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Answer:
The heat of the reaction is 105.308 kJ/mol.
Explanation:
Let the heat released during reaction be q.
Heat gained by water: Q
Mass of water ,m= 1kg = 1000 g
Heat capacity of water ,c= 4.184 J/g°C
Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C
Q=mcΔT
Heat gained by bomb calorimeter =Q'
Heat capacity of bomb calorimeter ,C= 4.643 J/g°C
Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C
Q'=CΔT'=CΔT
Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.
q= -(Q+Q')
q = -mcΔT - CΔT=-ΔT(mc+C)

Moles of propane =
0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.
The heat of the reaction will be:

Answer:
Statements Y and Z.
Explanation:
The Van der Waals equation is the next one:
(1)
The ideal gas law is the following:
(2)
<em>where n: is the moles of the gas, R: is the gas constant, T: is the temperature, P: is the measured pressure, V: is the volume of the container, and a and b: are measured constants for a specific gas. </em>
As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2),<u> by the incorporation of the intermolecular interaction between the gases and the gases volume</u>. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, <u>taking into account the volume occuppied by the gas in the total volume, which implies</u> a reduction of the total space available for the gas molecules.
So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.
Have a nice day!