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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Answer:
D
Explanation:
The source of the thermal energy in a combustion engine is the spark from the spark plug that ignites the fuel and air mixture converting chemical energy into thermal energy.
Says here the answer is <span>dipole-dipole</span>
Answer:
c.boron-11
Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.81u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u=ux
x=0.19
1−x=0.81
And thus the abundance of boron-11 is roughly 81%.