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2 years ago

A piece of copper at 10 °C is dropped into a glass of water at 35 °C. Which statement is correct?

Chemistry

2 answers:

kakasveta [241]2 years ago

8 0

I say B. Why? because the heat will keep on moving in molecules until it reaches same temperature

Sergeu [11.5K]2 years ago

5 0

Answer:

The correct answer is option B, that is, heat will flow from the water in the glass to the copper until both reach the same temperature.

Explanation:

Let us see a perfectly thermal insulation system with a glass of water at higher temperatures and a piece of copper at a lower temperature, and when both encounters each other than water will lose some of the heat and that heat will be gained by the copper. The procedure will take place until and unless both attain a similar temperature. After attaining a similar temperature, there will be no change in temperature or heat between the two. The change in heat is a function of temperature difference.

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2 years ago

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For each solution, determine the p-values for each ion indicated. A solution that is 0.493 M in NaCl and 0.314 M in NH 4 Cl .

miss Akunina [59]

Complete Question:

Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺

Answer:

pNa = 0.307

pCl = 0.093

pNH₄ = 0.503

Explanation:

The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.

Both substances are salts that solubilize completely, thus, by the solution reactions:

NaCl → Na⁺ + Cl⁻

NH₄Cl → NH₄⁺ + Cl⁻

So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.

[Na⁺] = 0.493 M

[Cl⁻] = 0.493 + 0.314 = 0.807 M

[NH₄⁺] = 0.314 M

The p-values are:

pNa = -log[Na⁺] = -log(0.493) = 0.307

pCl = -log[Cl⁻] = -log(0.807) = 0.093

pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503

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3 years ago

) 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!

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