Hydrocarbons are compounds that are composed of carbon and hydrogen. Which statement does the law of definite proportions suppor
1 answer:
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The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
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Answer:
The equilibrium constant is
The equilibrium pressure for oxygen gas is
Explanation:
From the question we are told that
The equation of the chemical reaction is
The Gibbs free energy for is
The Gibbs free energy for is
The Gibbs free energy for is
The Gibbs free energy of the reaction is mathematically represented as
Substituting values
From the balanced equation
The Gibbs free energy of the reaction can also be represented mathematically as
Where R is the gas constant with a value of
T is the temperature with a given value of
K is the equilibrium constant
Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state
The equilibrium constant for this chemical reaction is mathematically represented as
Where is the equilibrium pressure of oxygen
The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction
Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction
Substituting values
Now substituting this into the equation above to obtain the equilibrium of oxygen
multiplying through by