What type of energy does a ball have when it rolls down a hill?

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

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:

$pH = -log([H_{3}O^{+}])$

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

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

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:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0

3 years ago

Which of the following is not a reason why chemical bonds are important

velikii [3]

What are the followings?

4 0

3 years ago

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant (for benzene} = $5.12^0Ckg/mol$

m = molality

Putting in the values we get:

$0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}$

${\text{ Mass of solute in g}}=0.028g$

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0

3 years ago

What is the molarity of a solution in which 25g NaCl in a 2.00 L solution?

Paladinen [302]

Mass of sodium :- 23 grams

Mass of Chlorine :- 35.5 grams

Mass of NaCl :- 58.5 grams

mass given :- 25 grams

moles :- 0.432 ( given mass/ ionic mass)

$molarity = \frac{moles}{vol} \\ molarity = \frac{0.432}{2} \\ molarity = 0.216 \: mole{l}^{ - 1} or \: molar$

5 0

2 years ago

Landslides and slumps are forms of ?

masha68 [24]

Landslides and slumps are forms of mass movements. Types of erosion both move, in different ways. In slumps, land breaks off in one piece, or chunk. Landslides are when rocks and dirt rapidly move down a slide.

5 0

3 years ago