"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)
Answer: d
Explanation:it’s physics no so hard
Answer:
Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments
Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments
Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)
Explanation:
Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL
Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.
