Answer:
The enthlapy of solution is -55.23 kJ/mol.
Explanation:
Mass of water = m
Density of water = 1 g/mL
Volume of water = 50.0 mL
m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g
Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C
Heat capacity of water,c =4.186 J/g°C
Heat gained by the water when an unknown compound is dissolved be Q
Q= mcΔT
heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.
Q'=-Q
Q'= -983.71 J =-0.98371 kJ
Moles of unknown compound = 
The enthlapy of solution :
The enthlapy of solution is -55.23 kJ/mol.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
![ka = \frac{[H_3O^+][X^-]}{[HX]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BX%5E-%5D%7D%7B%5BHX%5D%7D)
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
![[H_3O^+]} = 10^{-2,40}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%7D%20%3D%2010%5E%7B-2%2C40%7D)
[H₃O⁺] = <em>0,004M</em>
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = <em>0,085M</em>
![ka = \frac{[0,004][0,004]}{[0,085]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5B0%2C004%5D%5B0%2C004%5D%7D%7B%5B0%2C085%5D%7D)
ka = 1,88x10⁻⁴
And <em>pka = 3,73</em>
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
<em>pkb = 10,27</em>
I hope it helps!
Hi there
Formula
M1V1=M2V2
(455*6)/2500
Answer
1.092 M
<h2>Selective & Differential Medium</h2>
Explanation:
- Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
- Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
- Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
