Answer:
2 should be potential energy and 3 should be A
Explanation:
Answer:
Writing down background information can help you to remember what you learned and can be used for notes in the future when taking an assessment.
Explanation:
Can I get brainliest? It's for a challenge
<u>Answer:</u> The enthalpy of the formation of
is coming out to be -65.3 kJ/mol
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\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)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times \Delta H^o_f_{(CO(g))})]-[(1\times \Delta H^o_f_{(SiO_2(s))})+(3\times \Delta H^o_f_{(C(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiO_2%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![624.6=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times (-110.5))]-[(1\times (-910.9))+(3\times (0))]\\\\\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol](https://tex.z-dn.net/?f=624.6%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%29%2B%282%5Ctimes%20%28-110.5%29%29%5D-%5B%281%5Ctimes%20%28-910.9%29%29%2B%283%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%3D-65.3kJ%2Fmol)
Hence, the enthalpy of the formation of
is coming out to be -65.3 kJ/mol.
Answer:
3564 = 3.564 x 10^3
Explanation:
i.e expresing the value in standard form.
“Aluminum” is most likely to