The question is not fully stated; here is the full version:
Using this data alongside the standard enthalpies of formation for 
, 
, and 
found in Appendix C, determine the standard enthalpy of formation for acetone.
The complete combustion of 1 mole of acetone 
releases 1790 kJ:
Answer: The standard enthalpy of formation for is calculated to be -247.9 kJ/mol
Explanation:
Enthalpy change represents the variation in enthalpy for all products and reactants based on their respective mole counts. This is denoted as 
The enthalpy change calculation for a chemical reaction follows this equation:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
Concerning the chemical reaction in question:
The equation reflecting the enthalpy change for this reaction is:
![\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_3H_6O%28l%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Provided data includes:
Substituting values from the equation gives us:
![-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol](https://tex.z-dn.net/?f=-1790%3D%5B%283%5Ctimes%20%7B%28-393.5%29%7D%29%2B%283%5Ctimes%20%28-285.8%29%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_3H_6O%28g%29%29%7D%29%2B%284%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28C_3H_6O%28g%29%29%7D%3D-247.9kJ%2Fmol)
Thus, the enthalpy of formation of 
computes to -247.9 kJ/mol.
Response:
FALSE
Rationale:
Sedimentary rocks are defined as rocks formed through the processes of compaction and cementation. Initially, sediments derived from various locations must accumulate. Over time, these deposited sediments undergo substantial compaction due to the weight of the layers above. This process converts loose sediments into solid rock. This is the process through which sedimentary rocks, comprised of sand-sized particles, are formed. For instance, examples include Shale, Sandstone, and Mudstone.
Thus, both compaction and cementation are crucial in the formation of sedimentary rocks.
Therefore, the statement above is False.
Answer:
Explanation:
A mole is defined as the number of molecules divided by 6.02×10^23
Mole = (2×10^19)/(6.02×10^23)
Mole = 3.32×10^-5 mole
Utilize the ideal gas law:
n = PV / RT
P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272
