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.
Learn more about mole here:-brainly.com/question/15374113
#SPJ4
Answer is (1) - no reaction.
<em>Explanation;
</em>
Some of you may think this reaction as a single replacement reaction which gives NaBr + F₂ as products.
But, according to the reactivity of the halogens, reactivity decreases from up to bottom of the group. F is placed above Br. Hence, F is more reactive than Br. Hence, Br can't replace F.
Their locations can vary depending on the molecule they are associated with but they are usually in a "cloud " that is on the outside of an atom/molecule and if the atom is unstable the electrons tend to be located farther away from the atom.
Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Balanced equation is
HBr + NaOH ----> NaBr + H2O
Using molar masses
80.912 g HBr reacts with 39.997 g of Naoh to give 18.007 g water
so 1 gram of NaOH reacts with 2.023 g of HBR
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction
Mass of water produced = (5.7 * 18.007 / 39.997 = 2.6 g to 2 sig figs
