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.
Answer: 3.28 mins
Explanation:
Here’s how it breaks down:
Conversions
74 mph = 33.08 m/s
6.5 km = 6500 m
(6500 m)/(33.08 m/s) = 196.5 seconds
196.5 seconds is equivalent to 3.28 minutes
<span>Using PV=nRT, which represents a universal constant for any state, we have:
P1V1/n1T1=R
and
P2V2/n2T2=R;
This implies that:
P1V1/n1T1=P2V2/n2T2
Thus we can express it as
V1/n1=V2/n2.
Rearranging yields:
V2=V1 x (n2/n1) = 750 mL x ((0.65+0.35)/(0.65)) = 1200 mL = 1.2 L... with 2 significant figures</span>
Every unicellular organism prospers by executing metabolic activities.
Metabolic activities encompass the set of chemical reactions essential for sustaining life.
Explanation:
Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.
These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.
Organisms acquire the energy necessary for these activities through food consumption.
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