Answer:
0!
Explanation:
- You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:
-COOH -NH3
pH 7------------------------------------------------------
Asn - +
Gly - +
Leu - +
- Now that we have our table we'll sketch our peptide's structure:
<em>HN-Asn-Gly-Leu-COOH</em>
This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:
+1 (HN-Asn)
-1 (Leu-COOH)
=0
The net charge is 0!
I hope you find this information useful and interesting! Good luck!
Filtration is a method for separating an insoluble solid from a liquid.
"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)
<u>Given information:</u>
Mass of NaCl (m) = 87.75 g
Volume of solution (V) = 500 ml = 0.5 L
Molar mass of NaCl (M) = 58.44 g/mol
<u>To determine:</u>
The molarity of NaCl solution
<u>Explanation:</u>
Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)
i.e. M = moles of solute/liters of solution = n/V
Moles of solute (n) = mass of solute (m)/molar mass (M)
moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles
Therefore,
Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M
<u>Ans: (D)</u>
