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 : The balanced chemical equation is,
Explanation :
Rules for the balanced chemical equation in basic solution are :
- First we have to write into the two half-reactions.
- Now balance the main atoms in the reaction.
- Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
- If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
- If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion
at that side where the less number of hydrogen are present. - Now balance the charge.
The half reactions in the basic solution are :
Reduction : 
......(1)
Oxidation : 
.......(2)
Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.
The balanced chemical equation in a basic solution will be,
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
The average speed is calculated using the following formula
Where
T = 298 K [room temperature]
m = mass of each molecule of oxygen
mass of one mole of oxygen molecules = 32 g / mol
mass of one molecule of oxygen will be = 
mass of one molecule of oxygen will be = 
putting values
average speed = 
Average speed = 482.19 m / s
average speed = 1735.8 km / h
so approx = 1700 km/h
8. b
9. c
10.a
all of those can be determined by units
