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
The answer will be Carbon Molecule
Answer:
ΔE = 150 J
Explanation:
From first law of thermodynamics, we know that;
ΔE = q + w
Where;
ΔE is change in internal energy
q is total amount of heat energy going in or coming out
w is total amount of work expended or received
From the question, the system receives 575 J of heat. Thus, q = +575 J
Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.
Thus;
ΔE = 575 + (-425)
ΔE = 575 - 425
ΔE = 150 J
Although the moon's distance from earth varies each month because of its eccentric orbit, the moon's mean distance from Earth is nonetheless increasing at the rate of about 3.8 centimeters (1.5 inches) per year
Answer:
Explanation:
E = (hc)/(λ)
E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /
(77.8x10^(-9)m)
E = 2.55 x 10^(-11) J
