<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
Bromine 35
Chlorine 17
Flourine 9
Iodine 53
Astatine 85
Answer:
2 L is the new volume
Explanation:
We can apply the Ideal Gases Law to solve the problem.
At STP, we kwow that 1 mol of gas occupy a volume of 22.4 L
Then, how many moles do we have in 1 L?
Let's do it by a rule of three:
(1L . 1 mol) / 22.4L = 0.0446 moles
These moles are at 1 atm and 273 K of temperature, so let's change our conditions.
P . V = n . R .T
1 atm . V = 0.0446 mol . 0.082 L.atm/mol K . 546 K
V = 2 L
If we pay attention, we can notice that, if we double temperature, we double the volume.
Answer:
We will produce 8.0 moles of HCl , this is 291.7 grams HCl
Explanation:
Step 1: Data given
Number moles of H2S = 3.0 moles
Step 2: The balanced equation
2HSbCl4 + 3H2S → Sb2S3 + 8HCl
Step 3: Calculate moles HCl
For 2 moles HSbCl4 we need 3 moles H2S to produce 1mol Sb2S3 and 8 moles HCl
For 3.0 moles H2S we'll have 8.0 moles HCl
Step 4: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
Mass HCl = 8.0 moles * 36.46 g/mol
Mass HCl = 291.7 grams
We will produce 8.0 moles of HCl , this is 291.7 grams HCl
Answer: A. or C. (most likely C.)
Explanation:
All molecules of a compound have the same properties. ... All molecules of a compound have one type of atom.
Jeez, they r both true.....go with A. or C.
Hope this helps!
