Answer
pH=8.5414
Procedure
The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.
pH = pKa + log₁₀ ([A⁻] / [HA])
Where
pH = acidity of a buffer solution
pKa = negative logarithm of Ka
Ka =acid disassociation constant
[HA]= concentration of an acid
[A⁻]= concentration of conjugate base
First, calculate the pKa
pKa=-log₁₀(Ka)= 8.6383
Then use the equation to get the pH (in this case the acid is HBrO)
Neon has filled its outer shells, therefore it is very stable and does not need to react with other elements and doesn’t form compounds.
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :

= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :

Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:



Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
When ketone is reacted with phosphorous pentachloride, chlorination takes place at the carbonyl carbon with substitution of the oxygen atom to give a geminal dichloride (with 2 Cl atoms on same carbon) according to the following equation:
so we can say that acetone is converted into 2,2-dichloropropane by action of PCl₅
<span />
0.125 g=(0.125 g)(1000 mg/1g)=125 mg.
Then, we need 125 mg of ampicillin.
5 ml of liquid suspension contains 250 mg of ampicilling , therefore:
5 ml----------------250 mg of ampicilling
x--------------------125 mg of ampicilling
x=(5 ml * 125 mg of ampicilling) / 250 mg of ampicilling=2.5 ml
Answer: we require 2.5 ml