOH, what we do is to divide the given amount by the molar mass of NaOH.
The molar mass of NaOH is 40g/mol.
So,

Therefore, the correct answer is A. 2.25moles