Answer: Magnesium
Explanation:
Galvanic cell is a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy.
The standard reduction potential for magnesium and zinc are as follows:
![E^0_{[Mg^{2+}/Mg]}= -2.37V](https://tex.z-dn.net/?f=E%5E0_%7B%5BMg%5E%7B2%2B%7D%2FMg%5D%7D%3D%20-2.37V)
![E^0_{[Zn^{2+}/Zn]}=-0.76V](https://tex.z-dn.net/?f=E%5E0_%7B%5BZn%5E%7B2%2B%7D%2FZn%5D%7D%3D-0.76V)
Reduction takes place easily if the standard reduction potential is higher (positive) and oxidation takes place easily if the standard reduction potential is less (more negative).
Here Mg undergoes oxidation by loss of electrons, thus act as anode. Zinc undergoes reduction by gain of electrons and thus act as cathode.
Thus magnesium gets oxidized.
Answer: The overall equation will be 
Explanation:
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.
Anode : 
Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.
Cathode : 
The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :
<u>Answer:</u> The enthalpy of the formation of 
is coming out to be -410.8 kJ/mol.Z
<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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%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_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol](https://tex.z-dn.net/?f=81.1%3D%5B%281%5Ctimes%20%28226.7%29%7D%29%2B%284%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%3D-410.8kJ%2Fmol)
Hence, the enthalpy of the formation of is coming out to be -410.8 kJ/mol.
Answer:
20 oxygen atoms
Explanation:
Li2SO4 has 4 oxygen atoms, so if there are 5 of that compound then there will be 20 oxygen atoms.
