The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.
The volume of the bottle = 4.7 L
Mass of ethanol = 0.33 g
Temperature (T1) = -11 oC = 273-11 = 262 K
P1 = 6.65 torr
Now we will calculate the mole by applying the ideal gas equation:-
PV = nRT
Or, n = PV/RT
Where P is the pressure
T is the temperature
R is the gas constant = 0.0821 L atm mol-1K-1
V is the volume
Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-
= 0.001913 mol C2H6
Conversion of the mole to gm
Molar mass of ethanol (M) = 46.07 g/mol
Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.
Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.
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Number #2 is joule
In the SI system the unit of heat is the joule. The calorie was defined so that the heat capacity of water was equal to one.The heat capacity of a substance is the amount of heat required to raise the temperature of a defined amount of pure substances by one degree.
Number #3 is Endothermic
Why? I can't explain. I Just left class where they were just explaining that.
The third shell has 3 subshells: the subshell, which has 1 orbital with 2 electrons, the subshell, which has 3 orbitals with 6 electrons, and the subshell, which has 5 orbitals with 10 electrons, for a total of 9 orbitals and 18 electrons.
