Can anyone solve this please. I need this fastly
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<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
Answer : The work done by the system is, 2.2722 J
Explanation :
The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 0.00100 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas =
= initial volume of gas = 25 mL
= final volume of gas = 75 mL
Now put all the given values in the above formula, we get:
Therefore, the work done by the system is, 2.2722 J
Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
Best regards.