Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
![pH = -log([H_{3}O^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20)
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
<u>Trimethyl ammonium</u>:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
![Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5B%28CH_%7B3%7D%29_%7B3%7DN%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5B%28CH_%7B3%7D%29_%7B3%7DNH%5E%7B%2B%7D%5D%7D)

By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
<u>Phenol</u>:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
![Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DOH%5D%7D)


Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%289.99%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.00%20)
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
Therefore, the difference in pH values is 4.95.
I hope it helps you!
Answer:
Examples of Chemical Changes
Burning wood.
Souring milk.
Mixing acid and base.
Digesting food.
Cooking an egg.
Heating sugar to form caramel.
Baking a cake.
Rusting of iron.
Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
value calculated by the student.
2.
The
value of ethylamine = 11.87


![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
of ethylamine value calculated by the student.
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
Answer: Carbon dioxide is a pure substance.
Explanation: A pure substance is defined when a substance has a single type of molecule. If more than 1 type of molecule is present in a substance, then it is considered as a mixture.
- Soda is basically a mixture of water and carbon dioxide. More than 1 type of molecule is present.
- Gasoline is a mixture of may gases. More than 1 type of molecule is present.
- Salt water contains salt and water molecules, hence it is considered as a mixture.
- Carbon dioxide has only 1 type of molecule which is
molecules. Hence, it is a pure substance.