Electron is negative
Proton is positive
Neutron is just nothing really
Answer:- B- 
Explanations:- Sharing of electrons takes place between non metals and the bond formed by the sharing of electrons is known as covalent bond. In first choice we only have a zinc metal and so no sharing of electrons would be possible here.
In third choice we have NaCl. Na is a metal and Cl is a non metal. The property of metals is to transfer its valence electrons to the non metal and the bond formed is known as ionic bond. So, third choice is also not correct.
In choice B we have sulfate ion that has sulfur and oxygen atoms and both of these are non metals. So, sharing of electrons is present here between the S and O atoms and covalent bonds are formed.
So, choice B-
is the right answer.
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So,
for the given reaction is -99.4 J/K
(3) loses one electron and becomes positively charged
Lithium has one valence electron and Bromine has seven. Therefore Lithium will give up its one to Bromine for both to have an octet
The solution for this problem would be:
We are looking for the grams of magnesium that would have
been used in the reaction if one gram of silver were created. The computation
would be:
1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) =
0.1127 grams of Magnesium