Compozitia procentuala a NaCl
1 answer:
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19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.
<h3>What is vapour pressure?</h3>Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2
Volume of (V) = 40 liter
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm
Moles of (n) can be calculated by ideal gas equation.
PV = nRT
n = 1.007 40 ÷ 0.0821 295 = 1.663
Balanced chemical reaction;
2 + 13
---> 8
+ 10
From reaction;
13 moles require 2 moles
So, 1.663 moles will require = 2 x 1.663 ÷13 = 0.256 moles of
Thus is a limiting reagent. So it will drive the yield of
.
Moles of produced = (8/2) 0.2 = 0.8 moles
Pressure of (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm
Applying the ideal gas equation for ,
PV = nRT
0.985 V = 0.8 0.0821 x 295
V = 19.7 liter
The volume of produced = 19.7 liter.
Learn more about the vapour pressure here:
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Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
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