The Correct is C₆H₅O⁻ : It is the most commonly bonded to the acidic hydrogen , and O2 is the most element bonded to the acidic hydrogen.
The explanation:
According to the table below we can see that the Ka ( the acid dissociation constant) for the Phenol is the smallest one.
when the Ka expression for this chemical reaction:
C₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺(aq)
is
Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH]
So, the water solution of the phenol has a very low equilibrium concentration of H ions [H+] , So C₆H₅O⁻ is the element which most commonly bonded to the acidic hydrogen.
Answer:
Rate =
Explanation:
Answer:
Given:
New pressure: approximately (
.) (Assuming that the gas is an ideal gas.)
Explanation:
Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:
By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)
For this gas, while
.
Let denote the pressure of this gas before the volume change (
.) Let
denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between
and
:
.
In other words, because the final volume is of the initial volume, the final pressure is
times the initial pressure. Therefore:
.
On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)
Convert the unit of the temperature of this gas to degrees Kelvins:
.
.
Let denote the pressure of this gas before this temperature change (
.) Let
denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at
.
Apply Amonton's Law to find the ratio between and
:
.
Calculate , the final pressure of this gas:
.
In other words, the pressure of this gas after the volume and the temperature changes would be approximately .
Answer : the correct answer for ksp = 1.59 * 10⁻⁹
Following are the steps to calculate the ksp of reaction
BaCO₃ →Ba ²⁺ + CO₃²⁻ :
Step 1 : To find ΔG° of reaction :
ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :
ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .
Given : ΔG° for Ba²⁺ ( product )= -560.7
ΔG° for CO₃²⁻ (product ) =- 528.1
ΔG° BaCO₃ ( reactant) = –1139
Plugging value in formula :
ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )
⁻ = ( -560.7 + 528.1
) - ( -1139
)
= ( -1088.8 ) - (-1139
)
= - 1088.8 + 1139
ΔG° of reaction = 50.2
Step 2: To calculate ksp from ΔG° of reaction .
The relation between Ksp and ΔG° is given as :
ΔG° = -RT ln ksp
Where ΔG° = Gibb's Free energy R = gas constant T = Temperature
Ksp = Solubility constant product .
Given : ΔG° of reaction = 50.2
T = 298 K R = 8.314
Plugging values in formula
((Converting 2477
Since , 1 KJ = 1000 J So , ))
Dividing both side by
ln ksp =
Removing ln :
ksp = 1. 59 * 10⁻⁹