Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
3). Chemical potential energy
1). lower in charcoal than in coal.
Explanation:
Chemical potential energy is defined as the energy that can be absorbed or stored in a substance's chemical bonds. It can be released when there is a change in the number of particles of the substance.
As per the question, coal releases more heat than charcoal because 'the chemical potential energy of charcoal is lower than the coal' <u><em>and hence, the latter would release more heat on burning i.e. the energy that was absorbed in the chemical bonds of the substance</em></u>. Hence, option 3 and 1 are the correct answers.
Answer: 3.45 L carbon dioxide are produced
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
According to stoichiometry :
5 moles of 
produce = 3 moles of 
Thus 0.257 moles of will produce=
of
Volume of 
Thus 3.45 L carbon dioxide are produced
<span>pv=nrt; Pressure and moles are constant.
p=nr(150k)/.5 L; Pressure initially
After temp change
pv=nrt; What is volume?
v=nr(350k)/p; p is constant so we can substitute from above
v=nr(350k)/(nr(150k)/.5 L))
v=350/150/.5 L
v=4.66 liters</span>
Answer:
The pressure remains constant
Explanation:
this is an example in charles law where as the temperature increases so does the volume.
