Answer:
Lewis structure of polyatomic formate anion.
Explanation:
To draw Lewis structure for any chemical species,
1)Count the total number of valence electrons present in it.
This can be obtained by adding valence electrons of each constituent atom.
2)Arrange those valence electrons in such a way that each atom should attain eight electrons around it to satisfy octet theory.
The structure of formate ion and its Lewis structure are shown below:
HCOO- is the formate ion.
It has total:
1+4+6+6+1 = 18 valence electrons.
Since, hydrogen has one, carbon has four and oxygen has six valence electrons and the charge of the anion is one.
Arrange this 18 electrons in such a way that each atom should get 8 electrons around it.
Resonance structures of formate ion:
Magnesium element exists as single atoms. Free oxygen is always diatomic (two atoms bonded together) as O2. Nitrogen gas is also diatomic N2.
2Mg + O2 → MgO
Mg forms a +2 ion, O forms a -2 ion, so they combine and balance in a 1:1 ratio. MgO, magnesium oxide, is a solid.
3Mg + N2→Mg3N2
Mg forms a +2 ion, but N forms a -3 ion. Hence the Mg3N2 configuration for magnesium nitride.
Do you know the difference between an extensive property verses an intensive property?
Answer:
2.5 g/cm3 .
Explanation:
Density is mass in grams over volume in cubic centimeters. So it is 25 g 10 cm 3 =2.5 g/cm 3 The unit is grams per cubic centimeter. NB - density can also be kilograms over cubic meters Hope this helps!
Answer: The value of 
for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
Its solubility product will be as follows.
![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
Reduction half-reaction: 
, 
Oxidation half-reaction: 
, 
= ?
Cell reaction: 
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product,
= 
Therefore, according to the Nernst equation
![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
At equilibrium, 
= 0.00 V
Putting the given values into the above formula as follows.
![0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=0.00%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
= 
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
= 0.222 V
Thus, we can conclude that value of for the half-cell reaction is 0.222 V.
