Answer:
0.1 is the retention factor.
Explanation:
Distance covered by solvent ,![d_s= 2.0 cm](https://tex.z-dn.net/?f=d_s%3D%202.0%20cm)
Distance covered by solute or ion,![d = 0.20 cm](https://tex.z-dn.net/?f=d%20%3D%200.20%20cm)
Retention factor
is defined as ratio of distance traveled by solute to the distance traveled by solvent.
![R_f=\frac{d}{d_s}](https://tex.z-dn.net/?f=R_f%3D%5Cfrac%7Bd%7D%7Bd_s%7D)
![R_f=\frac{0.20 cm}{2.0 cm}=0.1](https://tex.z-dn.net/?f=R_f%3D%5Cfrac%7B0.20%20cm%7D%7B2.0%20cm%7D%3D0.1)
0.1 is the retention factor.
Answer:
Neutrons = 11
Explanation:
To get the number of neurons an element has you have to subtract the number of protons from the atomic mass of that element.
The number of protons is equal to the atomic number of that element.
Atomic number = Number of Protons
So the number of protons is 25
Neutrons = Atomic mass - Number of Protons
Neutrons = 36 - 25
Neutrons = 11
<u>Answer:</u> The value of
of the reaction is 28.38 kJ/mol
<u>Explanation:</u>
For the given chemical reaction:
![SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)](https://tex.z-dn.net/?f=SO_2%28g%29%2BCl_2%28g%29%5Crightarrow%20SO_2Cl_2%28g%29)
- 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:
![\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%3D-364kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%3D-296.8kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%3D0kJ%2Fmol)
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:
![\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%3D311.9J%2FKmol%5C%5C%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%3D248.2J%2FKmol%5C%5C%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%3D223.0J%2FKmol)
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:
![\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5CDelta%20H%5Eo_%7Brxn%7D-T%5CDelta%20S%5Eo_%7Brxn%7D)
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:
![\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D-67200-%28600%5Ctimes%20%28-159.3%29%29%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D28380J%2Fmol%3D28.38kJ%2Fmol)
Hence, the value of
of the reaction is 28.38 kJ/mol
Answer:
4
Explanation:
Si has atomic number 14 so the electronic configuration 2,8,4
VII . A because it contains Br (l), Cl (g) and I (s)