Answer:it’s abc it’s just science you know
Explanation:
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!
Answer:
1. A state of balance in which the rates of the forward and reverse reactions are equal.
Explanation:
A dynamic equilibrium is like a cycle, the reactants change to products, but the products also change to reactants keeping the amount of each constant.
2. A state of balance in which the forward reaction stops but reverse reaction continues.
In this statement there isnt a equilibrium. The products will change to reactants until the reaction stops.
3. A state of balance in which the forward reaction continues but reverse reaction stops.
Here the reactants will change to products until the reaction stops.
4. A state of balance in which the forward and reverse reactions stop.
In this case the reaction has stopped.
Answer:
1.4 × 10² mL
Explanation:
There is some info missing. I looked at the question online.
<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>
Step 1: Given data
- Initial volume (V₁): 215 mL
- Initial pressure (P₁): 625 mmHg
- Final pressure (P₂): 1.3 atm
Step 2: Convert 625 mmHg to atm
We will use the conversion factor 1 atm = 760 mmHg.
625 mmHg × 1 atm/760 mmHg = 0.822 atm
Step 3: Calculate the final volume of the air
Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL
