Answer:
optical (optic is the word used for referring to the eye ie optic nerve)
Explanation:
Seas are smaller than oceans. Oceans are where land and water meet, and are usually bigger.
<u>Answer:</u> The value of 
of the reaction is 28.38 kJ/mol
<u>Explanation:</u>
For the given chemical reaction:
- 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)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-364%29%29%5D-%5B%281%5Ctimes%20%28-296.8%29%29%2B%281%5Ctimes%200%29%5D%3D-67.2kJ%2Fmol%3D-67200J%2Fmol)
- The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20311.9%29%5D-%5B%281%5Ctimes%20248.2%29%2B%281%5Ctimes%20223.0%29%5D%3D-159.3J%2FKmol)
To calculate the standard Gibbs's free energy of the reaction, we use the equation:
where,
= standard enthalpy change of the reaction =-67200 J/mol
= standard entropy change of the reaction =-159.3 J/Kmol
Temperature of the reaction = 600 K
Putting values in above equation, we get:
Hence, the value of of the reaction is 28.38 kJ/mol
Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation:
B. The study of matter and energy.
Because physical science is everything that doesn't include organic things.
