Answer:
In general an acid reacts with a carbonate or hydrogen-carbonate to produce a salt, carbon dioxide gas and water.
Answer:
fireproofing spray used in firefighting can be termed as one of the possible discovery that would be attributed to an inorganic molecule.
Explanation: • There a number of advancement been made each day to improve the different process either domestic, industrial, or related to research work. As a number of professionals are always in search to get the right set of information or data to get to a more proper conclusion.
Explanation:
Anions and cations have different atomic sizes than the neutral atoms.
When the bromine atom becomes an anion, it ______ in size.
Anions involves gaining extra electrons. This basically leads to increased repulsion between the electrons and thereby increasing the size. Anions are always larger than the neutral atoms.
This means the answer is A. Increase
When the strontium atom becomes a cation, it ______ in size.
Cations involves losing valence electrons. This basically leads to a decrease in electron shells and electron repulsion thereby reducing the size. Cations are always larger than the neutral atoms.
This means the answer is B. Decrease
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.
