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:
Matter:
toothpaste
gasoline
bacteria
cell
Not matter:
happiness
sound
Explanation:
Matter is something that has mass and volume. Happiness is just a feeling, it doesn't exist in 3d world. Sound is a wave or just the movement of matter (particles of matter). Others have masses and volumes.
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Here are some disadvantages, is that nitrogen dioxide is a toxic gas and it can still be harmful when ingested by human, also critics of hydrogen fuel cells argue that although these cells do not emit carbon after burning, they give out nitrogen dioxide and other emissions.
Hope this helps
Answer:
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
Explanation:
Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.
Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
Now, we have to balance Li by multiplying LiNO₂ by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
The equation is now balanced.
