Answer:
liquid's heat of vaporization = 38.4 kJ/mol
Explanation:
given data
vapor pressure P1 = 6.91 mmHg
at temperature = 0 °C = 273.15 K
boiling temperature = 105 °C
solution
for vapor pressure and temperature we get here
P2 = 760.0 mmHg
T2 = 68.73°C = 378.15 K
we use here the Clausius-Clapeyron Equation that is
ln = .................1
put here value
In
solve it we get
x = 38445 J/mol
liquid's heat of vaporization = 38.4 kJ/mol
Answer:wat do u think answer it and get back to me
Explanation:
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Answer:
39.3%
Explanation:
Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:
CaF2 + H2SO4 --> CaSO4 + 2HF
We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.
Number of moles of calcium fluoride reacted= mass of calcium fluoride reacted/ molar mass of calcium fluoride
Molar mass of calcium fluoride= 78.07 g/mol
Number of moles of calcium fluoride= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of 0.28 moles of hydrogen fluoride = number of moles× molar mass
Molar mass of hydrogen flouride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g this is the theoretical yield of HF
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%