One
Balance the Equation.
This has been done for you or it is given. Anyway this step is finished, but it must always be done.
Two
Find the molar mass of C6H12O6
6C = 6 * 12 = 72
12H = 12*1 = 12
CO = 6 * 16 = 96
Mol Mass = 180 grams / mol
Three
Find the mols of C6H12O6
n = ???
Molar Mass = 180 grams / mol
given mass = 13.2 grams.
n = given mass / molar mass
n = 13.2 / 180
n = 0.07333333 mols.
Four
Find the mols CO2
1 mol C6H12O6 will produce 6 mols CO2
0.0733333 mols will produce x
1/0.073333 = 6/x Cross multiply
x = 0.073333 * 6
x = 0.4399 moles.
Five
Find the volume given the conditions for temperature and pressure are STP conditions.
PV = nRT
R = 0.082057
n = 0.43999
P = 1 atmosphere
T = 0 + 273 = 273
V = ???
1 * V = 0.43999 * 082057 * 273
V = 10.2 L
Answer: B
Note if you give out Brainly awards I'd sure appreciate one. This was a lot of typing
Answer:
I think B no is correct orterwise Ano
Answer:
I would say about 250 meters.
Explanation:
Answer:
the correct answer is B) that each element has a different of protons
Explanation:
<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
