"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.
Option: b
<u>Explanation</u>:
As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant 
for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.
To calculate firstly molarity of ions are needed to be found with formula: 
Then at equilibrium cations and anions concentration is considered same hence:
![\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%3D%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D%3D%5Ctext%20%7B%20molarity%20of%20ions%20%7D)
Hence from above data can be calculated by: = ![\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%20%5Ccdot%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D)
The equation that scientists could use to find the wavelength of the emission lines of the hydrogen atom would be that of Balmer.
The wavelength of the emission lines of the hydrogen atom can be derived using the Balmer series:
1/λ 
Where λ = wavelength, 
= Rydberg constant, and n = level of the original orbital.
The equation becomes applicable in getting the wavelength of emitted light when electrons in hydrogen atoms transition from higher (n) orbital to lower orbital (2) levels.
More on the Balmer series can be found here: brainly.com/question/5295294
Answer:
Should be a 15 imo
Explanation:
N is one, 4H = 4, S 2 because of the 2 and O 8 because of the 4 and the 2 before the S.
There is 9.375 g of nitrogen. This is because when a half-life passes, the mass becomes 2 times less. Since two half-lives have passed, you need to divide by 4.
