Answer:
(1). The vapor pressure is 91 mmHg at 20°C.
(2). No, benzene will not boil at sea level.
Explanation:
Benzene, C6H6 is an aromatic, liquid compound with with molar mass of 78.11 g/mol and Melting point of 5.5 °C. One of the importance or the uses of benzene is in the making of fibres and plastics.
The vapour pressure of benzene can be gotten from the table showing the vapor pressure of different liquids.
Boiling point can simply be defined as the point or the temperature in which the vapor pressure is the same with the atmospheric pressure.
The atmospheric pressure is 760mmHg, while the vapor pressure at sea level is at the temperature of 15°C which is equal to 71 mmHg( from the table showing the vapor pressure of different liquids).
71 mmHg is not equal to 760 mmHg, thus, at sea level Benzene will not boil.
I'm actually not sure... but maybe this will help!
http://www.dummies.com/education/science/chemistry/how-to-identify-chiral-centers-in-a-molecule/
Answer:
121 g/mol
Explanation:
To find the molar mass, you first need to calculate the number of moles. For this, you need to use the Ideal Gas Law. The equation looks like this:
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = constant (0.0821 L*atm/mol*K)
-----> T = temperature (K)
Because density is comparing the mass per 1 liter, I am assuming that the system has a volume of 1 L. Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.
P = 1.00 atm R = 0.0821 L*atm/mol*K
V = 1.00 L T = 25.0. °C + 273.15 = 298.15 K
n = ? moles
PV = nRT
(1.00 atm)(1.00L) = n(0.0821 L*atm/mol*K)(298.15 K)
1.00 = n(0.0821 L*atm/mol*K)(298.15 K)
1.00 = (24.478115)n
0.0409 = n
Now, we need to find the molar mass using the number of moles per liter (calculated) and the density.
0.0409 moles ? grams 4.95 grams
---------------------- x ------------------ = ------------------
1 L 1 mole 1 L
? g/mol = 121 g/mol
**note: I am not 100% confident on this answer
Answer:
if im not mistaken that's a jetstream
