Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
The ionic character of any compound depend on the lattice energy as well as the electronegativity of element present in that compound.
More would be the lattice energy more would be ionic nature of that compound.
The lattice energy of any compound is inversely proportional to the ionic radii cation and anion.
In given case the ionic radii of oxide in both oxides would be equal therefore the lattice energy only depend on the ionic radii of cation.
As the radii of Magnesium less then radii of lithium therefore lattice energy of Magnesium oxide would be more than lithium oxide.
Hence, MgO would be more ionic in nature than 
Predators decreased, food source increased, migration patterns, natural disaster or threat in normal habitat.
I think the correct answer among the choices listed above is option D. Polar covalent bond is seen in molecules where <span>electrons are not shared equally between atoms. This makes the molecule to have a partially positive and partially negative sides.</span>
Answer: 
Explanation:

Here Mn undergoes oxidation by loss of electrons, thus act as anode. silver undergoes reduction by gain of electrons and thus act as cathode.

Where both
are standard reduction potentials.
![E^0_{[Mn^{2+}/Mn]}= -1.18V](https://tex.z-dn.net/?f=E%5E0_%7B%5BMn%5E%7B2%2B%7D%2FMn%5D%7D%3D%20-1.18V)
![E^0_{[Ag^{2+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B2%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Mn^{2+}/Mn]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-%20E%5E0_%7B%5BMn%5E%7B2%2B%7D%2FMn%5D%7D)

The standard emf of a cell is related to Gibbs free energy by following relation:

= gibbs free energy
n= no of electrons gained or lost = 2
F= faraday's constant
= standard emf = 1.98V

Thus the value of
is 