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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(a) The heat generated in the process is 28 kJ.
(b) The work done in the process is determined as -28 kJ.
(c) The change in the internal energy is 0.
<h3>
Heat of the isothermal compression </h3>
The heat generated in the process is negative done in the process.
W = -PΔV
W = -P(V₂ - V₁)
<h3>From A to B</h3>
W = -P(VB - VA)
W = -11(7 - 12.5)
W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J
<h3>From C to D</h3>
W = -25(20.5 - 7)
W = -337.5 L.atm = -34,197.18 J
Total work , w = -34,197.18 J + 6,130.16 J = -28 kJ
q = - w
q = 28 kJ
<h3>Change in internal energy</h3>
ΔE = q + w
ΔE = 28 kJ - 28 kJ = 0
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Answer:
The bronsted- Lowry acid is H₂PO₄⁻
Explanation:
Bronsted-Lowry acid donates a proton (H⁺)
H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O
In the reaction above, H₂PO₄⁻ is donating the proton to OH⁻ resulting in H₂O and the deprotonated species. This makes it a bronsted-Lowry acid.
