Answer:
C is the reaction intermediate.
Explanation:
A reaction intermediate is a molecular structure that is formed during the reaction but then is converted in the final products.
Usually, these reaction intermediates are unestable and, for that reason, the lifetime of these structures is low.
In the reaction, you can see in the first step C is produced, but also, in the second step reacts producing D. As is produced and, immediately consumed,
<h3>C is the reaction intermediate.</h3>
Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³
Answer:
The main difference between evaporation and boiling are : 1. Evaporation takes place at all temperatures, while boiling occurs at a particular temperature. 2. Evaporation takes place from the surface, whereas the entire liquid boils. 3. Evaporation can occur using the internal energy of the system, while boiling requires an external source of heat.
Answer:
(a) 
(b) 
(c) 
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant
should be computed for air as an ideal gas by:


Next, we compute the final temperature:

Thus, the work is computed by:

(b) In this case, since
is given, we compute the final temperature as well:

And the isentropic work:

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

Regards.